pages.uwc.edupages.uwc.edu/shubhangi.stalder/MAT 110/College...Page iii About the authors Dr....

COLLEGE ALGEBRA

Stalder & Martin

Page i

Dedication

This book is dedicated to

All students who want to learn

And

All teachers who want to teach.

Page ii

College Algebra

Dr. Shubhangi Stalder

University of Wisconsin Waukesha

[email protected]

&

Dr. Paul Martin

University of Wisconsin Marathon

[email protected]

All material in this book is copyrighted to the authors and can only be used for non-profit educational

purposes only. Please contact the authors for any other use.

First Edition, 2016

mailto:[email protected]

mailto:[email protected]

Page iii

About the authors Dr. Shubhangi Stalder is a full professor of mathematics at the University of Wisconsin

Waukesha. She has received her doctoral degree in mathematics from the University of

Wisconsin Milwaukee in 1993. She has decades of teaching experience and her focus has

always been to reach out to those who are struggling in

mathematics. Her main belief is that everyone can learn basic

mathematics if they tried. The key is to understand the “Why”

and the “How, and to be able to see the patterns across different

mathematical processes. She believes that in the long run rote

memorization does not work to learning mathematics. She uses

yoga and meditation techniques with her students who

experience math and test anxiety and continues to include

mindfulness practice in her teaching of mathematics. She has received the UW System Board of

Regents Teaching in Excellence Award (the state of Wisconsin’s highest teaching award), the

UW Colleges Chancellor’s Excellence in Teaching Award, and the UW Colleges Kaplan Teacher

Award.

Dr. Paul Martin is a full professor of mathematics at the University of Wisconsin Marathon. He received his doctoral degree in mathematics from the University of Wisconsin Madison in 1994. He has decades of teaching experience and his focus has always been to help his students see how mathematics connects to the real world. He does this through building 3-dimensional models, modeling physical processes such as heat loss through the attic of a house, to connecting mathematics to his students’ other classes from chemistry to music. As a teacher, Martin stresses the importance of reasoning over memorization. He has received several teaching awards over his career including the prestigious UW Colleges Chancellor’s Excellence in Teaching Award.

Page iv

Acknowledgements This book is the first edition. The material in this book is continuation of the developmental and

Intermediate Algebra text by the same authors.

The authors would like to thank their families who gave their professional opinions, their

constant support and put up with long hours on this project (without their support it would

have been difficult to put this book together in a such a short time).

We also would like thank Arman Banimahd, Amanuel Teweldemedhin, and Adam Figarsky for

their willingness to discuss the content and try it out in the classroom.

We also want to thank Kent Kromarek who was our sounding board and input into some of the

History of Mathematics presented in the book and for reviewing and editing the book. In

addition, we want to thank all the countless brave souls (students and teachers) who were

willing to give us their time in trying this material, and giving us their feedback to make this

project better. We hope their work and ours has made it possible to get a product that we hope

will be useful to other students and teachers.

Table of Contents

CHAPTER 1: ALL ABOUT FUNCTIONS AND RELATIONS ............................................................ 1

1.1 INTRODUCTION AND DOMAIN AND RANGE OF FUNCTIONS AND RELATIONS ....................................................1

WORKSHEET SEC 1.1 ....................................................................................................................................... 16

EXERCISES 1.1 ................................................................................................................................................ 17

1.2 INVERSE FUNCTIONS AND A BRIEF LIBRARY OF FUNCTION TYPES .............................................................. 23

OTHER TYPES OF FUNCTIONS ............................................................................................................................. 27

WORKSHEET SEC 1.2 ....................................................................................................................................... 36

EXERCISES 1.2 ................................................................................................................................................ 37

1.3 EXPONENTIAL AND LOGARITHMIC FUNCTIONS .................................................................................... 50

COMPOUND INTEREST AND THE NUMBER 𝒆 .......................................................................................................... 52

EVALUATING LOGARITHMIC FUNCTIONS .............................................................................................................. 56

PROPERTIES OF LOGARITHMS ............................................................................................................................ 59

WORKSHEET SEC 1.3A ..................................................................................................................................... 64

EXERCISES 1.3A .............................................................................................................................................. 65

ANOTHER CLASS OF FUNCTIONS: SEQUENCES ...................................................................................................... 69

FIBONACCI SEQUENCE AND THE GOLDEN RATIO .................................................................................................... 68

WORKSHEET SEC 1.3B...................................................................................................................................... 73

EXERCISES 1.3B ............................................................................................................................................... 75

EVEN AND ODD FUNCTIONS .............................................................................................................................. 77

WORKSHEET SEC 1.3C...................................................................................................................................... 80

EXERCISES 1.3C ............................................................................................................................................... 81

1.4 ARITHMETIC AND COMPOSITION OF FUNCTIONS ..................................................................................... 83

WORKSHEET SECTION 1.4A ............................................................................................................................... 87

EXERCISES 1.4A ............................................................................................................................................... 88

COMPOSITION OF FUNCTIONS ........................................................................................................................... 89

WORKSHEET SECTION 1.4B ............................................................................................................................... 94

EXERCISES 1.4B .............................................................................................................................................. 95

CHAPTER 2: GRAPHING FUNCTIONS AND RELATIONS ...................................................................... 101

2.1 LIBRARY OF FUNCTIONS .................................................................................................................. 101

HORIZONTAL AND VERTICAL SHIFT OF FUNCTION GRAPHS .................................................................................... 106

HORIZONTAL AND VERTICAL STRETCH/COMPRESSION AND REFLECTIONS ................................................................ 110

COMBINATION OF TRANSLATIONS, STRETCHES AND REFLECTIONS .......................................................................... 117

WORKSHEET SEC 2.1 ..................................................................................................................................... 121

EXERCISES 2.1 .............................................................................................................................................. 122

2.2 GRAPHING CONIC SECTIONS ............................................................................................................ 129

2.2A PARABOLAS .......................................................................................................................................... 130

WORKSHEET SEC 2.2A ................................................................................................................................... 138

EXERCISES 2.2A ............................................................................................................................................. 139

2.2B CIRCLES AND ELLIPSES ............................................................................................................................. 140

CIRCLES........................................................................................................................................................ 144

WORKSHEET SEC 2.2B.................................................................................................................................... 146

EXERCISES 2.2B ............................................................................................................................................. 147

2.2C HYPERBOLAS .............................................................................................................................. 148

SUMMARY AND TIPS ON IDENTIFYING CONIC SECTIONS ........................................................................................ 151

WORKSHEET SEC 2.2C.................................................................................................................................... 154

EXERCISES 2.2C ............................................................................................................................................. 155

2.3 GRAPHING POLYNOMIAL FUNCTIONS ......................................................................................................... 159

WORKSHEET SEC 2.3 ..................................................................................................................................... 182

EXERCISES 2.3 .............................................................................................................................................. 183

2.4 GRAPHING RATIONAL FUNCTIONS ..................................................................................................... 190

LINEAR ASYMPTOTES OF RATIONAL FUNCTIONS .................................................................................................. 190

RECIPROCAL OF A FUNCTION ........................................................................................................................... 197

ZEROS OF THE NUMERATOR AND DENOMINATOR ............................................................................................... 201

END BEHAVIOR OF A GRAPH OF A RATIONAL FUNCTION ....................................................................................... 206

WORKSHEET SEC 2.4 ................................................................................................................................... 214

EXERCISES 2.4 .............................................................................................................................................. 215

CHAPTER 3 SOLVING EQUATIONS AND INEQUALITIES ..................................................................... 222

3.1 QUADRATIC EQUATIONS ................................................................................................................. 222

REVIEW OF COMPLEX NUMBERS ...................................................................................................................... 224

QUADRATIC EQUATIONS ................................................................................................................................. 227

COMPLETING THE SQUARE .............................................................................................................................. 228

QUADRATIC FORMULA ................................................................................................................................... 230

WORKSHEET SEC 3.1 ..................................................................................................................................... 234

EXERCISES 3.1 .............................................................................................................................................. 235

3.2 POLYNOMIAL EQUATIONS OF DEGREE THREE OR HIGHER ........................................................................ 241

RATIONAL ZEROS THEOREM: ........................................................................................................................... 241

DIVISION AND FACTOR THEOREMS ................................................................................................................... 242

COMPLEX ZEROS OF POLYNOMIALS WITH REAL COEFFICIENTS .............................................................................. 242

USING DIVISION TO FIND ZEROS....................................................................................................................... 245

WORKSHEET SEC 3.2 ..................................................................................................................................... 249

EXERCISES 3.2 .............................................................................................................................................. 250

3.3 EXPONENTIAL AND LOGARITHMIC EQUATIONS ..................................................................................... 258

LOGARITHMIC EQUATIONS AND INEQUALITIES .................................................................................................... 258

EXPONENTIAL EQUATIONS AND INEQUALITIES .................................................................................................... 260

WORKSHEET SEC 3.3 ..................................................................................................................................... 266

EXERCISES 3.3 .............................................................................................................................................. 267

3.4 SYSTEMS OF EQUATIONS ................................................................................................................ 276

WORKSHEET SEC 3.4 ..................................................................................................................................... 281

EXERCISES 3.4 .............................................................................................................................................. 282

3.5 INTRODUCTION TO MATRICES AND GAUSS ELIMINATION METHOD ........................................................... 284

ELEMENTARY ROW OPERATIONS ON A MATRIX OF NUMBERS. .............................................................................. 285

WORKSHEET SEC 3.5 ..................................................................................................................................... 289

EXERCISES 3.5 .............................................................................................................................................. 290

Functions and Relations Page 1

Chapter 1: All about Functions and Relations

1.1 Introduction and Domain and Range of Functions and Relations Lecture

Functions and Relations Part 1 (13.15 min) https://www.youtube.com/watch?v=2HCxvI_S-WQ

We very often consider how different things are related in many situations in every-day life. Some

examples include: each person is related to their biological mother; the price of gas in dollars per gallon

is related to time; temperature in Fahrenheit degrees is related to temperature in Celsius degrees; your

height in inches is related to your age in years; the profit of an automaker is related to the annual car

sales; the number of movie ticket sales at a theater is related to the ticket price. Many of these

relationships can be quantified using numerical measures of each related quantity, e.g., the height of a

person in inches on his birthday is related to his age in years, or the relationship between Fahrenheit

and Celsius temperatures is given by the equation ℉ =9

5℃+ 32 for any value of ℃.

The connections between different quantities can be referred to as relations.

Oftentimes we think of a direction in the relation in that when the value of one quantity (called an

input) is known, this determines a value of the second quantity (called an output).

A function is a directed relation where every individual input has a unique output.

We will focus on relations between two items or characteristics that are connected to each other. For

example, think of the students in a class and their height in inches. We could represent these data as a

set of ordered pairs or as a table.

A={(Robert, 72 ), (Sarah, 61 ), (Matt, 70 ), (Robin, 61 ), (Sarah, 65)}

Name of Student Height in Inches

Robert 72

Sarah 61

Matt 70

Robin 61

Sarah 65

In the data pairs above, two people could have different names and have the same height. Also two

people can have the same name but different heights, or two could have the same height and the same

name with a larger class. The way this data is presented suggests that the name of the person is the

“input”, while the height of that person is the “output”.

Clearly the relation above is not a function, since the single input “Sarah” has two different outputs 61’’

and 65’’. Note that if the input column included middle initial,

https://www.youtube.com/watch?v=2HCxvI_S-WQ


e.g., B={(Robert A, 72 ),(Sarah K, 61 ), (Matt L, 70 ), (Robin P, 61 ),(Sarah M, 65)}, then each input has a

single output and 𝐵 would be a functional relation of the form (𝑛𝑎𝑚𝑒, ℎ𝑒𝑖𝑔ℎ𝑡). However the relation

𝐵 in the other direction from height to name would not be a function since in that direction, the input of

61” yields two different names!

Consider the sister relationship where any person with one or more sisters is the input and that person’s

sister(s) is the output. Since there are people who have more than one sister, this is not a function

relationship. Next, consider the biological mother relationship where any person is an input and the

output is that person’s biological mother. Explain why this is a function. Note if we reverse the direction

of this mother relation, it is not a function, i.e., starting with an input of a mother with more than one

child, that mother would give rise to at least two different outputs (her children).

Another way to express some relations and functions is through an equation with two variables. For

example, 𝑦 = 3𝑥 + 1, where the input 𝑥 is any real number, and 𝑦 is the corresponding output number.

This is a relation between 𝑥 and 𝑦 that is a function since for any value of the input 𝑥, there is only one

output, namely 3𝑥 + 1. We’d say this equation represents a function from 𝑥 to 𝑦. In the relation 𝑥 =

𝑦2, where input 𝑥 is any nonnegative real number, 𝑦 is not a function of 𝑥, since the input of 𝑥 = 4 will

yield two different outputs 𝑦 = 2, 𝑦 = −2.

In order to distinguish two different functions like 𝑦 = 3𝑥 + 1 and 𝑦 = 𝑥 − 5, we need a notation that

clearly describes the two “𝑦”s as being different. We can use the notation 𝑓(𝑥) = 3𝑥 + 1 and g(𝑥) =

𝑥 − 5 where we are giving names of 𝑓 and 𝑔 to these two different functional relations. We would read

that as “f of 𝑥 is 3𝑥 + 1” and “g of 𝑥 is 𝑥 − 5”. We could also write this as e.g., 𝑦 = 𝑓(𝑥) = 3𝑥 + 1

which identifies the function equation 𝑦 = 3𝑥 + 1 with the function name 𝑓. It is very important to

make sure you learn to read and process this notation correctly. Many students who make mistakes in

working with functions confuse the notation 𝑓(𝑥) with 𝑓 times 𝑥. Remember that the notation 𝑓(𝑥) is

NOT read as 𝑓 times 𝑥. You can think of it as 𝑓( ____ ), and the blank can be filled in with many different

inputs.

In the notation 𝑓(𝑥) = 3𝑥 + 1 , 𝑓 is the name of the function, and 𝑥 is called the input or argument,

𝑓(𝑥) refers to the output. This statement simply says that the function 𝑓 computes an output by

multiplying an input by 3 and then adding 1.

The collection of all the input 𝑥 values we can have in a function so that the output is well

defined is called the domain of the function. The collection of all the output values of a function

is called the range of the function.

Working with Domain and Range of a Function or a Relation

Playing

Remember there cannot be a function or a relation without having a domain and range. In mathematics we are always trying to use our imagination to see what all possibilities are in every new scenario.

Some questions to ask are


1. What kinds of sets of objects we can be used for domains and ranges. Try to come up with functions for each of the domains: {All the houses in the city you live in}, {All the employees at the school you are attending}, {All the days in time since you were born}.

2. Do functions have to have one input variable and one output variable, or could we have multiple inputs and one unique output? For example can you think of a function perhaps given by an equation where the output depends on two separate real numbers as inputs?

Before you continue reading, spend some time just thinking and trying to answer these questions. There are many different answers. Try being creative and see what you can come up with.

A function of one variable means the input can be described by just one variable, e.g., 𝑓(𝑥) = 3𝑥 + 1. When a function has multiple inputs and one output we call it a function of several variables. For example, the volume of a box is a function of three variables, its length 𝑙, its width 𝑤 and its height ℎ and we might write this as 𝑉 = 𝑓(𝑙, 𝑤, ℎ) = 𝑙 ⋅ 𝑤 ⋅ ℎ. The domain of a function can be anything you want it to be from a finite set of objects to infinite sets. We will mostly work with functions of a single variable in this book.

Next we develop some notation and conventions on how work with functions efficiently. Any letter really can represent the single input. For example, we may use letters like 𝑥, 𝑜𝑟 𝑡 to represent an input to a function. Similarly we use a different letter to represent the corresponding single output, e.g., 𝑦, or 𝑧 might represent the output of a function. The functional relationship itself will often be denoted by the letter 𝑓 or some other appropriate letter. A statement like 𝑦 = 𝑓(𝑥) simply says that 𝑦 represents an output, 𝑥 an input and they are related by the function called 𝑓. (Remember that this is not saying to multiply 𝑓 by 𝑥!) To define what this relationship is, we’d have to elaborate on the meaning of 𝑦 and 𝑥 and how 𝑦 and 𝑥 are related perhaps by some equation. We can also look at the function notation as follows where the object 𝑥 is transformed by the application of the function 𝑓 to become a new object called 𝑓(𝑥) (read as 𝑓 of 𝑥)

𝑥𝑓→ 𝑓(𝑥)

This notation is credited to the mathematician Leonard Euler and was developed around 1734.

To denote what kind of function we are working with mathematicians may use the notation shown

below. Writing 𝑓: R→R to means we have a real number as an input and also a real number as the

output. Such a function is called a real valued function of one variable. Writing 𝑓: R×R→ R to mean the function 𝑓 is taking an ordered pair of real numbers and transforming them into a real number. Such a function is called a real valued function of two variables. Functions can have as many input variables as

needed. Writing 𝑓: Rn→R to means we have an n-tuple as an input and output is a real number.

We can represent functions in many ways. Below are some of the ways.

a) Using function notation as described above. 1 𝑓(𝑥) = 𝑥2

Here the domain is set of all real numbers and range is set of all non-negative real numbers.

2 𝑓(𝑥, 𝑦) = 𝑥2 + 𝑦2 Here the domain is the set of ordered pairs of the type (𝑥, 𝑦) with 𝑥 and 𝑦 both real numbers and the range is the set of all non-negative real numbers.

3 𝑓(𝑡) = 2𝑡 − 5 Here the domain is set of all real numbers and range is the set of all real numbers.


b) As a set of objects. For example, 1. 𝑓: {(2,−1), (4,2), (3,2), (0,4)}.

This means the input 2 to 𝑓 produces the output −1. The input 0 produces the output 4 and so on. Here the domain is the set {2,4,3,0} and the range is the set {−1,2,4}. A set is just a group of objects and we don’t need to list the “2” twice.

2. 𝑓: {(𝑥, 𝑦)|𝑦 = 𝑥2, 𝑥 ∈ R}. Read this as 𝑓 is the function that is a collection of all ordered pairs of the type (𝑥, 𝑦) in which 𝑦 = 𝑥2 and 𝑥 is a real number. Here the domain is the set of all real numbers and range is set of all non-negative real numbers.

c) As formulas or equations in two or more variables. For example,

1. 𝑦 = 𝑥2, Here the domain is the set of all real numbers and range is set of all non-negative real numbers.

2. 𝑧 = 𝑥2 + 𝑦2, Here the domain is the set of ordered pairs of the type (𝑥, 𝑦) with 𝑥 and 𝑦 both real numbers and the range is the set of all non-negative real numbers.

d) As English sentences modeling scenarios. For example, 1. Linda’s parents loaned her $12,480 interest free for her college tuition and books. Linda

promised to pay her parents back $80 a week until the loan is completely paid off.

Write a function equation that tell Linda how much money she still owes 𝑥 weeks after

she began paying on the loan.

The equation would look like: Amount owed by Linda = 𝑦 = 12480 − 80 ⋅ 𝑥 The

domain is all whole numbers less the number of weeks it takes to pay off the loan, i.e., 12480

80= 156. The range is all whole numbers counting down by 80’s from 12480 to 0.

2. The period of a pendulum is proportional to the square root of its length. Here domain is set of all positive real numbers and range is also set of positive real numbers.

e) As graphs.

1. Input is the 𝑥 coordinate of the point and output is the 𝑦 coordinate of the point. For example input of−2 will give you output of 5

since (−2,5) is a point plotted above. Here the domain is {−5,−3,−2,0,1,2,3,4,7} and range is {−1,0,1,2,3,4,5}.

2. Input is the 𝑥 coordinate of a point on the graph of the function 𝑓(𝑥) =𝑥2, then output is the 𝑦 coordinate of that point. For

example input of−2 will give you output of 4 since (−2,4) is a point on the graph above. Here the domain is the set of all real numbers and range is set of all non-negative real numbers.


Below are more examples of functions.

1. F= 𝑇(𝐶) =9

5𝐶 + 32 . This statement is a description of the degrees centigrade to Fahrenheit

conversion function. The input is the temperature in centigrade denoted as 𝐶, the output is the temperature in Fahrenheit 𝐹, and the name of the function is 𝑇. In this course our functions usually will be defined in this way, i.e., 𝑦 = 𝑓(𝑥) =some algebraic formula that involves 𝑥.

Also note that a function is the relation between input and output and that 𝑓(𝑥) =9

5𝑥 + 32 is

the same function as 𝑇(𝐶) =9

5𝐶 + 32 since they both have exactly the same set of

(input, output) ordered pairs.

2. 𝑉 =4

3𝜋𝑟3 is the formula for volume V of a sphere where the radius= 𝑟 is known. This is a

polynomial function. We might call this function simply 𝑉 in which case we’d have 𝑉 = 𝑉(𝑟) =4

3𝜋𝑟3 where we use the same variable to denote an output and the name of the relationship.

Likewise, the area of a circle function can be expressed as 𝐴 = 𝐴(𝑟) = 𝜋𝑟2 where 𝑟 =radius of the circle.

3. The height of a baseball at a given time might be given by 𝑦 = ℎ(𝑡) = 2 + 160𝑡 − 16𝑡2 where 𝑦 is the height above the ground in feet and 𝑡 is the number of seconds after the ball is hit by the bat and ℎ is the name of this relationship between time and height. This is also an example of a polynomial function. This notation allows us to easily describe the height at several different times, e.g.,ℎ(2) = 258 says that when the input is 2 seconds, the height is 258 feet. Other questions can also be stated easily using this notation, e.g., “determine when ℎ(𝑡) =100” says that the output of the height function is 100 and we are to find the input 𝑡 when this happens. You could try guessing and checking to find out that this happens at 𝑡 ≈0.66𝑠𝑒𝑐, and 𝑡 ≈ 9.34 𝑠𝑒𝑐. You can use the quadratic formula to find exact answers to 100 =2 + 160𝑡 − 16𝑡2. We will review this in a later section.

4. Another type of function that is often useful is called an exponential function where the input is actually in the exponent. For example 𝑃(𝑡) = 7.4(1.012)𝑡 might be used as a model of the world’s population 𝑡 years after 2016 where the output is in billions of people. Thus 𝑃(10) =7.4(1.012)10 ≈ 8.34 billion. This number 8.34 billion is the projection of the world population in 2026.


Playing with function notation

If we say 𝑓(𝑥) = 3𝑥 + 1, then that means that we really have

𝑓( ______) = 3 × (______) + 1

That means if you are asked to evaluate the following you must replace the blank above with whatever

takes its place in the notation

𝑓(2) = 3 × (2) + 1 = 6 + 1 = 7

𝑓(100) = 3 × (100) + 1 = 300 + 1 = 301

𝑓(𝑎) = 3(𝑎) + 1 = 3𝑎 + 1

𝑓(𝑎 + ℎ) = 3(𝑎 + ℎ) + 1 = 3𝑎 + 3ℎ + 1

See if you can use these examples to work on the following practice problems.

Practice Problems

1. Let 𝑓(𝑥) = 7𝑥2 + 1 and 𝑔(𝑡) = √𝑡 + 7. Find the values of

a. 𝑓(4)

b. 𝑓(𝑎 + ℎ)

c. 𝑓(−2)

d. 𝑔(𝑎 + 1)

e. 𝑔(4)

f. 𝑓(3) + 𝑔(𝑎)

2. Let 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒(𝑡) = |𝑡|. Find the values of

a. 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒(−5.1)

b. 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒(10)

c. 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒(−1200)

d. 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒(1400)

𝑓( ______) = 3 × (______) + 1

𝑓(𝐶𝑂𝐿𝐷) = 3 × (𝐶𝑂𝐿𝐷) + 1


Solutions

1. Let 𝑓(𝑥) = 7𝑥2 + 1 and 𝑔(𝑡) = √𝑡 + 7. Find the values of

a. 𝑓(4) = 7 ⋅ 42 + 1 = 7 ⋅ 16 + 1 = 113

b. 𝑓(𝑎 + ℎ) = 7 ⋅ (𝑎 + ℎ)2 + 1

c. 𝑓(−2) = 7 ⋅ (−2)2 + 1 = 7 ⋅ 4 + 1 = 29

d. 𝑔(𝑎 + 1) = √𝑎 + 1 + 7

e. 𝑔(4) = √4 + 7 = 9

f. 𝑓(3) + 𝑔(𝑎) = 7 ⋅ (3)2 + 1 + √𝑎 + 7 = 63 + 1 + √𝑎 + 7 = 71 + √𝑎

2. Let 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒(𝑡) = |𝑡|. Find the values of

a. 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒(−5.1) = |−5.1| = 5.1

b. 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒(10) = |10| = 10

c. 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒(−1200) = |−1200| = 1200

d. 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒(1400) = |1400| = 1400

Vertical Line Test:

The graph of a relation is a function from 𝑥 to 𝑦 if and only if every vertical line intersects the

graph at no more than one point.

Examples

1. Below 𝑦 is a function of 𝑥 as it passes the vertical line test.

2. Below 𝑦 is not a function of 𝑥 as it does not pass the vertical line test.

3. Below 𝑦 is a function of 𝑥 as it passes the vertical line test.

As you can see from the graph above, any vertical line will intersects the graph at only one point.

As you can see from the graph above, any vertical line will intersects the graph at only one point for the domain 𝑥 ≥ 0.

As you can see on the graph above, the vertical line intersects the graph in two points.


Playing

If you look at examples 1 and 3 above, a natural question that we can ask is what information would a

horizontal line test provide? We can see that if a horizontal line is drawn in example 1, it intersects the

graph at two points. This means that a single 𝑦-value where the horizontal line is at is related to two

points or two 𝑥-values. Thus one 𝑦 → (𝑡𝑤𝑜 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠) and 𝑥 is not a function of 𝑦. As an equation

the relation 𝑦 = 𝑥2, has two 𝑥-solutions for each 𝑦 and thus 𝑥 is not a function of 𝑦.

However, in example 3, all horizontal lines hit the graph at most one time and thus 𝑥 is a function of 𝑦

for graph 3. The relation in 3. has equation 𝑦 = √𝑥 which can also be converted to 𝑦2 = 𝑥, with 𝑦 ≥

0. Thus each 𝑦 gives rise to a single 𝑥. Note that we have the same set of ordered pairs here, but the

function directions are in opposite directions thus 𝑦 = √𝑥 starts with 𝑥 to produce the square root of

𝑥 and this is the square root function. However, 𝑥 = 𝑦2 takes as input a 𝑦-value and the output is the

square of this input, thus in this direction from 𝑦 → 𝑥 we have the squaring function.

A function that satisfies the horizontal line test is called a one-one-to function.

One-to-One Function: A function in which each output comes from only one input is called a

one-to-one function.

Horizontal Line Test:

A function from 𝑥 to 𝑦 is one-to-one if and only if every horizontal line intersects the graph at

no more than one point.

To check algebraically whether a function is one-to-one, we need to make sure that the outputs of a

function 𝑓 are equal, i.e., 𝑓(𝑎) = 𝑓(𝑏) , only when 𝑎 = 𝑏.


Examples

Determine if the functions below given by a graph or an equation are one-to-one.

1. Below 𝑦 is not a one-to-one function of 𝑥 as it does not pass the horizontal line test.

2. Below 𝑦 is not a one-to-one function of 𝑥 as it does not pass the horizontal (or vertical) line test.

3. Below 𝑦 is a one-to-one function of 𝑥 as it passes the horizontal line test.

As you can see from the graph above any horizontal line intersects the graph in only one point.

As you can see from the graph above the horizontal line will intersect the graph in two points.

As you can see from the graph above the horizontal line intersects the graph in two points.

4. 𝑦 = 3𝑥 − 1 We could graph the function and use horizontal line test as above or check algebraically whether the function is one-to-one. If 𝑥 = 𝑎, and 𝑥 = 𝑏 are two real values, then the outputs 𝑓(𝑎) = 3𝑎 −1 = 3𝑏 − 1 = 𝑓(𝑏) , solving the equations for 𝑏 we get 3𝑎 = 3𝑏, or 𝑎 = 𝑏. That is the function output values are only equal if 𝑎 = 𝑏 implying the function is one-to-one.

5. 𝑦3 = 𝑥 Checking algebraically whether the function is one-to-one: For two real values 𝑦 = 𝑎, and 𝑦 = 𝑏 If 𝑓(𝑎) = 𝑎3 = 𝑏3 = 𝑓(𝑏), solving the equations for 𝑏 we get 𝑎 = 𝑏 Thus the function values are only equal if 𝑎 = 𝑏 implying the function is one-to-one.

6. 𝑦 + 3𝑥2 = 4 or 𝑦 = 4 − 3𝑥2 Checking algebraically whether the function is one-to-one: For two real values 𝑥 = 𝑎, and 𝑥 = 𝑏, 𝑓(𝑎) = 4 −3𝑎2 = 4 − 3 = 𝑓(𝑏), we get 𝑎2 = 𝑏2 or that 𝑎 = ±𝑏. That implies the function is not one-to-one. For example 𝑥 =2 = 𝑎 and 𝑥 = −2 = 𝑏 both give the same output 𝑦 = 4 −12 = −8.


Practice Problems

1. Determine the following

i. Is the relation a function or not.

ii. Is the relation a one-to-one function

iii. Domain and Range

A. Domain Range

E Bus F

G

H

o Function o One-to-One

o Not a Function o Not One-to-One

Domain: Range:

B. Domain Range

0 𝑎 −10 𝑏 4 𝑐 34 𝑑



Domain: Range:

C. {(𝑎, 𝑏), (𝑎, 2), (𝑏, 2), (𝑐, 4)}



Domain: Range:

D. {(𝑎, −1), (𝑏, 2), (𝑐, 2), (𝑑, 4)}



Domain: Range:

E. An Olympic size swimming pool has a capacity of 2,500,000 liters of water. The pool currently holds 100,000 liters of water, and water is being pumped at 400,000 liters/hour into the pool. Write a relation that represents the amount of water 𝑊 (liters) in the pool after 𝑡 hours.



Domain: Range:


F.



Domain: Range:

G.



Domain: Range:

H. 𝑓(𝑥) = |𝑥|



Domain: Range:

I. 𝑥2 + 𝑦2 = 25



Domain: Range:

J. 𝑇𝑒𝑚𝑝(𝑡) =5

9(𝑡 − 32)



Domain: Range:

K. 𝑦3 = 𝑥



Domain: Range:


L.



Domain: Range:

M.



Domain: Range:

N.



Domain: Range:

O.



Domain: Range:


Practice Problems Solutions

1. Determine the following

i. Is the relation a function or not.

ii. Is the relation a one-to-one function

iii. Domain and Range

A. Domain Range

E Bus F

G

H

✓ Function o One-to-One

o Not a Function ✓ Not One-to-One

Domain: {E,F,G,H} Range: {Bus}

B. Domain Range

0 𝑎 −10 𝑏 4 𝑐 34 𝑑


✓ Not a Function o Not One-to-One

Domain:{0,−10,4,34} Range: {𝑎, 𝑏, 𝑐, 𝑑}

C. {(𝑎, 𝑏), (𝑎, 2), (𝑏, 2), (𝑐, 4)}



Domain: {𝑎, 𝑏, 𝑐} Range: {𝑏, 2,4}

D. {(𝑎, −1), (𝑏, 2), (𝑐, 2), (𝑑, 4)}



Domain: {𝑎, 𝑏, 𝑐, 𝑑} Range: {−1,2,4}

E. An Olympic size swimming pool holds 2,500,000 liters of water. The pool currently holds 100,000 liters of water, and water is being pumped at 400,000 liters/hour into the pool. Write a function that represents the amount of water 𝑊 (liters) in the pool at 𝑡 hours. Find the domain and range of this function.

We will represent the amount of water time 𝑡 hours as 𝐴(𝑡). Since there is already 100,000 liters of water in the pool and then we are adding more water per hour we have 𝑊 = 𝐴(𝑡) = 100,000 + 400,000𝑡. To find domain and range we have to think about what 𝑡 values we can have. Note that the water in the pool cannot exceed the pool size of 2,500,000 liters. So we have 100000 ≤ 100000 + 400000𝑡 ≤ 2500000 solving the inequality we will get that 0 ≤400000𝑡 ≤ 2500000 − 100000 0 ≤ 400000𝑡 ≤ 2400000 or

0 ≤ 𝑡 ≤2400000

400000 or 0 ≤ 𝑡 ≤ 6, also note that when time 𝑡 = 6 hours the pool will be at full

capacity of 2,500,000𝑙 and at time 𝑡 = 0 the pool would have 100,000 liters. Domain = [0,6] , Range [100000,2500000] The function is one-to-one as the graph is a line with slope 𝑚 = 100,000 and thus passes the horizontal (and vertical) line tests.


F.

𝑦 is a function of 𝑥 and vice versa.

✓ Function ✓ One-to-One


Domain: (−∞,∞) Range: (−∞,∞)

G.

Only 𝑥 is a function of 𝑦



Domain: [0,∞) Range: (−∞,∞)

H. 𝑓(𝑥) = |𝑥| For every real number 𝑥 = 𝑎, we have 𝑦 =|𝑎| which results in a unique real number.



Domain: (−∞,∞) Range: [0,∞)

I. 𝑥2 + 𝑦2 = 25 For every real number 𝑥 = 𝑎, we have 𝑦 =

±√25 − 𝑎2 which are two real numbers for any 𝑎 that satisfies −5 ≤ 𝑎 ≤ 5.



Domain: [−5,5] Range: [−5,5]

J. 𝑇𝑒𝑚𝑝(𝑡) =5

9(𝑡 − 32)

For every real number 𝑡 we will get a unique

real number 5

9(𝑡 − 32).

For any two values of = 𝑎 , and 𝑡 = 𝑏 we

have 5

9(𝑎 − 32) =

5

9(𝑏 − 32) or 𝑎 − 32 =

𝑏 − 32, or 𝑎 = 𝑏 which means the function is one-to-one.



Domain: (−∞,∞) Range: (−∞,∞)

K. 𝑦3 = 𝑥 𝑦 is a function of 𝑥 and vice versa. For every real number 𝑥 = 𝑎 we have one

and only one output 𝑦 = √𝑎3

. For any two values of = 𝑎 , and 𝑦 = 𝑏 we

√𝑎3

= √𝑏3

or that 𝑎 = 𝑏 which means the function is one-to-one.



Domain:(−∞,∞) Range: (−∞,∞)


L.

Domain: [−6,−1] ∪ [2,7] Range: {𝟓, 𝟐}

Each 𝑥 coordinate has a unique 𝑦 coordinate, so 𝑦 is a function of 𝑥. But many 𝑥 coordinates give the same output, e.g., 𝑦 = 2, so 𝑦 is not a one-to-one function of 𝑥.



M.

Domain:[−6,2] Range: {𝟓, 𝟏} Each 𝑥 coordinate between −3 and −1 has two 𝑦 coordinates associated to it on the graph making 𝑦 not a function of 𝑥.



N.

Each 𝑥 coordinate has a unique 𝑦 coordinate, so 𝑦 is a function of 𝑥. But many 𝑥 coordinates give the single 𝑦-coordinate 𝑦 = 5, so 𝑦 is not a one-to-one function of 𝑥.


o Not a Function

✓ Not One-to-One

Domain:[−𝟒, 𝟔] Range: {𝟑, 𝟓}

O.

Each 𝑥 coordinate has a unique 𝑦 coordinate, so 𝑦 is a function of 𝑥. But several 𝑥 coordinates have the same output, so 𝑦 is not a one-to-one function of 𝑥.



Domain:[−𝟑, 𝟔] Range: [−𝟏, 𝟒]

From above examples you can see that sometimes even when the domain is an interval, the range can

be discrete points and other times it is an interval of 𝑦-values. So there is no fixed pattern, just pay

careful attention to the function and remember the 𝑦-coordinates are the range values and the 𝑥-

coordinates are the domain values.

Range

Domain

Range

Domain

Range

Domain

Domain

Range


Section1.1 Worksheet Date:______________ Name:__________________________

Concept Meaning in words and/or examples as required

1. Relation

2. Function

3. Domain of a function

4. Range of a function

5. All the different ways to represent a function (give examples you make up not the ones that appear in the book)

6. One-to-one function

7. Give an example of a relation that is not a function

8. Given an example of a function that is not one-to-one.

Difficulties encountered in the section:


Exercises 1.1

1. Determine the following. (For graphs, the direction of the relation is from 𝑥 to 𝑦.) i. Is the relation a function or not.

ii. If the relation is a function, is it one-to-one? iii. Domain and Range of the relation.

A. Domain Range

d Math F

I

j



Domain: Range:

B. Domain Range

5 𝑅𝑒𝑑 −20 𝑃𝑢𝑟𝑝𝑙𝑒 13 𝑐𝑜𝑤 7 𝑑𝑜𝑔



Domain: Range:

C. {(−1, 𝑏), (2,8), (5, −3), (2, −3)}



Domain: Range:

D. {(2,−1), (2,1), (𝑐, 2), (𝑑, 4)}



Domain: Range:

E.



Domain: Range:

F.



Domain: Range:


G. 𝑓(𝑥) = 3√𝑥 − 1



Domain: Range:

H. 𝑥2 + 𝑦2 = 9 (Imagine the graph!)



Domain: Range:

I. 𝑇𝑒𝑚𝑝(𝑡) =5

9(𝑡 − 32)



Domain: Range:

J. 𝑦5 = 𝑥



Domain: Range:

K.



Domain: Range:

L.



Domain: Range:


M.



Domain: Range:

N.



Domain: Range:

2. Draw the graph of 𝑦 = 𝑓(𝑥) where the domain is −3 ≤ 𝑥 ≤ 5 and the Range is −1 ≤ 𝑦 ≤ 3 and 𝑓(0) = 2 and the function is one-to-one.

3. Draw the graph of 𝑦 = 𝑔(𝑥) where the domain is −3 ≤ 𝑥 ≤ −1 ∪ 1 ≤ 𝑥 ≤ 3 and the Range is 𝑦 ∈ {−2, 3} and 𝑓(2) = −2. Is your function one-to-one?

6. For each if the following equations, determine whether:

In column 1, is 𝑦 a function of 𝑥? In column 2, is 𝑥 a function of 𝑦?

A. 𝑦2 = 2𝑥 Function Not a function

B. 7𝑥 − 5𝑦 = 10 Function Not a function

C. 2𝑥 − 7𝑦 = 9 Function Not a function

D. 𝑥 = 3𝑦2 Function Not a function

E. 6𝑥 + |𝑦| = 3 Function Not a function

F. 𝑥3 = 𝑦 Function Not a function

G. 𝑥2 + 2 = 𝑦 Function Not a function

H. 𝑥2𝑦2 = 9 Function Not a function

I. 𝑥2 + 2𝑦2 = 8 Function Not a function

J. 𝑥𝑦 = 5 Function Not a function


7. The function ℎ is defined by the following rule: ℎ(𝑥) = 4𝑥 + 5. Complete the following table.

𝑥 ℎ(𝑥) −4

−2

2

4

5

8. The functions 𝑓 and 𝑔 are defined as follows:

𝑓(𝑥) = −3𝑥 + 2, 𝑔(𝑥) = 3𝑥3 + 5. Find 𝑓(3) and 𝑔(−3). Simplify your answers as much as possible. 𝑓(3) = 𝑔(−3) =

9. The function 𝑓 is defined as follows: 𝑓(𝑥) =4𝑥

3𝑥−15.

Find 𝑓(4). Simplify your answer as much as possible.

10. The function ℎ is defined as follows:

ℎ(𝑥) =𝑥2−3𝑥−10

𝑥2−14𝑥+45.

Find ℎ(6). Simplify your answer as much as possible.

11. Fill the table using the function rule

𝑓(𝑥) = √𝑥 + 6. Simplify your answers as much as possible.

𝑥 𝑓(𝑥) −9

0

1/4

4

25

12. The function ℎ is defined as follows:

ℎ(𝑥) = √𝑥 − 13

. Evaluate each output value and simplify your answers as much as possible. ℎ(9) = ℎ(126) =

ℎ(−26) =


13. Simplify the expressions below for 𝑓(𝑥) =2𝑥2 + 3𝑥 − 5. 𝑓(−6) =

𝑓(𝑎) = 𝑓(𝑎 + ℎ) =

14. With 𝑀(𝑥) and 𝐹(𝑥) being the mother and father of person 𝑥, try to evaluate: 𝑀(𝑦𝑜𝑢𝑟𝑠𝑒𝑙𝑓) =

𝐹(𝑀(𝑦𝑜𝑢𝑟𝑠𝑒𝑙𝑓)) =

𝑀(𝐹(𝑦𝑜𝑢𝑟𝑠𝑒𝑙𝑓) =

𝑀(𝐹(𝑀(𝑦𝑜𝑢𝑟𝑠𝑒𝑙𝑓))) =

15. Describe one or more functions and what the relations mean through using each of four methods: by a formula; by a table; by a graph; and through a sentence or two of words. State the domain and range of your function. (You could do this all for a single function or you can come up with different functions for each method.)

16. Create a function 𝑦 = 𝑓(𝑥) that satisfies the following criteria. a. Is One-to-one b. Has domain all real numbers c. 𝑓(2) = −1

Questions about your function- i) What is the domain of 𝑓(𝑥)? ii) What is the range of 𝑓(𝑥)?


17. Create a function 𝑦 = 𝑓(𝑥) that satisfies the following criteria. a. Has domain [−2, 1] b. For at least two 𝑥’s in the domain of 𝑓, 𝑓(𝑥) = 4

Questions about your function- i) What is the domain of 𝑓(𝑥)? ii) What is the range of 𝑓(𝑥)? iii) Is your function one-to-one?

18. Create a function 𝑦 = 𝑓(𝑥) that satisfies the following criteria. a. Is One-to-one b. Has domain [−2, 1] ∪ [3,10] c. 𝑓(−2) = 1 d. 𝑓(3) = 0 e. 𝑓(0) = 5 f. For at least one 𝑥 in the domain of 𝑓, 𝑓(𝑥) = −4

Questions about your function- i) What is the domain of 𝑓(𝑥)?

ii) What is the range of 𝑓(𝑥)?


1.2 Inverse Functions and a Brief Library of Function Types

Functions and Relations Part 2 (10:21 min) https://www.youtube.com/watch?v=RZwGOfAeqp4

Functions and Relations Part 3 (10:41 min)https://www.youtube.com/watch?v=gPBE_QqVwbk

In section one we developed the basic concept of a function and the domain and range sets as well as

what it means for a function to be one-to-one. In this section we’ll start by switching the direction of

functional relations to obtain what is called the inverse relation. Sometimes these inverse relations will

also be functions and it turns out that will be the case exactly when a function is one-to one. We’ll also

get familiar with a collection of basic types of functions that come up frequently in modeling in the

natural and social sciences. Even some very simple functions can lead to very beautiful images. As an

example, the simple function 𝑓(𝑥) = 𝑥2 + 𝑐 can be used to create Fractal Images called “Julia Sets” Julia

Set video. The kinds of functions you can create is limited only by your imagination.

Recall in the previous section our definition of one-to-one functions.

One-to-One Function: A function in which each output 𝑦 comes from only one input 𝑥 is called

a one-to-one function.

We define the inverse of a function to be the same pairing as the original function, but with the direction reversed.

Inverse Function: If 𝑦 = 𝑓(𝑥) is a one-to-one function, then the function denoted by 𝑓−1 is

called the inverse of 𝑓 function and for every 𝑦 in the range of 𝑓, 𝑓−1 takes this 𝑦 as an input

and gives the one 𝑥-value in the domain of 𝑓 as its output. Thus when 𝑦 = 𝑓(𝑥), we also have

𝑥 = 𝑓−1(𝑦). In other words the inverse function simply maps 𝑦 back to 𝑥. Also 𝑓 being one-to-

one is exactly what is required for 𝑓−1 to be a function, i.e., only one 𝑥-value to go back to.

Note: The notation 𝑓−1 is the name of the inverse function and the −1 is not an exponent.

𝑓−1 reverses the direction of the 𝑓 relation. The roles of the domain and range for 𝑓−1 are reversed from that for 𝑓. Thus the range of 𝑓 is the domain of 𝑓−1 and the domain of 𝑓 is the range of 𝑓−1.

https://www.youtube.com/watch?v=RZwGOfAeqp4

https://www.youtube.com/watch?v=gPBE_QqVwbk

https://en.wikipedia.org/wiki/File:Julia_circling.ogg

https://en.wikipedia.org/wiki/File:Julia_circling.ogg


The examples below illustrate these properties of 𝑓 and 𝑓−1 and the one-to-one necessity of 𝑓 for 𝑓−1 to be a function.

1.

𝑥𝑓→ 𝑦

Domain Range

−1 1

2

0 1 1 2 2 4 3 8

𝑥𝑓−1

← 𝑦

Range Domain

−1 1

2

0 1 1 2 2 4 3 8

𝑥𝑔→ 𝑦

Domain Range

−1 1

0 1 0

2 2

−2

𝑥𝑔−1

← 𝑦

Range Domain

−1 1

0 1 0

2 2

-2

As you can see the function 𝑓 is a one-to-one function. You can also see that if we reverse the direction it remains a function. This is the concept of the inverse function. You can see in the inverse function the roles that 𝑥 and 𝑦 play are reversed. Domain of 𝑓 = {−1,0,1,2,3} = Range of 𝑓−1

Range of 𝑓 = {1

2, 1,2,4,8} = Domain of 𝑓−1

As you can see the relation 𝑔−1 is not a function. For example 𝑦 = 1 gets mapped back to two 𝑥-values. We can still look at the inverse relation but 𝑔−1 is not a function. We will concentrate on one-to-one functions when working with the concept of inverse functions.

2. If 𝑓: {(1, 𝑎), (−2, 𝑡), (4, 𝑔), (8, 𝑝)} (which is a one-to-one function), then the function where the

domain values become the range values and vice versa for each of the pairs is the inverse function. So 𝑓−1: {(𝑎, 1), (𝑡, −2), (𝑔, 4), (𝑝, 8)} this new relation is also a function. Domain of 𝑓 = {1,−2,4,8} = Range of 𝑓−1

Range of 𝑓 = {𝑎, 𝑡, 𝑔, 𝑝} = Domain of 𝑓−1 3. If 𝑦 = 𝑓(𝑥) is the graph shown below in red, then the purple graph would represent its inverse

function. Since all points (𝑥, 𝑦) on the original graph would become (𝑦, 𝑥) in the inverse function graph, you can see that all points where 𝑦 = 𝑥 will remain the same. So essentially it means if we graph the function 𝑦 = 𝑓(𝑥) its inverse function is the same graph reflected across the line 𝑦 = 𝑥.


Steps to graph inverse functions: To sketch the graph of 𝑦 = 𝑓−1(𝑥), switch the 𝑥,𝑦-roles of all the points on the graph of 𝑦 = 𝑓(𝑥). If(𝑥 = 𝑎, 𝑦 = 𝑓(𝑎) = 𝑏), or (𝑎, 𝑏), is on the graph of 𝑓, then (𝑥 = 𝑏, 𝑦 = 𝑎) is on the graph of 𝑦 = 𝑓−1(𝑥). You may notice that each of the point pairs (𝑎, 𝑏) and (𝑏, 𝑎) lie directly across the diagonal line 𝑦 = 𝑥 from each other. In fact the complete graphs of 𝑦 = 𝑓(𝑥) and 𝑦 = 𝑓−1(𝑥) are reflections of each over the line 𝑦 = 𝑥. Finding formulas for the inverse function when given a formula for the function:

Consider the function 𝑦 = 𝑓(𝑥) = 𝑥³. The relationship from 𝑥 to 𝑦 is given by 𝑦 = 𝑥³. When

we want the inverse function, we want to start with a 𝑦-value from the output of 𝑓 and

somehow determine what input 𝑥 the function 𝑓 uses to get to that 𝑦. To do this we take the

equation for 𝑓, i.e., 𝑦 = 𝑥³ and solve it for 𝑥 to see how to get 𝑥 when 𝑦 is known.

Details: Solve the equation 𝑦 = 𝑓(𝑥) = 𝑥³ for 𝑥 to get 𝑥 = √𝑦3 .

Thus 𝑥 = 𝑓⁻¹(𝑦) = √𝑦3 is the formula that tells what 𝑓−1 does to the input 𝑦 to get 𝑥.

We usually (but not always) use 𝑥 to indicate the input and 𝑦 for the output of a function. Doing

this for the function 𝑓⁻¹, means we switch the 𝑥 and 𝑦 roles to:

𝑦 = 𝑓⁻¹(𝑥) = √𝑥3

.

Typically we switch the 𝑥, 𝑦 roles first as 𝑥 = 𝑓(𝑦) and then solve for 𝑦 to obtain 𝑦 = 𝑓⁻¹(𝑥).

This way, we'd start with 𝑥 = 𝑓(𝑦) 𝑜𝑟 𝑥 = 𝑦³ → √𝑥3

= 𝑦 → 𝑦 = 𝑓⁻¹(𝑥) = √𝑥3

.

Steps to finding a formula for 𝒚 = 𝒇−𝟏(𝒙) when given a formula for 𝒚 = 𝒇(𝒙) : Step 1: Write the original one-to-one function as 𝑦 = 𝑓(𝑥) Step 2: Switch the roles of 𝑥 and 𝑦 so we have 𝑥 = 𝑓(𝑦) Step 3: Solving for 𝑦 will give us the inverse function 𝑦 = 𝑓−1(𝑥).

Examples: Find the inverse function to each of the functions below.

1. 𝑓(𝑥) = 2𝑥 − 3 Step 1: 𝑦 = 2𝑥 − 3 Step 2: 𝑥 = 2𝑦 − 3 Step 3: 𝑥 + 3 = 2𝑦

𝑥+3

2= 𝑦

Solution: 𝑓−1(𝑥) =𝑥+3

2

Since both the function and its inverses are lines Domain of 𝑓 = (−∞,∞) = Range of 𝑓−1 Range of 𝑓 = (−∞,∞) = Domain of 𝑓−1

2. 𝑓(𝑥) = 𝑥2, 𝑓𝑜𝑟 𝑥 ≥ 0 Step 1: 𝑦 = 𝑥2 𝑓𝑜𝑟 𝑥 ≥ 0. Keeping the domain at 𝑥 ≥ 0 is makes the function one-to-one. Step 2: 𝑥 = 𝑦2 𝑓𝑜𝑟 𝑦 ≥ 0 Here 𝑥 ≥ 0 too, so taking square roots will give a real 𝑦-value.

Step 3: √𝑥 = 𝑦 𝑓𝑜𝑟 𝑥 ≥ 0

Solution: 𝑓−1(𝑥) = √𝑥, 𝑥 ≥ 0 Domain of 𝑓 = [0,∞) = Range of 𝑓−1 Range of 𝑓 = [0,∞) = Domain of 𝑓−1


3. Sketch the graph of the inverse function for each function shown below. Also state the domain

and range of the original function and of its inverse.

a. 𝑦 = 𝑓(𝑥).

Domain of 𝑓 = {−2,2,4,5} Range of 𝑓 = {−1,2,3,0}

Domain of 𝑓−1 = {−1,2,3,0}

Range of 𝑓−1 = {−2,2,4,5}

b. 𝑦 = 𝑔(𝑥)

Domain of 𝑔 = [−5,0] Range of 𝑔 = [−2,2]

Domain 𝑔−1 = [−2,2] Range of 𝑔−1 = [−5,0]

𝑦 = 𝑓(𝑥) 𝑦 = 𝑓−1(𝑥)

(−5,−2)

(−2.5,0)

(0,2)

(−2,−5)

(0, −2.5)

(2, 0)

𝑦 = 𝑔−1(𝑥)

𝑦 = 𝑔(𝑥)


Other Types of Functions

1. Square Root Function: A function defined as 𝑔(𝑥) = √𝑥, for all real numbers 𝑥 ≥ 0.

➢ Graph

➢ One-to-One: Yes ➢ Inverse function 𝑔−1(𝑥) = 𝑥2, 𝑓𝑜𝑟 𝑥 ≥

0

Step 1: 𝑦 = √𝑥, 𝑓𝑜𝑟 𝑥 ≥ 0

Step 2: 𝑥 = √𝑦, 𝑓𝑜𝑟 𝑦 ≥ 0

Step 3: 𝑥2 = 𝑦 𝑓𝑜𝑟 𝑦 ≥ 0, and therefore 𝑥 ≥0 ➢ Domain of 𝑔 = [0,∞) = Range of 𝑔−1 ➢ Range of 𝑔 = [0,∞) = Domain of 𝑔−1

Examples: For 𝑔(𝑥) = √𝑥 find the values of

a. 𝑔 (2

3)

b. 𝑔(100) c. 𝑔(3456)

Solution:

a. 𝑔 (2

3) = √

2

3≈0.816

b. 𝑔(100) = √100 = 10

c. 𝑔(3456) = √3456 ≈ 58.79

2. Polynomial function: A function defined as 𝑃(𝑥) = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + 𝑎3𝑥

3 +⋯+ 𝑎𝑛𝑥𝑛,

where 𝑎0, 𝑎1, 𝑎2, 𝑎3,…𝑎𝑛 are all real numbers and n≥ 0 is a whole number. The domain of these functions is all real numbers. The range is all real numbers when 𝑛 is odd, but is more difficult to determine when 𝑛 is even.

Example

i) Constant Function: A function defined as 𝑓(𝑥) = 𝑎, where 𝑎 is any real number. A special case of the polynomial function of degree zero.

➢ Generic Graph

➢ Domain: All real numbers ➢ Range: {𝑎} ➢ One-to-One: NO

𝑔(𝑥) = √𝑥

𝑔−1(𝑥) = 𝑥2, 𝑥 ≥ 0

(4,2)

(2,4)

𝑦 = 𝑎


Example If 𝑓(𝑥) = 5, then find

a. 𝑓(3) b. 𝑓(−2) c. 𝑓(−3456) d. 𝑓(𝑎 + ℎ) e. Sketch the graph

Solution: a. 𝑓(3) = 5 b. 𝑓(−2) = 5 c. 𝑓(−3456) = 5 d. 𝑓(𝑎 + ℎ) = 5 e. As you can see from the parts a,b,and c, no matter what 𝑥-coordinate you plot the 𝑦-coordinate is always 5 so it is a horizontal line as shown above.

ii) Linear Function: A function defined as (𝑥) = 𝑚𝑥 + 𝑏 . This is a polynomial function with degree one. Remember 𝑚 =slope of the line, 𝑏 = 𝑦-intercept.

Example 1. If 𝑓(𝑥) = 2𝑥 − 3, then

a. Find 𝑓(−1) b. Find 𝑓(0) c. Find 𝑓(4) d. Find the inverse function and its graph. e. Sketch the graph of 𝑦 = 𝑓(𝑥)

Solution: a. 𝑓(−1) = 2(−1) − 3 = −2 − 3 = −5 b. 𝑓(0) = −3 c. 𝑓(4) = 2(4) − 3 = 8 − 3 = 5 d. Inverse function

𝑦 = 2𝑥 − 3 𝑥 = 2𝑦 − 3 𝑥 + 3 = 2𝑦 𝑥 + 3

2= 𝑦

Inverse function is 𝑓−1(𝑥) =𝑥+3

2, slope is

1

2 and 𝑦-intercept is at

3

2.

e. Graph of the function 𝑦 = 𝑓(𝑥) is to the right with slope of 2 and 𝑦-intercept of −3.

2. An Olympic size swimming pool holds 2,500,000 liters of water. If the pool currently holds 100,000 liters of water, and water is being pumped at 400,000 liters/hour into the pool.

a. Write a function that represents the number of liters 𝑊 of water in the pool at 𝑡 hours. b. Find the domain and range of this function. c. Sketch the graph of this function. d. Find the inverse of this function, but don’t switch the variable names.


Solution: a. 𝑊 = 𝐴(𝑡) = 100000 + 400000𝑡 Liters,

for 0 ≤ 𝑡 ≤ 6 The domain starts at 𝑡 = 0 and after 6 hours the pool will be full since it holds a maximum of 2,500,000 liters of water.

b. Domain of 𝐴 = [0,6] and Range of 𝐴 =[100000,2500000]

c. Graph is to the right. The scale is each tick mark represents 1 million liters on the 𝑦-axis and 1 hour on the 𝑡-axis.

d. We solve the original equation for 𝑡 to get

𝑡 =𝑊−100,000

400,000= 𝐴−1(𝑊) ,

Domain is 100,000 ≤ 𝑊 ≤ 2,500,000 Range is 0 ≤ 𝑡 ≤ 6

iii) Square Function: 𝑓(𝑥) = 𝑥2 This is a special case of a general second degree polynomial 𝑝(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 .

With 𝑓(𝑥) = 𝑥2:

a. Find 𝑓 (2

3)

b. Find𝑓(−2) c. Find 𝑓(2) d. Find 𝑓(𝑎 + ℎ) e. Sketch the graph the function 𝑦 = 𝑥2 f. Is the function one-to-one?

Solution:

a. 𝑓 (2

3) = (

2

3)2=

4

9

b. 𝑓(−2) = (−2)2 = 4 c. 𝑓(2) = (2)2 = 4 d. 𝑓(𝑎 + ℎ) = (𝑎 + ℎ)2 = 𝑎2 + 2𝑎ℎ + ℎ2 e. Plot a few points to sketch the graph of the

function. See to the left f. No. The function is not one-to-one as it does

not pass the horizontal line test.

𝑥 𝑦 = 𝑥2 −2 (−2)2 = 4 −1 (−1)2 = 1 0 (0)2 = 0 1 (1)2 = 1 2 (2)2 = 4

iv) Cube Function: 𝑓(𝑥) = 𝑥3 This is a special case of a general third degree polynomial function 𝑓(𝑥) = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑

With 𝑓(𝑥) = 𝑥3:

a. Find 𝑓 (2

3)

b. Find𝑓(−2) c. Find 𝑓(2) d. Sketch the graph the function 𝑦 = 𝑥3 e. Is the function one-to-one? f. Find the inverse function if it is one-to-one.

𝑥 𝑦 = 𝑥3 −2 (−2)3 = −8 −1 (−1)3 = −1 0 (0)3 = 0 1 (1)3 = 1 2 (2)3 = 8

Mill

ion

s o

f Li

ters

Number of Hours


Solution:

a. 𝑓 (2

3) = (

2

3)3=

8

27

b. 𝑓(−2) = (−2)3 = −8 c. 𝑓(2) = (2)3 = 8 d. Plot a few points to sketch the graph of the

function. See to the left e. Yes. The function is one-to-one as it does pass

the horizontal line test. f. To find the inverse function to 𝑦 = 𝑓(𝑥) = 𝑥3,

switch 𝑥, 𝑦 roles so: = 𝑦3 . Then solving for 𝑦

we get that inverse function is 𝑓−1(𝑥) = √𝑥3

4. Piecewise Defined Function: A piecewise defined function is exactly what its name suggests. It is defined in pieces by two or more equations for different parts of the domain.

Examples:

I. For 𝑓(𝑥) = {1, for 0 ≤ 𝑥 < 10, for − 1 ≤ 𝑥 < 0−1, for − 2 ≤ 𝑥 < −1

a) Graph the function 𝑓 b) State the domain of 𝑓 c) State the range of 𝑓 d) Evaluate 𝑓(−0.5) e) Evaluate 𝑓(0.5) f) Evaluate 𝑓(−1.5)

Solution: a) Our function can be thought of as being made up

of three separate functions. So to plot the graph we have plot each of the three separate functions in one graph. You can do so by plotting some points until you know how it looks like. In this case all three pieces are constant functions in the given interval so the graph would be as shown below.

𝑓(𝑥) = {

1, 𝑓𝑜𝑟 0 ≤ 𝑥 < 10, 𝑓𝑜𝑟 − 1 ≤ 𝑥 < 0−1, 𝑓𝑜𝑟 − 2 ≤ 𝑥 < −1

b) Domain of 𝑓 = [−2, 1) c) Range of 𝑓 = {−1,0,1} d) 𝑓(−0.5) = 0 since 𝑥 = −0.5 falls in

the interval −1 ≤ 𝑥 < 0 e) 𝑓(0.5) = 1 since 𝑥 = 0.5 falls in the

interval 0 ≤ 𝑥 < 1 f) 𝑓(−1.5) = −1 since 𝑥 = −1.5 falls in

the interval −2 ≤ 𝑥 < −1

Range


II. For 𝑓(𝑥) = {

−1

3𝑥 − 1, 𝑓𝑜𝑟 − 3 ≤ 𝑥 < 0

−2, 𝑓𝑜𝑟 0 ≤ 𝑥 < 22𝑥 − 4, 𝑓𝑜𝑟 3 ≤ 𝑥

a) State the domain of 𝑓 b) State the range of 𝑓 c) Graph the function 𝑓 d) Evaluate 𝑓(−3)

e) Evaluate 𝑓 (−1

2)

f) Evaluate 𝑓(1.5) g) Evaluate 𝑓(5)

Solution: c) Our function can be thought of as being made

up of three separate functions. So to plot the graph we have plot each of the three separate functions in one graph. The first and the last piece are lines so plotting the two end points of the line segments it is made up of will suffice, and the middle is a constant function like we have seen before.

𝑥 𝑦 = −

1

3𝑥 − 1

𝑥 𝑦 = 2𝑥 − 4

−3 −1

3(−3) − 1

= 1 − 1 = 0

3 2(3) − 4= 6 − 4 = 2

0 −1

3(0) − 1

= 0 − 1 = −1

5 2(5) − 4= 10 − 4 = 6

Solution:

𝑓(𝑥) = {−1

3𝑥 − 1, 𝑓𝑜𝑟 − 3 ≤ 𝑥 < 0

−2, 𝑓𝑜𝑟 0 ≤ 𝑥 < 22𝑥 − 4, 𝑓𝑜𝑟 3 ≤ 𝑥

a) Domain of 𝑓 is [−3,0) ∪ [0,2) ∪ [3,∞) We see that this is [−3,2) ∪ [3,∞) which also be seen by observing the totality of the 𝑥-cordinates on the graph.

b) The range of 𝑓 is {−2} ∪ (−1,0] ∪[2,∞). Look at the graph to see what 𝑦-cordinates are involved in the graph.

d) 𝑓(−3) = −1

3(−3) − 1 = 1 − 1 = 0

since 𝑥 = −3 falls in the interval −3 ≤𝑥 < 0

e) 𝑓 (−1

2) −

1

3(−

1

2) − 1 =

1

6− 1 = −

5

6

since 𝑥 = −1

2 falls in the interval 3 ≤

𝑥 < 0 f) 𝑓(1.5) = −2 since 𝑥 = −1.5 falls in

the interval 0 ≤ 𝑥 < 2 g) 𝑓(5) = 2(5) − 4 = 10 − 4 = 6

Before we get into the next few types of functions we need to get familiar with a few definitions.

Range


5. Rational Function: A rational function is defined as the ratio of two polynomial functions denoted as

𝑅(𝑥) =𝑃(𝑥)

𝑄(𝑥), for polynomial functions 𝑃(𝑥), 𝑄(𝑥). The domain of 𝑅(𝑥) = is all 𝑥 such that 𝑄(𝑥) ≠ 0

Examples

I. 𝑅(𝑥) =1

𝑥

a. Find 𝑅(2) b. Domain of 𝑅(𝑥) c. Sketch the graph of 𝑅(𝑥) d. Is 𝑅(𝑥) one-to-one? e. What is the inverse

function of 𝑅(𝑥) Solution:

a. 𝑅(2) =1

2

b. Domain of 𝑅(𝑥) = all real numbers 𝑥 ≠ 0

= (−∞, 0) ∪(0,∞)

c. We can’t have 𝑥 = 0, but we see the 𝑦′𝑠 get very large as 𝑥 → 0. As you can see from the graph, as 𝑥 approaches zero from the positive side the 𝑦 coordinate shoots to infinity, and as 𝑥 approaches zero from the negative side the 𝑦 coordinate shoots to negative infinity. Also when 𝑥 → +∞, the 𝑦 coordinate approaches zero from the positive side and as 𝑥 → −∞, the 𝑦 coordinate approaches zero from the negative side. The lines 𝑥 = 0 and 𝑦 = 0 are called horizontal and vertical asymptotes respectively. We will study these in-depth later.

d. Yes. 𝑅(𝑥) passes the horizontal line test.

e. Inverse function of 𝑦 =1

𝑥 𝑥 =

1

𝑦 solving for 𝑦 we get

𝑦 =1

𝑥 or 𝑅−1(𝑥) =

1

𝑥

For now, we will plot a bunch of points to get a sense of the shape of the graph. We will deal with rational functions in more detail later. We must avoid 𝑥 = 0 as the denominator cannot be zero.

𝑥 𝑦 =

1

𝑥

𝑥 𝑦 =

1

𝑥

−0.1 1

−0.1= −10

−10 1

−10= −0.1

−0.01 1

−0.01= −100

−100 1

−100= −0.01

−0.001 1

−0.001= −1000

−1000 1

−1000= −0. .001

0.1 1

0.1= 10

10 1

10= 0.1

0.01 1

0.01= 100

100 1

100= 0.01

0.001 1

0.001= 1000

1000 1

1000= 0. .001

From the chart the closer to zero the 𝑥 values come from left or the right of zero, the 𝑦 values either shoot to ∞, or −∞ respectively. In other words, if we were to zoom in closer and closer to 𝑥 = 0, the graph resembles the vertical line 𝑥 = 0. Such a line is then denoted by the dotted line you see and is called a horizontal asymptote. Similarly if we zoom out so that 𝑥 values shoot to either ∞ or −∞ then the 𝑦 coordinate seems to get closer to zero from either above or below respectively. In other words, we were to zoom out a lot, the graph would resemble the line 𝑦 = 0. Such a line is called the horizontal asymptote.


II. 𝑅(𝑥) =2𝑥−1

𝑥2−4𝑥+3=

2𝑥−1

(𝑥−3)(𝑥−1)

a. Find 𝑅(5) b. Domain of 𝑅(𝑥)

Solution:

a. 𝑅(5) =2(5)−1

(5−3)(5−1)=

10−1

(2)(4)=

9

8

b. Domain of 𝑅(𝑥) = All real numbers for which denominator is non-zero, i.e., (𝑥 − 3)(𝑥 − 1) ≠ 0 = All real numbers so that 𝑥 ≠ 3 or 𝑥 ≠ 1 = (−∞, 1) ∪ (1,3) ∪ (3,∞)

6. Exponential Function: An exponential function is defined as 𝑓(𝑥) = 𝑎𝑥, where 𝑎 is a positive real number not equal to 1.

Properties of an exponential function ➢ Domain: All real

numbers ➢ Range: (0,∞) ➢ One-to-One: Yes! ➢ Points on the

graph to plot are

𝒙 𝒚 = 𝒂𝒙

−2 𝑎−2 =

1

𝑎2

−1 𝑎−1 =

1

𝑎

0 𝑎0 = 1

1 𝑎1 = 𝑎

2 𝑎2

Graph of 𝑓(𝑥) = 𝑎𝑥, for 𝑎 > 1

Graph of 𝑓(𝑥) = 𝑎𝑥, for 𝑎 < 1

For exponential functions we can see that for all 𝑎 > 0 the negative powers of

𝑎 gives us values of the type 1

𝑎𝑛 where 𝑛 > 0. The value of

1

𝑎𝑛 will keep on

getting smaller and smaller taking the 𝑦-coordinate to zero either as 𝑥 goes to ∞, or −∞, depending on whether 𝑎 > 1 or 0 < 𝑎 < 1 respectively. Therefore the line 𝑦 = 0 is called the horizontal asymptote. In other words, the graph resembles the line 𝑦 = 0 when we zoom out sufficiently.

Horizontal Asymptote 𝑦 = 0 Horizontal Asymptote 𝑦 = 0


Example I. If 𝑓(𝑥) = 2𝑥, then

a. Find 𝑓(−1) b. Find 𝑓(3) c. Find 𝑓(−3) d. Find 𝑓(𝑎 + ℎ) e. Sketch the graph of the

function. Solution:

a. 𝑓(−1) = (2)−1 =1

2

b. 𝑓(3) = 23 = 8

c. 𝑓(−3) = 2−3 =1

8

d. 𝑓(𝑎 + ℎ) = 2𝑎+ℎ e. Sketch the graph of the function.

See graph to the right. You can see from the chart that as 𝑥 approaches −∞ the 𝑦-coordinate approaches zero. This makes 𝑦 = 0 the horizontal asymptote.

Graph

𝒙 𝒀 = 𝟐𝒙

−2 2−2 =

1

22=1

4

−1 2−1 =

1

21=1

2

0 20 = 1

1 21 = 2

2 22 = 4

Example

II. If 𝑓(𝑥) = (1

3)𝑥

, then

a. Find 𝑓(−1) b. Find 𝑓(2) c. Find 𝑓(−2) d. Sketch the graph of the

function. Solution:

a. 𝑓(−1) = (1

3)−1

= 3

b. 𝑓(2) = (1

3)2=

1

9

c. 𝑓(−2) = (1

3)−2

= 32 = 9

d. Sketch the graph of the function. See graph to the right. You can see from the chart that as 𝑥 approaches ∞ the 𝑦-coordinate approaches zero. This makes 𝑦 = 0 the horizontal asymptote.

Graph

𝒙 𝒚 = (

1

3)𝑥

−2 (1

3)−2

= 32 = 9

−1 (1

3)−1

= 3

0 (1

3)0

= 1

1 (1

3)1

=1

3

2 (1

3)2

=1

9

Horizontal Asymptote 𝑦 = 0

Horizontal Asymptote 𝑦 = 0


Playing

As you can see all the exponential functions discussed here are one-to-one and therefore have inverses.

If we follow our steps to find an inverse function of say 𝑓(𝑥) = 2𝑥 we get the following

𝑦 = 2𝑥 and then 𝑥 = 2𝑦. At this step we try to solve the equation for 𝑦 but with 𝑦 in the exponent, we

are unable to isolate 𝑦 using basic tools for solving equations. To get around this problem,

mathematicians created a new function to represent this inverse function and called it a logarithmic

function. The inverse function to 𝑓(𝑥) = 2𝑥 must solve 𝑥 = 2𝑦for 𝑦. Thus we seek the exponent 𝑦 that

the base 2 must be raised to so that this will equal 𝑥. We call this inverse function the base-2 logarithm

and denote it as 𝑓−1(𝑥) = 𝑙𝑜𝑔2(𝑥) read as log base 2 of 𝑥. What does it mean to evaluate values of this

inverse function?

First note that 𝑥 = 2𝑦 and 𝑦 = 𝑙𝑜𝑔2(𝑥) are equivalent equations. So if we wanted to find the value of

𝑦 = 𝑙𝑜𝑔2(8) then we need to find the exponent 𝑦 for which 8 = 2𝑦. This means 𝑦 = 3 or that

log2(8) = 3. Similarly you can play and see that 𝑙𝑜𝑔2 (1

2) = −1 since

1

2= 2−1.

Recall that 𝑦 = 0 is the horizontal asymptote to the function 𝑦 = 2𝑥. Also since 𝑦 = log2 𝑥 is the

inverse of 𝑦 = 2𝑥 , and the 𝑥, 𝑦 roles get reversed, then the line 𝑥 = 0 will be the vertical asymptote to

the function 𝑦 = 𝑙𝑜𝑔2(𝑥). Also, with inverse functions’ reversal of the domain and range rolls, The

Range of 𝑦 = 2𝑥 being all positive numbers is the domain of 𝑦 = log2 𝑥. And the range of 𝑦 = log2 𝑥 is

the domain of 𝑦 = 2𝑥 which is all real numbers.

7. Logarithmic Function: A function denoted as 𝑦 = 𝑓(𝑥) = 𝑙𝑜𝑔𝑎𝑥, where 𝑎 > 0, 𝑎 ≠ 1. This means 𝑦 is the exponent such that 𝑎𝑦 = 𝑥.

Properties of the logarithmic function: ➢ To compute the value of

𝑓(𝑥) = log𝑎 𝑥 means answering: 𝑎? = 𝑥

➢ Graph: ➢ Domain:(0,∞) ➢ Range: All real numbers ➢ One-to-One: Yes! ➢ Inverse function 𝑓−1(𝑥) = 𝑎𝑥

Graph of 𝑓(𝑥) = 𝑙𝑜𝑔𝑎𝑥, for 𝑎 > 1

Graph of 𝑓(𝑥) = 𝑙𝑜𝑔𝑎𝑥, for 0 <𝑎 < 1

Vertical Asymptote 𝑥 = 0 Vertical Asymptote 𝑥 = 0




1. One-to-One function

2. Inverse function

3. Relationship of domain and range of inverse functions to the original function

4. Polynomial functions

5. Rational functions

6. Exponential functions

7. Logarithmic function

8. Describe the difference between the exponential and logarithmic functions and the relationship between them.



Exercises 1.2

1. The function 𝑓 is defined by the following rule: 𝑓(𝑥) = 8𝑥. Find 𝑓(𝑥) for each 𝑥-value in the table.

𝑥 𝑓(𝑥) = 8𝑥 −2

−1

0

1

2

2. The tables below give exponential functions in the form 𝑦 = 𝑎𝑥. Write the equation for each of the functions.

𝑥 𝑓(𝑥) 𝑥 𝑔(𝑥)

−2 1

4

−2 1

𝑒2

−1 1

2

−1 1

𝑒

0 1

0 1

1 2

1 𝑒

2 4

2 𝑒2

𝑓(𝑥) =________ 𝑔(𝑥) =_________

3. Evaluate each logarithmic function below. a. log2 16 =

b. log5 25 =

c. log21

8=

d. log3 √3 =

e. log6 1 =

f. log9 3 =

g. log10 0.00001 =

4. Convert each logarithmic expression to its exponential format and then evaluate the unknown value 𝑥.

a. log2 32 = 𝑥

b. log16 𝑥 = 1.5

c. log5 𝑥 = −3

d. log6 𝑥 = 3

e. logx 16 = 2

f. logx 25 = 1/2


5. The function 𝑓 is defined as follows:

𝑓(𝑥) = {

3

4𝑥 + 2 𝑖𝑓 𝑥 ≠ −1

2 𝑖𝑓 𝑥 = −1

Find the following. 𝑓(−3) = 𝑓(−1) = 𝑓(−2) =

6. The function 𝑔 is defined by

𝑔(𝑥) =3𝑥−4

𝑥+5 .

Find the following: 𝑔(−3) = 𝑔(𝑥 + 5) = The domain of 𝑔.

7. 𝑓(𝑥) = √𝑥 − 2 a. Find 𝑓(6) b. Find 𝑓(2) c. Find domain and range of the function. d. Is 𝑓 one-to-one? e. If you answered yes to part d, find the inverse function.


8. Find the domain and range of all the relations below either in interval notation or set notation as appropriate.

a) 𝑓(𝑥) =1

𝑥−4

Domain: Range:

b) 𝑔(𝑥) = √𝑥 − 4 Domain: Range:

c) ℎ(𝑥) = √4 − 𝑥 Domain: Range:

d) 𝑟(𝑥) = √𝑥 − 4 + 2 Domain: Range:

e)

Domain of 𝑇: Range of 𝑇:

f)

Domain of ℎ: Range of ℎ:

g)

Domain of 𝑆: Range of 𝑆:

h)

Domain of 𝑓: Range of 𝑓: Evaluate 𝑓(0) =___________ Find the values of 𝑥, for which

𝑓(𝑥) = 0 _____ , ______

i)

𝑔(𝑥) = {

2 𝑖𝑓 𝑥 = 1−2 𝑖𝑓 𝑥 < 13 𝑖𝑓 𝑥 > 1

Domain of 𝑔: Range of 𝑔:

j)

𝑝(𝑥) = {3√𝑥 𝑖𝑓 𝑥 > 4

2𝑥 − 1 𝑖𝑓 𝑥 < −4

Domain of 𝑝: Range of 𝑝:


9. A species of bacteria doubles every 30 minutes at room temperature. If you initially started with 30000 bacteria at 𝑡 = 0: a. Find the number of bacteria after 2 hours.

b. Find the number of bacteria at 4 hours.

c. See if you can come up with a formula for an exponential function that describes this where 𝑡

is the input in minutes and 𝑦 is the output being the number of bacteria present at time 𝑡.

10. The radioactive substance Uranium-240 has a half-life of 14 hours. The amount 𝐴(𝑡) of a sample of Uranium-240 remaining in grams after 𝑡 hours is given by the exponential function

𝐴(𝑡) = 3000(1

2)

𝑡

14.

a. Find the initial amount in the sample.

b. Find the amount remaining after 30 hours. Round your answer to the nearest gram. c. Determine how many grams remain after two years.

11. Suppose Rahul places $2000 in an account that pays 5% interest compounded each year. Assume that no withdrawals are made from the account. Round to the nearest cent. a. Find the amount in the account at the end of 1 year.

b. Find the amount in the account at the end of 2 years.

c. Try to find a formula for the account value after 𝑡 years. 𝐴(𝑡) =


12. A car is purchased for $26,500. After each year, the resale value decreases by 15%. a. What will the resale value be after 5 years? Round your answer to the nearest dollar.

b. Try to find a formula for the value after 𝑡 years. 𝑉(𝑡) =

13. Devon deposited $4000 into an account with 4.4% interest compounded quarterly. a. Assuming the no withdrawals are made, how much will she have in the account after 10

years? Round your answer to the nearest dollar.

b. Try to find a formula for the value after 𝑡 years.

𝑉(𝑡) =

14. At the beginning of a population study, a city had 360,000 people. Each year since, the population has grown by 2.6%. Let 𝑡 be the number of years since the start of the study. Let 𝑃(𝑡) represent the city’s population at time 𝑡. a. Predict the population 5 years after the start of the study.

b. Try to write an exponential function for the population 𝑃(𝑡), 𝑡 years after the study starts.

15. Compare the function 𝑓(𝑥) = 3𝑥 and 𝑔(𝑥) = 50𝑥2 by completing parts a. and b. a. Fill in the table below. Note that the table is already filled in for 𝑥 = 6

𝑥 𝑓(𝑥) = 3𝑥 𝑔(𝑥) = 50𝑥2

6 36 = 729 50(62) = 1800

7

8

9

10

b. For all 𝑥, 𝑥 ≥ 8 please check which of the following are true statements 𝑓(𝑥) ≥ 𝑔(𝑥) 𝑓(𝑥) = 𝑔(𝑥) 𝑓(𝑥) ≤ 𝑔(𝑥)

c. Explain your answer in part b.


16. An organic farmer raises free-range chickens and will deliver eggs to your home for a fee. The farmer sells a one-year contract for delivery of 25 to 100 dozen eggs to your home. The annual cost in dollars for the service is given by 𝑦 = 𝐶(𝑛) = 25 + 4𝑛, where 𝑛 is the number of dozen eggs you will use over the year.

Domain: a. State the domain variable and what it

stands for.

b. What is the domain here, assuming eggs are available only in full dozen quantities?

c. Determine the value 𝑦 = 𝐶(50) and what it means.

Range a. State the Range variable and what it

stands for.

b. What is the Range here, assuming eggs are available only in full dozen quantities?

c. Determine the value of 𝑛 if your contract payment was $325.

17. A construction crew needs to pave a road that is 204 miles long. The crew paves 9 miles of the road each day. The length, 𝐿 (in miles), that remains to be paved after 𝑑 days is given by the following function. 𝐿(𝑑) = 204 − 9𝑑.

Answer the following questions. a. How many miles of the road the crew have left to pave after 13 days?

b. After how many days will there be only 114 miles left to pave?

18. Find the difference quotient 𝑓(𝑥+ℎ)−𝑓(𝑥)

ℎ

where ℎ ≠ 0 for the function 𝑓(𝑥) =5𝑥2 −6. The graph of this function is given as well. Simplify your answer as much as possible. 𝑓(𝑥+ℎ)−𝑓(𝑥)

ℎ =

Try to explain what the quotient represents.


19. Sketch the graph of the functions and relations below. Explain clearly how you decided the graph was the shape you drew. Can you determine the domain and range of the functions and relations that you graphed based on the graphs.

a. 𝑦 = 2𝑥

𝑥 𝑦

b. 𝑦 = √𝑥

𝑥 𝑦

c. 𝑦 = 𝑥3

𝑥 𝑦

d. 𝑦 = 𝑥4 𝑥 𝑦


e. 𝑦 = 4𝑥

𝑥 𝑦

f. 𝑦 = (1

4)𝑥

𝑥 𝑦

g. 𝑦 =3

𝑥

𝑥 𝑦

h. 𝑦 = log4(𝑥) 𝑥 𝑦

4

16

1/4

1

1/16


i. 𝑦 = log10 (𝑥)

𝑥 𝑦

j. 𝑦 = 𝑙𝑜𝑔3(𝑥)

𝑥 𝑦

3

9

1/3

1/9

1


k. 𝑦 = {

2, 𝑖𝑓 𝑥 > 1−2, 𝑖𝑓 𝑥 < 1 0, 𝑖𝑓 𝑥 = 1

𝑥 𝑦

l. 𝑦 = {𝑥 + 2, 𝑖𝑓 𝑥 > 2

𝑥2, 𝑖𝑓 𝑥 ≤ 2

𝑥 𝑦

20. Evaluate the following for the given one-to-one functions below.

a. 𝑓(−5) b. 𝑓−1(−5)

c. 𝑓(4)

d. 𝑓−1(4)

a. 𝑓(2) b. 𝑓−1(0)

c. 𝑓(−5)

d. 𝑓−1(−2)


21. Find the inverses of the following one-to-one functions. Then find the domains and ranges of the functions and their inverses.

a. 𝑓(𝑥) =7𝑥+1

2𝑥−1

b. 𝑔(𝑥) = √2𝑥 − 1 for 𝑥 ≥1

2

Domain of 𝑓

Range of 𝑓−1

Domain of 𝑔

Range of 𝑔−1

Domain of 𝑓−1

Range of 𝑓

Domain of 𝑔−1

Range of 𝑔

c. ℎ(𝑥) = 4𝑥 d. 𝑝(𝑥) = 𝑙𝑜𝑔4𝑥

Domain of ℎ

Range of ℎ−1

Domain of 𝑝

Range of 𝑝−1

Domain of ℎ−1

Range of ℎ

Domain of 𝑝−1

Range of 𝑝

22. Given that 𝑓−1(𝑥) = 3𝑥3 + 5 find the formula for 𝑦 = 𝑓(𝑥).


23. Obtain the piece-wise formula for the functions whose graphs are given below. If the functions below are one-to-one, please also find the formulas for their inverse functions and sketch their graphs.

a.

b.

c.

d.


24. Find the original function whose inverse function is given below.

a. 𝑓−1(𝑥) =7𝑥+1

2𝑥−1

b. 𝑔−1(𝑥) = √1 − 𝑥, for 𝑥 ≤ 1

c.

d.

25. Create functions with the given properties below and then answer the questions.

A. Create a function 𝑦 = 𝑓(𝑥) that satisfies the following criteria. a. Is One-to-one b. Has domain [−2, 1] ∪ [3,10] c. 𝑓(−2) = 1 d. 𝑓(3) = 0 e. 𝑓(0) = 5 f. For at least one 𝑥 in the domain of 𝑓, 𝑓(𝑥) = −4

Questions about your function- iii) What is the domain of 𝑓(𝑥)? iv) What is the range of 𝑓(𝑥)? v) What are the 𝑥-intercepts of 𝑦 = 𝑓(𝑥)? vi) What is the 𝑦-intercepts of 𝑦 = 𝑓(𝑥)? vii) What is 𝑓−1(𝑥)?


1.3 Exponential and Logarithmic Functions In section two, we introduced the basic concept of the exponential and logarithmic functions as inverses

of each other.

Review

a. Exponential Function: An exponential function is defined by 𝑓(𝑥) = 𝑎𝑥, where 𝑎 is a positive real number not equal to 1.

Properties of an exponential function ➢ Domain: All real

numbers ➢ Range: (0,∞) ➢ One-to-One: Yes! ➢ Points on the

graph to plot are

𝒙 𝒚 = 𝒂𝒙

−2 𝑎−2 =

1

𝑎2

−1 𝑎−1 =

1

𝑎

0 𝑎0 = 1

1 𝑎1 = 𝑎

2 𝑎2

Graph of 𝑓(𝑥) = 𝑎𝑥, for 𝑎 > 1

Graph of 𝑓(𝑥) = 𝑎𝑥, for 𝑎 < 1

b. Logarithmic Function: A logarithmic function is denoted by 𝑦 = 𝑓(𝑥) = 𝑙𝑜𝑔𝑎𝑥, where 𝑎 > 0, 𝑎 ≠1. This means 𝑦 is the exponent such that 𝑎𝑦 = 𝑥.

Properties of the logarithmic function: ➢ To compute the value of 𝑓(𝑥)

means answering: 𝑎? = 𝑥 ➢ Graph: ➢ Domain:(0,∞) ➢ Range: All real numbers ➢ One-to-One: Yes! ➢ Inverse function 𝑓−1(𝑥) = 𝑎𝑥

Graph of 𝑓(𝑥) = 𝑙𝑜𝑔𝑎𝑥, for 𝑎 > 1

Graph of 𝑓(𝑥) = 𝑙𝑜𝑔𝑎𝑥, for 0 <𝑎 < 1

We will now explore these functions more fully and learn their properties so we can use them in many

different applications.

Horizontal Asymptote 𝑦 = 0 Horizontal Asymptote 𝑦 = 0

Vertical Asymptote 𝑥 = 0 Vertical Asymptote 𝑥 = 0


Remember the fact that the logarithmic function is the inverse function of the exponential function. This

fact means the equations below are equivalent.

Exponential Logarithmic Equation Equation

𝒚 = 𝒂𝒙 𝒍𝒐𝒈𝒂 𝒚 = 𝒙

There are two logarithm bases that are commonly used and show up as functions on scientific

calculators. One of these is logarithm base 10 which we write as 𝑙𝑜𝑔10𝑥 = log 𝑥 where the base

subscript is left out. Check out your calculators, you should see a 𝑙𝑜𝑔 button on it. This represents the

logarithm base ten also called the “Common Logartihm”. Check that it works by using the calculator to

take log 1000 and see that the answer is 3.

So: 𝑦 = log 𝑥 𝑥 = 10𝑦 , (A logarithm is always the value of an exponent!)

This notation for common logarithms originated ~1675 about 60 years after they were invented to provide an efficient method for multiplying and dividing numbers and taking powers and roots of numbers. Another motivation for working with logarithms is that it allows us to more easily express quantities that vary over many orders of magnitude. Below are examples where in real life logarithms base ten are used. 1. In chemistry pH of a substance is measured by the formula 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] where [𝐻+] =

hydrogen ions concentration in units of moles/liter. A substance having pH of less than 7 is characterized as being acidic and above 7 to be basic or alkaline. Water has a pH of 7. That means 7 = −𝑙𝑜𝑔[𝐻+] or that 1 × 10−7 moles/liter of hydrogen ions are present in water. The pH of a orange juice is between 3 and 4, while sweet corn has a pH of about 8. Nutrition experts talk about the link of alkaline diets [Foods for which the ash that remains after combustion produces a pH greater than 7 when dissolved in water.] to a reduced risk of cancer. Fruits and vegetables are typically alkaline foods.

2. Seismologist’s measure earthquakes using the Richter scale. The formula for this is given by 𝑀 =

𝑙𝑜𝑔 (𝐼

𝐼0) where 𝐼 = Intensity or amplitude of a seismograph reading in mm as measured by a

seismograph usually located within 100 kilometers from the epicenter, and 𝐼0 = a baseline reading of 0.001 mm. To get a sense of this scale, a Richter scale reading of a 3 is similar to the vibrations felt when a large truck is driving by. A Richter reading of 5, however is 100 times more intense. Also, the energy released by a magnitude 5 earthquake is equivalent to a 500 ton TNT explosion whereas a magnitude 9 earthquake that hit Japan in 2011 was equivalent to a 500,000,000 ton of TNT explosion. This gives you a sense of how logarithms are useful for relating quantities that range over many powers of ten in size. That 2011 earthquake that hit Japan was so powerful it nudged the earth’s axis so that the days on earth shortened by about 1.8 microseconds.

3. Physicists use a similar formula to calculate the decibel level of sound as 𝑑𝐵 = 10 𝑙𝑜𝑔 (𝐼

𝐼0) where

𝐼 = Intensity of the sound measured in 𝑊𝑎𝑡𝑡𝑠/𝑚𝑒𝑡𝑒𝑟2, and 𝐼0 = lowest threshold for sound that can be heard and is about 10−12 𝑊𝑎𝑡𝑡𝑠/𝑚𝑒𝑡𝑒𝑟2 (like a whisper). A 60 decibel sound is like when you are in a restaurant and you hear the noises around you. A jet taking off can expose airport


workers to sounds of 130-150 decibels. (This kind of high intensity noise exposure can lead to hearing loss.)

The other common base for logarithms is the constant "𝑒" where 𝑒 ≈ 2.718…. This is sometimes to

referred to as Euler’s number (or Napier’s constant) and therefore the letter 𝑒 is used to denote it. This

special number was first discovered in the mid 1600s’ and was first noticed by Jacob Bernoulli in 1683

when working with compound interest. Let’s play with compound interest formulas to see how Bernoulli

came to this important mathematical constant 𝑒.

Compound interest and the number 𝒆

Suppose that you invested a 1000$ at an interest rate of 5%/yr. Compound interest means that the

each time interest is paid it gets added to the principal. There are many options for how often the

interest earnings are computed and added to the account. If we compute interest 𝑛 times a year, then

annual compounding is when 𝑛 = 1, Quarterly compounding for 𝑛 = 4 , Monthly for 𝑛 = 12, and

Weekly compounding for 𝑛 = 52, and so on …

With annual compounding, the principal remains at $1000 until the end of the year when the 5%

interest is added to the principal.

For Annual Compounding:

After 1 year we will have our principal + interest

= 1000 + 1000(0.05) = 1000(1 + 0.05) = 1000(1.05) = 1050$. This will then be the

principal for the interest that is earned during the second year.

After 2 years we will have our principal of 1050 plus 5% interest = 1050 + 1050(0.05)

= 1050(1 + 0.05) = 1050(1.05) = 1000(1.05)(1.05) = 1000(1.05)2 = 1102.50$

We see each 5% interest addition corresponds to multiplying by (1.05) and the value after 𝑡 years is

given by 𝑉(𝑡) = 1000(1.05)𝑡 dollars.

Annual Compounding Formula 𝑽(𝒕) = 𝑷(𝟏 + 𝒓)𝒕

For Quarterly Compounding:

When interest is paid quarterly, only ¼ of the 5%/yr is added after each ¼ year.

After the first quarter, the value will be: 1000 + 1000(0.05

4) = 1000(1 +

0.05

4)1= 1012.50 .

After two quarters, the value becomes: 1000 (1 +0.05

4)2= 1025.16 .

At one year, after 4 interest payments, the value will be: 1000(1 +0.05

4)4= 1050.95 . This is slightly

more money than with annual compounding where the one-year value was $1050.

At two years with 8 interest additions, the value becomes: 1000 (1 +0.05

4)4×2

= 1104.49$.

This idea extends to interest being added 𝑛 times per year as:


Compounding Formula with 𝒏 interest additions per year 𝑽(𝒕) = 𝑷(𝟏 +𝒓

𝒏)𝒏𝒕

Using this formula the chart below will follow what happens to 1000 dollars when we change the

number of pay periods, and the number of years.

𝑛 pay periods in a year 𝑡 years later

1 2 3 10 15

Yearly 1 1050 1102.50 1157.63 1628.89 2078.93

Semi annually 2 1050.63 1103.81 1159.69 1638.62 2097.57

Quarterly 4 1050.95 1104.49 1160.75 1643.62 2107.18

Monthly 12 1051.16 1104.94 1161.47 1647.01 2113.70

Weekly 52 1051.25 1105.12 1161.75 1648.32 2116.24

Per day 365 1051.26 1105.16 1161.82 1648.66 2116.89

Per hour 𝑛 = (365 × 24) 8760 1051.26 1105.16 1161.82 1648.66 2116.89

Per minute 𝑛 =(365 × 24 × 60)

525600 1051.27 1105.17 1161.83 1648.72 2117.00

Per second 𝑛 =(365 × 24 × 60 × 60)

31536000 1051.27 1105.17 1161.83 1648.72 2117.00

As a mathematician this table should make you ponder a bit as we would have expected our values to

keep getting higher and higher when the number of pay periods increase. We wonder what happens as

𝑛 goes to infinity or 𝑛 → ∞. We see that the benefit of adding the interest more and more frequently

seems to level off as 𝑛 gets large. This would suggest that the numbers probably will start to accumulate

closer and closer to the same number no matter how many times interest is paid in a year.

This is the question that Bernoulli pondered. I.e., is there a nice way to describe what happens as the

frequency of interest additions goes to infinity?

Looking at 𝑉(𝑡) = 1000(1 +0.05

𝑛)𝑛𝑡 as 𝑛 → ∞

Bernoulli was a bit of a genius, and manipulated this formula as:

𝑉(𝑡) = 1000 (1 +0.05

𝑛)𝑛𝑡

= 1000(1 +1

(𝑛

0.05))

𝑛0.05

⋅0.05𝑡

= 1000 [(1 +1

𝑚)𝑚

]

0.05𝑡

, 𝑚 =𝑛

0.05

In this way, the question comes down to what happens to (1 +1

𝑚)𝑚

as 𝑚 → ∞.

Use your calculator and evaluate this expression for 𝑚 = 10, 100, 1000, 1000000.

Once you have done so without looking it up on your own make a prediction of what you think this

number is below (1 +1

𝑚)𝑚 𝑚→∞→ ___________________ (Do this before continuing down the page.)


What you should have seen is that the number (1 +1

𝑚)𝑚 , as 𝑚 gets very large seems to settle down at

about (1 +1

𝑚)𝑚→≈ 2.7182.. In fact the limiting value is an irrational number mathematicians called 𝑒.

The first six digits of 𝑒 are 2.71828…. See some of the digits for particular 𝑛 values below.

𝑛 (1 +

1

𝑛)𝑛

100 2.704813829

1000 2.716923932

10000 2.718145927

100000 2.718268237

1,000,000 2.718280469

10,000,000 2.718281692

This number is used in many applications and it is a special kind of irrational number called a

transcendental number.

A transcendental number is a real or complex number that is not a solution of a polynomial equation

with rational coefficients. Numbers that are solutions of such polynomials are called “Algebraic”.

You already know another transcendental number 𝜋. The complex number 𝑖 however is an algebraic

number, since 𝑖 is a solution to the equation 𝑥2 + 1 = 0. The number √2 is also an algebraic number.

Can you figure out why?

If we let the number of pay periods in the compound interest problem above go to infinity we see that

in the following formula 1000(1 +0.05

𝑛)𝑛𝑡= 1000 [(1 +

1

𝑚)𝑚]0.05𝑡

, 𝑚→∞→ 1000𝑒0.05𝑡.

Continuous Compounding Formula 𝑽(𝒕) = 𝑷𝒆𝒓𝒕 .

In general, the formula 𝐴 = 𝑃𝑒±𝑟𝑡 can be used to describe exponential growth and decay problems

where 𝑟 is the per unit time rate of increase or decrease of some quantity. The sign for the exponent is

negative in case of decay. We will work in these applications in more detail later.

The “natural logarithm” denoted as 𝑙𝑛𝑥 = 𝑙𝑜𝑔𝑒𝑥 is the inverse of the base-𝑒 exponential function

𝑦 = 𝑒𝑥. Often this is read as “ell en of x” or “the natural log of x”.

Review of equivalent logarithmic and exponential equations

Exponential Logarithmic Equation Equation

𝒚 = 𝒂𝒙 𝒍𝒐𝒈𝒂𝒚 = 𝒙

𝒚 = 𝟏𝟎𝒙 𝒍𝒐𝒈𝒚 = 𝒙

𝒚 = 𝒆𝒙 𝒍𝒏𝒚 = 𝒙


Practice Problems

1. Convert the exponential equations below into their equivalent logarithmic forms.

Exponential Equation Equivalent Logarithmic Equation

a. 8 = 23

b. 1

8= 2−3

c. 10 = 2𝑥

d. 5𝑥 = 25

e. (1

5)𝑥= 125

f. 32𝑥−4 = 81

g. 0.4𝑥 = 5

h. 𝑒𝑥 = 6

i. 10𝑥 = 0.0001

j. 𝑒𝑥+1 = 0.3

2. Convert the logarithmic equations below into their equivalent exponential forms.

Logarithmic Equation Equivalent Exponential Equation

a. 𝑙𝑜𝑔3𝑥 = −2

b. 𝑙𝑛𝑥 = 5

c. log(𝑥 + 1) = 5

d. −3 = 𝑙𝑜𝑔𝑥

e. 1.5 = 𝑙𝑛𝑥

f. 5 = 𝑙𝑜𝑔2𝑥

g. 4 = 𝑙𝑜𝑔1

2

(𝑥)

h. 𝑙𝑜𝑔525 = 2

i. 𝑙𝑜𝑔3 (1

81) = −4



It is helpful when doing these problems to identify the base in the problem. In the exponential version,

the base and exponent are easily identified and the exponent is the input. Also since logarithms are

inverses to exponential functions, the output of the logarithm is the input of the exponential and is the

exponent on the base in the exponential form of an equation.

In general then, 𝑩𝒂𝒔𝒆𝑬𝒙𝒑𝒐𝒏𝒆𝒏𝒕 = 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑬𝒙𝒑𝒐𝒏𝒆𝒏𝒕 = 𝑙𝑜𝑔𝑩𝒂𝒔𝒆(𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦) and

1. Convert the exponential equations below into their equivalent logarithmic forms.

Exponential Equation Equivalent Logarithmic Equation

a. 8 = 23

Base is 𝟐

Exponent is 𝟑

a. 𝑙𝑜𝑔28 = 𝟑

b. 1

8= 2−3

Base is 𝟐

Exponent is −𝟑

b. 𝑙𝑜𝑔2 (1

8) = −3

c. 10 = 2𝑥

Base is 𝟐

Exponent is 𝟏𝟎

c. 𝑙𝑜𝑔2 10 = 𝑥

d. 5𝑥 = 25

Base is 𝟓

Exponent is 𝒙

d. 𝑥 = 𝑙𝑜𝑔5(25)

e. (1

5)𝑥= 125

Base is 𝟏

𝟓

Exponent is 𝒙

e. 𝑥 = 𝑙𝑜𝑔1

5

(125)

f. 32𝑥−4 = 81 Base is 𝟑

Exponent 𝟐𝒙 − 𝟒

f. 2𝑥 − 4 = 𝑙𝑜𝑔381

g. 0.4𝑥 = 5 Base is 𝟎. 𝟒

Exponent is 𝒙

g. 𝑥 = 𝑙𝑜𝑔0.4(5)

h. 𝑒𝑥 = 6 Base is 𝒆

Exponent is 𝒙

h. 𝑥 = ln 6

i. 10𝑥 = 0.0001 Base is 𝟏𝟎

Exponent is 𝒙

i. 𝑥 = log (0.0001)

j. 𝑒𝑥+1 = 0.3 Base is 𝒆

Exponent is 𝒙 + 𝟏

j. 𝑥 + 1 = ln (0.3)

2. Convert the logarithmic equations below into their equivalent exponential forms.


In general here we will use 𝑬𝒙𝒑𝒐𝒏𝒆𝒏𝒕 = 𝑙𝑜𝑔𝑩𝒂𝒔𝒆(𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦) 𝑩𝒂𝒔𝒆𝑬𝒙𝒑𝒐𝒏𝒆𝒏𝒕 = 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦

Logarithmic Equation Equivalent Exponential Equation

a. 𝑙𝑜𝑔3𝑥 = −2 Base is 𝟑

a. 𝑥 = 𝟑−2 =1

9

Exponent is −𝟐 b. ln 𝑥 = 5

Base is 𝒆 b. 𝑥 = 𝒆𝟓

Exponent is 𝟓 c. log(𝑥 + 1) = 5

Base is 𝟏𝟎

c. 𝑥 + 1 = 𝟏𝟎𝟓 Exponent is 𝟓 or 𝑥 = 9999

d. −3 = log 𝑥 Base is 𝟏𝟎

d. 𝟏𝟎−𝟑 = 𝑥 or 𝑥 = 0.001 Exponent is −𝟑

e. 1.5 = ln 𝑥

Base is 𝒆 e. 𝒆𝟏.𝟓 = 𝑥

Exponent is 𝟏. 𝟓 f. 5 = 𝑙𝑜𝑔2𝑥

Base is 𝟐 f. 𝟐𝟓 = 𝑥 or 32 = 𝑥

Exponent is 𝟓 g. 4 = 𝑙𝑜𝑔1

2

(𝑥)

Base is 𝟏

𝟐

g. (𝟏

𝟐)𝟒

= 𝑥 or 1

16= 𝑥

Exponent is 𝟒 h. 𝑙𝑜𝑔525 = 2

Base is 𝟓 h. 25 = 𝟓𝟐

Exponent is 𝟐

i. 𝑙𝑜𝑔3 (1

81) = −4

Base is 𝟑

i. 1

81= 𝟑−𝟒

Exponent is −𝟒

Given the fact that the output of an exponential function 𝑦 = 𝑎𝑥 always produces a positive real

number, the domain of a logarithmic functions is just positive real numbers.

Practice Problems

1. Find the domain of the functions below.

a. 𝑓(𝑥) = 𝑙𝑜𝑔2(𝑥 − 1)

b. 𝑔(𝑥) = ln (2 − 𝑥)

c. ℎ(𝑥) = log (1 + 2𝑥)

Attempt these first before looking on the next page for solutions.

Solutions to Practice Problems


a. 𝑓(𝑥) = 𝑙𝑜𝑔2(𝑥 − 1)

Since the input has to be positive we must have 𝑥 − 1 > 0 or 𝑥 > 1

Domain is (1,∞)


b. 𝑔(𝑥) = ln (2 − 𝑥)

Since the input has to be positive we must have 2 − 𝑥 > 0 or 2 > 𝑥

Domain is (−∞, 2)

c. ℎ(𝑥) = log (1 + 2𝑥)

Since the input has to be positive we must have 1 + 2𝑥 > 0 or 1> −2𝑥 or −1

2< 𝑥

Domain is (−1

2, ∞)

Evaluating Logarithmic Functions

Evaluating 𝑙𝑜𝑔𝑎𝑥 means asking the question 𝑎? = 𝑥.

Practice Problems

Evaluate the following

1. 𝑙𝑜𝑔2(16)

2. 𝑙𝑜𝑔1

5

(25)

3. 𝑙𝑜𝑔1000

4. ln 𝑒3

5. 𝑙𝑜𝑔93

Attempt these problems yourself and then look at the answers.



Again, in evaluating logarithms the output is the exponent you are seeking and the base is indicated by

the notation of the logarithm.

Evaluate the following

1. 𝑙𝑜𝑔2(16)

Base = 2 and so we are looking for 2? = 16 and we know that 16 = 24 so we have 2? = 24 Since exponential functions are one-to-one the only way this can happen is if we have ?= 4.

Therefore 𝑙𝑜𝑔2(16) = 4

2. 𝑙𝑜𝑔1

5

(25)

3. Base =1

5 and so we are looking for (

1

5)?= 25 and we know that 25 = 52 = (

1

5)−2

so we have

(1

5)?= (

1

5)−2

Since exponential functions are one-to-one the only way this can happen is if we

have ?= −2. Therefore, 𝑙𝑜𝑔1

5

(25) = −2.

4. 𝑙𝑜𝑔1000 = 3 since 103 = 1000

5. 𝑙𝑛𝑒3 = 3 since 𝑒3 = 𝑒3

6. 𝑙𝑜𝑔93 =1

2 since 9

1

2 = 3

Since the output of logarithm functions are exponents, all the rules of exponents give rise to

corresponding rules of logarithms.

Properties of Logarithms

Playing

Now that we are more comfortable with logarithms and see them as exponents we may wonder how

the laws of exponents give rise to properties of logarithms. Let us play and see what happens.

Let 𝑥 = 𝑙𝑜𝑔𝑎𝑢, 𝑦 = 𝑙𝑜𝑔𝑎𝑣 and 𝑎 > 0, 𝑎 ≠ 1. We know from changing logarithmic equations to

exponential equations that we have 𝑎𝑥 = 𝑢, and 𝑎𝑦 = 𝑣.

1. Now the product property for exponents says 𝑢 ⋅ 𝑣 = 𝑎𝑥𝑎𝑦 = 𝑎𝑥+𝑦. Therefore we have

𝑢𝑣 = 𝑎𝑥+𝑦. Now if we put this into its logarithmic form, we get 𝑙𝑜𝑔𝑎(𝑢𝑣) = 𝑥 + 𝑦. Repacing 𝑥

and 𝑦, we obtain the product rule for logartihms!

𝑙𝑜𝑔𝑎(𝑢𝑣) = 𝑙𝑜𝑔𝑎(𝑢) + 𝑙𝑜𝑔𝑎(𝑣)

2. Now the quotient property for exponents says 𝑢

𝑣=

𝑎𝑥

𝑎𝑦= 𝑎𝑥−𝑦. Therefor we have

𝑢

𝑣= 𝑎𝑥−𝑦. Now if we put this into its logarithmic form, we get 𝑙𝑜𝑔𝑎 (

𝑢

𝑣) = 𝑥 − 𝑦. Repacing 𝑥

and 𝑦, we obtain the product rule for logartihms!

𝑙𝑜𝑔𝑎 (𝑢

𝑣) = 𝑙𝑜𝑔𝑎(𝑢) − 𝑙𝑜𝑔𝑎(𝑣)


3. When we convert the exponential equation 𝑎0 = 1, we get 0 = log𝑎 1 . That is the power that

𝑎 must be raised to so that 𝑎?=1 is 0 since 𝑎0 = 1.

𝑙𝑜𝑔𝑎(1) = 0

4. The power rule for exponents looks like (𝑎𝑥)𝑛 = 𝑎𝑛𝑥 . Putting this into logarithmic form, we

have loga(𝑎𝑥)𝑛 = loga 𝑎

𝑛𝑥 with 𝑎𝑥 = 𝑢 and log𝑎 𝑎𝑛𝑥 = 𝑛𝑥, we get:

loga(𝑢)𝑛 = 𝑛𝑥 and since 𝑥 = 𝑙𝑜𝑔𝑎𝑢

𝑙𝑜𝑔𝑎(𝑢𝑛) = 𝑛𝑙𝑜𝑔𝑎𝑢

5. One final property of logarithms deals with the relationship between the logarithms with

different bases of some number 𝑢. Let 𝑥 = loga 𝑢 and 𝑧 = logb 𝑢. Is there a relationship

between these two numbers? Can we find the base-𝑏 logarithm in terms of the base-

𝑎 logarithm? If so, then we can use the base-ten or natural logarithm functions on our

calculators to evaluate logarithms to any base.

We do have that 𝑢 = 𝑎𝑥 = 𝑏𝑧. If we take the base-𝑎 logarithm on both sides and use property

(4.) above, we get: loga 𝑢 = loga 𝑏𝑧 → loga 𝑢 = 𝑧 loga 𝑏 . Solving for 𝑧 which is logb 𝑢 we

get the useful result 𝑧 = logb 𝑢 =loga 𝑢

loga 𝑏.

logb 𝑢 =loga 𝑢

loga 𝑏

We can use this to calculate log3 5 by using the common or natural logarithm function on our

calculators. We have: log3 5 =𝑙𝑜𝑔5

𝑙𝑜𝑔3≈

0.6990

0.4771≈ 1.465, 𝑜𝑟 log3 5 =

𝑙𝑛5

𝑙𝑛3≈

1.6094

1.0986≈ 1.465 . The

individual numerator and denominator will be different values but the ratio is the same. Check

these numbers for yourself.

We can use these properties to decompose products, quotients and powers and radicals that are inputs

to logarithm functions. In the other direction, we can combine the sum and differences of logarithms

that are to the same base.


Practice Problems

1. Use the properties of logarithms above to either combine the log terms or to decompose the logarithm terms into a sum or product of simpler log terms.

A. 𝑙𝑜𝑔27 + 𝑙𝑜𝑔29 = ________________

B. 𝑙𝑜𝑔𝑎𝑟 − 𝑙𝑜𝑔𝑎𝑠 = ____________________

C. 𝑙𝑜𝑔𝑎(𝑠𝑡) = ____________________

D. 𝑙𝑜𝑔𝑎 (𝑠

𝑡) = _____________________

E. 𝑙𝑜𝑔𝑎(𝑟𝑛) = ____________________

F. 𝑛𝑙𝑜𝑔𝑎(𝑟) = ____________________

G. Change of base formula. Write in terms of the natural or common logarithm.

𝑙𝑜𝑔𝑎(𝑟) =

2. Use properties of logarithms to do the problems below.

A. Fill in the missing values to make the statement a true statement.

i. 𝑙𝑜𝑔712 − 𝑙𝑜𝑔7(_______) = 𝑙𝑜𝑔74

ii. 𝑙𝑜𝑔65 + 𝑙𝑜𝑔68 = 𝑙𝑜𝑔6(______)

iii. −2𝑙𝑜𝑔53 = 𝑙𝑜𝑔5(_____)

iv. 𝑙𝑜𝑔316 = (_______)𝑙𝑜𝑔32

v. 𝑙𝑛5

𝑙𝑛6= 𝑙𝑜𝑔6(_______)

B. Expand the following. Each logarithm in your answer should involve only one variable. Assume that all variables are positive.

I. log(𝑥5𝑦7) =____________________

II. log3 (𝑥7𝑦3

√𝑧3 ) =____________________

III. log (𝑥5

√𝑧7𝑦3) =____________________

IV. ln((4 − 𝑥)(𝑥 + 2)) =____________________

V. ln (𝑥6√7

5𝑧2) = ____________________


C. Write the following as one term. i. 6𝑙𝑜𝑔3𝑥 + 3𝑙𝑜𝑔3𝑦 = ______________

ii. 1

2𝑙𝑜𝑔𝑥 − 3𝑙𝑜𝑔𝑦 − 2𝑙𝑜𝑔𝑧 = _______________

D. Compute the values below exactly

i.𝑙𝑜𝑔2 (1

8) = ______________

ii.log (0.000001) = ______________

iii.ln(𝑒6) = _____________

iv.− ln(√𝑒) = __________

v.𝑙𝑜𝑔3 (1

27) = ____________

E. Use a scientific calculator evaluate the following, and round your answers to 2 digits.

i. 2000

log (1+0.04

5)= ________________

ii. 𝑙𝑜𝑔5

𝑙𝑜𝑔3= ____________

iii. ln (0.03)

2= ____________

iv. 𝑙𝑜𝑔54 = _____________


1. Properties of logarithms: Please fill out the missing values.

A. 𝑙𝑜𝑔27 + 𝑙𝑜𝑔29 = 𝑙𝑜𝑔2(7 × 9) = 𝑙𝑜𝑔263 B. 𝑙𝑜𝑔𝑎𝑟 − 𝑙𝑜𝑔𝑎𝑠 = 𝑙𝑜𝑔𝑎 (𝑟

𝑠)

C. 𝑙𝑜𝑔𝑎(𝑠𝑡) = 𝑙𝑜𝑔𝑎𝑠 + 𝑙𝑜𝑔𝑎𝑡

D. 𝑙𝑜𝑔𝑎 (𝑠

𝑡) = 𝑙𝑜𝑔𝑎𝑠 − 𝑙𝑜𝑔𝑎𝑡

E. 𝑙𝑜𝑔𝑎(𝑟𝑛) = 𝑛𝑙𝑜𝑔𝑎(𝑟) F. 𝑛𝑙𝑜𝑔𝑎(𝑟) = 𝑙𝑜𝑔𝑎(𝑟

𝑛)

G. Change of base formula: 𝑙𝑜𝑔𝑎(𝑟) =𝑙𝑜𝑔𝑟

𝑙𝑜𝑔𝑎=

𝑙𝑛𝑟

𝑙𝑛𝑎


A. Fill in the missing values to make the statement a true statement.

i. 𝑙𝑜𝑔712 − 𝑙𝑜𝑔7(3) = 𝑙𝑜𝑔74

ii. 𝑙𝑜𝑔65 + 𝑙𝑜𝑔68 = 𝑙𝑜𝑔6(40)

iii. −2𝑙𝑜𝑔53 = 𝑙𝑜𝑔5 (1

32) = 𝑙𝑜𝑔5 (

1

9)

iv. 𝑙𝑜𝑔316 = (4)𝑙𝑜𝑔32

v. 𝑙𝑛5

𝑙𝑛6= 𝑙𝑜𝑔6(5)

B. Expand the following. Each logarithm in your answer should involve only one variable. Assume that all variables are positive.

I. log(𝑥5𝑦7) = 5𝑙𝑜𝑔𝑥 + 7𝑙𝑜𝑔𝑦

II. log3 (𝑥7𝑦3

√𝑧3 ) = 7log3𝑥 + 3log3𝑦 −

1

3log3𝑧

III. log (𝑥5

√𝑧7𝑦3) = 5log3𝑥 −

1

2(7log3𝑧 + 3log37)

= 5log3𝑥 −7

2log3𝑧 −

3

2log3𝑦

IV. ln((4 − 𝑥)(𝑥 + 2)) = 𝑙𝑛(4 − 𝑥) + 𝑙𝑛(𝑥 + 2)

V. ln (𝑥6√7

5𝑧2) = 6ln𝑥 +

1

2𝑙𝑛7 − (𝑙𝑛5 + 2𝑙𝑛𝑧)


= 6ln𝑥 +1

2𝑙𝑛7 − 𝑙𝑛5 − 2𝑙𝑛𝑧

C. Write the following as one term.

i. 6𝑙𝑜𝑔3𝑥 + 3𝑙𝑜𝑔3𝑦 = 𝑙𝑜𝑔3(𝑥6𝑦3)

ii. 1

2𝑙𝑜𝑔𝑥 − 3𝑙𝑜𝑔𝑦 − 2𝑙𝑜𝑔𝑧 = 𝑙𝑜𝑔 (

√𝑥

𝑦3𝑧2)

D. Compute the values below exactly

i. 𝑙𝑜𝑔2 (1

8) = −3

ii.log(0.000001) = −6

iii.ln(𝑒6) = 6

iv.− ln(√𝑒) = −1

2

v.𝑙𝑜𝑔3 (1

27) = −3

E. Use a scientific calculator evaluate the following, and round your answers to 2 digits.

i. 2000

log(1+0.04

5)= 250998. 67

ii. 𝑙𝑜𝑔5

𝑙𝑜𝑔3= 1. 46

iii. ln (0.03)

2= −1. 75

iv. 𝑙𝑜𝑔54 =𝑙𝑜𝑔4

𝑙𝑜𝑔5= 0.86


Worksheet Section1.3a Date:______________ Name:__________________________


1. State the Properties of logarithms

2. List two uses each for exponential functions and logarithmic functions.

3. How are exponential and logarithmic functions related to each other?

4. How do we find domains of logarithmic functions?

5. In your own words describe how to change logarithmic equations into an exponential equation.

6. In your own words describe how to change exponential equations into logarithmic equations.



Exercises 1.3a

1. Rewrite the exponential equations in logarithmic form and logarithmic equations in exponential form. If possible simplify your answers.

Exponential Equation

Logarithmic Equation

𝑒𝑥 = 5

2𝑥+1 = 8

𝑙𝑜𝑔2(𝑥) = −1

log(𝑥 + 1) = 2

ln(𝑥 + 1) = 3

51−𝑥 = 3

𝑙𝑜𝑔12

(𝑥) = −3

Exponential Equation

Logarithmic Equation

𝑙𝑜𝑔13

(81) = −4

52 = 25

𝑙𝑜𝑔3 (1

9) = −2

𝑒3 = 𝑥

ln(𝑒2) = 2

10−2 = 0.01

𝑙𝑜𝑔1000 = 3


A. 𝑓(𝑥) = log (𝑥 + 1) Domain:

B. 𝑔(𝑥) = 𝑙𝑜𝑔 (3

𝑥−4)

Domain:

C. ℎ(𝑥) = 3𝑥−1 Domain:

D. 𝑘(𝑥) = 𝑙𝑛(1 − 𝑥)

Domain:

3. Properties of logarithms: Please fill out the missing values.

A. 𝑙𝑜𝑔𝑎𝑥 + 𝑙𝑜𝑔𝑎𝑦 = ________________

B. 𝑙𝑜𝑔𝑎𝑥 − 𝑙𝑜𝑔𝑎𝑦 = ____________________

C. 𝑙𝑜𝑔𝑎(𝑥𝑦) = ____________________

D. 𝑙𝑜𝑔𝑎 (𝑥

𝑦) = _____________________

E. 𝑙𝑜𝑔𝑎(𝑥𝑛) = ____________________

F. 𝑛𝑙𝑜𝑔𝑎(𝑥) = ____________________

G. Change of base formula: Write in terms of the common or natural logarithm function. 𝑙𝑜𝑔𝑎(𝑥) =



F. Fill in the missing values to make the statement a true statement.

vi. 𝑙𝑜𝑔58 − 𝑙𝑜𝑔5(_______) = 𝑙𝑜𝑔54

vii. 𝑙𝑜𝑔23 + 𝑙𝑜𝑔25 = 𝑙𝑜𝑔2(______)

viii. 3𝑙𝑜𝑔72 = 𝑙𝑜𝑔7(_____)

ix. 𝑙𝑜𝑔549 = (_______)𝑙𝑜𝑔57

x. 𝑙𝑛5

𝑙𝑛4= 𝑙𝑜𝑔4(_______)

G. Expand the following. Each logarithm in your answer should involve only one variable. Assume that all variables are positive.

VI. log(𝑥3𝑦2) =____________________

VII. log2 (𝑥3𝑦2

√𝑧) =____________________

VIII. log (𝑥3

√𝑧5𝑦) =____________________

IX. ln((4 + 𝑥)(𝑥 − 2)) =____________________

X. ln (𝑥5 √𝑦

3

3𝑧) = ____________________

H. Write the following as one term. i. 4𝑙𝑜𝑔2𝑥 + 2𝑙𝑜𝑔2𝑦 = ______________

ii. 1

3𝑙𝑜𝑔𝑥 − 2𝑙𝑜𝑔𝑦 + 3𝑙𝑜𝑔𝑧 = _______________

I. Compute the values below exactly

i. 𝑙𝑜𝑔2(8) = ______________

ii.log (0.000001) = ______________

iii.ln(𝑒5) = _____________

iv.ln(√𝑒) = __________

v.𝑙𝑜𝑔5 (1

25) = ____________

J. Use a scientific calculator to evaluate the following, and round your answers to 2 digits.

i. 20000

log (1+0.02

5)= ________________

ii. 𝑙𝑜𝑔3

𝑙𝑜𝑔2= ____________

iii. ln (0.5)

3= ____________

iv. 𝑙𝑜𝑔311 = _____________

5. State a formula for the value of each compound interest account after 𝑡 years.

a. An account starts with $5000 and earns 6.2%/yr interest compounded annually.


b. An investment is initially worth $20,000 and earns 6% interest/yr compounded

quarterly.

c. An investment of $2000 earns interest at 7.5%/yr compounded continuously.

d. A credit card debt has an interest rate of 18%/yr and is compounded monthly. Find the

function for the debt after 𝑡 years if it is initially at $10,000 and no payments are made.

e. You purchase a pair of shoes for $200 with your credit card but make no payments. The

interest rate then gets moved to 24% (2%/month). State a formula for the balance due

after 𝑛 years and state list the balance due at 1

2 , 1, 2, 3 and 10 years.

6. Population growth is modeled with exponential functions in several equivalent formats where

𝑃0 is the population at 𝑡 = 0.

• If the percentage growth over a one-year period is known, then we treat it like annual

compounding and get 𝑃(𝑡) = 𝑃0(1 + 𝑟)𝑡.

• If the continuous growth rate is known, then we use 𝑃(𝑡) = 𝑃0𝑒𝑟𝑡 just like continuous

compounding interest problems.

• If the time it takes a population to double is known, then we write the exponential function

in terms of the base-two exponential as 𝑃(𝑡) = 𝑃0 ⋅ 2𝑡

𝑇𝑑

State formulas for each population growth problem below.

a. Wisconsin’s population was 5,000,000 at 𝑡 = 0 in 2010 and is growing 0.35% each year.

b. The U.S. population was 310,000,000 in 2010 at 𝑡 = 0 and has a continuous growth rate

of 0.97% per year.

c. A blood infecting bacteria grows rapidly and has a doubling time of 30 minutes. Find the

exponential function that gives the number of bacteria 𝑡 minutes after a puncture

wound introduces 50 of the microbes into the blood stream.

7. Exponential decay models are modeled much like growth problems above but with the 𝑟 being

negative.

• If the percentage decline over a one-year period is known, then we treat it like annual

compounding and get 𝑃(𝑡) = 𝑃0(1 − 𝑟)𝑡. E.g., 10% decline per year in the value of a car

leads to the value function of a car originally worth $25,000 as 𝑉(𝑡) = 25000(0.9)𝑡

• If the continuous decay rate is known, then we use 𝑃(𝑡) = 𝑃0𝑒−𝑟𝑡 much like continuous

compounding interest problems. E.g., A 20 gram sample of radioactive tritium declines

continuously by 5.6% per year. The Amount after 𝑡 years is 𝐴(𝑡) = 20𝑒−0.056𝑡.


• If the time it takes for a quantity to decline to half its original amount (𝑇ℎ), then we write

the exponential function in terms of the base-(1

2) exponential as 𝐴(𝑡) = 𝐴0 ⋅ (

1

2)𝑡

𝑇ℎ

State formulas for each exponential decay problem below.

a. Ukraine’s population was 42,690,000 in 2016 and decreasing by 8.4% each year. State

𝑃(𝑡) =________________ where 𝑡 is years after 2016.

b. The light intensity under water in Lake Superior decreases continuously at 3% per foot

of depth. If the surface intensity (𝑥 = 0) is 180𝑊

𝑚2. State the intensity function as a

function of the number of feet below the surface. 𝐼(𝑥) =

Also predict the light intensity at a depth of 100 feet.

c. Radioactive Carbon-14 occurs naturally in all plant-derived materials. Once the plant

dies, the carbon-14 atoms in the plant are slowly converted back to nitrogen-14. The

half-life for this conversion is about 5,730 years. Give the exponential function that

describes the amount of carbon-14 (𝑡) years after the plant died if it started with 5

trillion atoms of carbon-14.

Also predict how many atoms would remain after 200,000 years.

8. Consider the population growth function for the world 𝑃(𝑡) = 7.4𝑒0.011𝑡 where 𝑡 = 0

corresponds to 2015.

a. What is the continuous growth rate?

b. Determine by what percent the population increases over the course of one year. Then give

a formula for 𝑃(𝑡) in the “annual compounding” format.

c. By guess and check or other methods, try to use 𝑃(𝑡) = 7.4𝑒0.011𝑡 to determine how many

years it takes for the population to double and give the formula for 𝑃(𝑡) in the “doubling

time” format.

9. Evaluate the measures of acidity, sound intensity and earthquake magnitude using the

appropriate logarithmic definitions of these measures.

a. Find the pH of blueberries which have a [𝐻+] concentration of 0.00076 𝑚𝑜𝑙

𝑙𝑖𝑡𝑒𝑟

b. Most foods have pH less than 7 and are thus acidic. Tofu has a pH of 7.2. Write down the

pH statement for this and convert it to its exponential form to obtain the [𝐻+]

concentration of tofu.

c. A very loud speaker outputs sound that you can feel a 5 feet from the speaker. The sound

intensity there is at 5 𝑤𝑎𝑡𝑡𝑠

𝑚2 . Determine the decibel level of this sound.

d. A weed whipper produces noise at 95 dB for the operator. Determine the sound intensity

level in 𝑊𝑎𝑡𝑡𝑠

𝑚2 .


e. Oklahoma has seen experienced unprecedented earthquakes since the advent of Fracking

for natural gas. In 2015 there were more than 30 of magnitude 4 or greater on the Richter

scale. Determine the seismograph reading in 𝑚𝑚 of a magnitude 4 and the most recent

magnitude 9 quake that occurred 6 years ago in Japan. What is the ratio of the amount of

movement between a level 4 and 9 earthquake?

10. Create an exponential function with 𝑓(0) = 100 and 𝑓(1) = 200. Do you think your function is

unique? Explain your answer.

11. Create a logarithmic function with 𝑓(1) = 100, and 𝑓(2) = 200. Do you think your function is

unique? Explain your answer.

Another Class of Functions: Sequences

A sequence is an ordered collection of objects. There is a first object, a second object, third

object, … etc. The list can be finite or infinite. A class roster of the first names would be a finite

sequence that may have repeats in the list. Sequences can be interpreted as functions from the

natural numbers to the set of objects where e.g., the natural number 3 as the input

corresponds to the third item in the ordered list.

If we take a real function and restrict the domain to the natural numbers 𝑁 = {1, 2, 3,… } the ordered

list of outputs 𝑓(1), 𝑓(2), 𝑓(3), ….is one way to generate interesting sequences. In the study of

sequences, we typically don’t use the function notation above, but instead use subscript notation. E.g.,

the sequence of squares of natural numbers might be denoted by {𝑎𝑛} = {𝑛2} = 𝑎1, 𝑎2, 𝑎3, … =

1, 4, 9, 16,… Here we’d typically define the sequence using 𝑎𝑛 = 𝑛2 instead of 𝑓(𝑛) = 𝑛2, for 𝑛 ∈ 𝑁.

A sequence {𝑎𝑛} = 𝑎1, 𝑎2, 𝑎3, … where 𝑎𝑛 is often given by some formula in terms of 𝑛

Examples of Sequences generated by functions with domain restricted to the natural numbers 𝑵.

1. The constant function 𝑓(𝑥) = 50 has all elements of the sequence being equal to 50. With

𝑎𝑛 = 𝑓(𝑛) = 50, we have {𝑎𝑛} = 50, 50, 50, 50, … for 𝑛 = 1,2,3,4,… is a sequence of numbers

with all outputs being 50. This is called a constant sequence. So when we write 𝑎43, it is

referring to the term in the sequence that is the 43rd number in the sequence. In our case it

would be 50 since all terms are 50.

2. Consider a linear function restricted to the domain of natural numbers say 𝑎𝑛 = 3 + 7𝑛.

Starting with 𝑛 = 1, this sequence is: 𝑎1 = 10, 𝑎2 = 17, 𝑎3 = 24, 𝑎4 = 31,…. The fiftieth

element in this sequence is 𝑎50 = 3 + 7(50) = 353.

Sometimes we start a sequence with 𝑎0 instead of 𝑎1. If asked to find the 50th element in the

sequence {3 + 7𝑛}𝑛=0 ∞ = 𝑎0, 𝑎1, 𝑎2, … = 3, 10, 17,… since we started at zero the fiftieth term

would actually correspond to 𝑛 = 49 thus the fiftieth element would be 𝑎49 = 3 + 49(7) =

346.


Sequences obtained from linear functions are called arithmetic sequences. 𝑎𝑛 = 𝑑 ⋅ 𝑛 + 𝑎0

The common difference in the sequence is the slope and usually denoted as 𝑑 and 𝑎0 is the

𝑧𝑒𝑟𝑜𝑡ℎ element.

3. A sequence obtained from exponential function restricted to the domain of whole numbers say

𝑎𝑛 = 3(2)𝑛 is called a geometric sequence. In this example, increasing 𝑛 by one, means we

have one more 2 bieng multiplied and thus each element in the sequence is 2 times larger than

the one before. 𝑎0 = 3, 𝑎1 = 6, 𝑎2 = 12, 𝑎3 = 24, 𝑎4 = 48,….

If asked to find the 50th term since we started at zero we would have 𝑎49 = 3(2)49 =

56294953421312.

Whenever, each successive term is obtained by multiplying by the same constant ratio (𝑟), the

sequence is really just an exponential function with domain restricted to 𝑁 or 𝑊 . You can

recognize this by observing that the ratio of any two consecutive elements is the same.

Sequences obtained from exponential functions are called geometric sequences. 𝑎𝑛 = 𝑎0𝑟𝑑

The common ratio 𝑟 is the base of the exponential function and 𝑎0 is the 𝑧𝑒𝑟𝑜𝑡ℎ element.

Practice Problems

1. Find the first 4 terms of the sequences given below.

𝑎𝑛 = 𝑛𝑡ℎ term of the sequence

First term

Second term

Third term

Fourth term

𝑎𝑛 = 5(1

2)2𝑛+1

, 𝑛 =

0,1,2,…

Arithmetic Geometric Neither

𝑎𝑛 = 4𝑛 + 3, 𝑛 = 4,5,… Arithmetic Geometric Neither

𝑎𝑛 =2𝑛−1

𝑛+2, 𝑛 = 1,2,3,… Arithmetic

Geometric Neither

𝑎𝑛 =(−1)𝑛

2𝑛, 𝑛 = 1,2, … Arithmetic

Geometric Neither

2. For the sequences below determine if they are arithmetic or geometric. Then find a formula for the 𝑎𝑛 and evaluate: 𝑎5 = , and 𝑎10 = .

Sequence Type 𝑓(𝑛) = 𝑎𝑛 = 𝑛𝑡ℎ term

Evaluate these elements.

{13,17,21,25,… . } Arithmetic Geometric Neither

𝑎5 =

𝑎10 =

{7,21,63,189, . . . } Arithmetic Geometric Neither

𝑎5 =

𝑎10 =

{12,10,7,1,… . } Arithmetic 𝑎5 =


Geometric Neither

𝑎10 =

{1,−11,1,−11,… . } Arithmetic Geometric Neither

𝑎5 =

𝑎10 =

3. For a given arithmetic sequence, the 82𝑛𝑑 term,

𝑎82 = −373 and the 6𝑡ℎ term, 𝑎6 = 7. Find the 42nd term 𝑎42. (One way is to think of this as a linear function, so find the slope and use the point (𝑛 = 6, 𝑦 = 7) to find the intercept. Thus 𝑎𝑛 = 𝑚 ⋅ 𝑛 + 𝑎0 ))

4. For a given geometric sequence, the 7𝑡ℎ

term, 𝑎7 = 15 and the 9𝑡ℎ term, 𝑎9 =

135. Find the 10𝑡ℎ term 𝑎10. (Recall that geometric sequences are exponential functions of (𝑛) and the formula for 𝑎𝑛 can be written in the form: 𝑎𝑛 = 𝑎0 ⋅ 𝑟

𝑛.)




First term

Second term

Third term

Fourth term

𝑎𝑛 = 5(1

2)2𝑛+1

, 𝑛 =

0,1,2,…

𝑎0 =5

2 𝑎1 =

5

8 𝑎2 =

5

32 𝑎3 =

5

128

X Arithmetic Geometric Neither

𝑎𝑛 = 4𝑛 + 3, 𝑛 = 4,5,… 𝑎4 = 19 𝑎5 = 23 𝑎6 = 27 𝑎7 = 31 Arithmetic X Geometric Neither

𝑎𝑛 =2𝑛−1

𝑛+2, 𝑛 = 1,2,3,… 𝑎1 =

1

3 𝑎2 =

3

4 𝑎3 =

5

5= 1

𝑎4 =7

6

Arithmetic Geometric X Neither

𝑎𝑛 =(−1)𝑛

2𝑛, 𝑛 = 1,2, … 𝑎1

= −1

2

𝑎2 =1

4

𝑎3

= −1

6

𝑎4 =1

8

Arithmetic Geometric X Neither

2. For the sequences below determine if they are arithmetic or geometric. Then find a formula for the 𝑎𝑛 and evaluate: 𝑎5 = , and 𝑎10 = .

Sequence Type 𝑓(𝑛) = 𝑎𝑛 = 𝑛𝑡ℎ term Evaluate these elements.

{13,17,21,25,… . } Arithmetic

If we start at 𝑎0 = 13, with 𝑑 = 𝑚 = 4, we have:

𝑎𝑛 = 4𝑛 + 13

𝑎5 = 33

𝑎10 = 53

{7,21,63,189, . . . } Geometric

If we start at 𝑎0 = 7 and the ratio is 𝑟 = 3, we have: 𝑎𝑛 = 7 ⋅ 3𝑛

𝑎5 = 7 ⋅ 35 = 1701

𝑎10 = 7 ⋅ 310 = 413343

{12,10,7,1,… . } Neither Many possibilities for continuing a patter here.

𝑎5 =

𝑎10 =

{1,−11,1,−11,… . } Neither It seems to just alternate, so we could just say 𝑎𝑜𝑑𝑑 =1 and 𝑎𝑒𝑣𝑒𝑛 = −11

𝑎5 = 1

𝑎10 = −11


3. For a given arithmetic sequence, the 82𝑛𝑑

element, 𝑎82 = −373 and the 6𝑡ℎ element, 𝑎6 =7. Find the 42nd element 𝑎42.

“Arithmetic” means a 𝑎𝑛 is a linear function of 𝑛. First we compute the slope which is the common

difference as = 𝑑 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑜𝑢𝑡𝑝𝑢𝑡

𝑐ℎ𝑎𝑛𝑔 𝑖𝑛 𝑖𝑛𝑝𝑢𝑡 𝑛=

𝑎82−𝑎6

82−6=

−373−7

76= −5.

Then using the element (𝑛 = 6, 𝑎6 = 7 we have: 7 = −5 ⋅ 6 + 𝑎0 → 𝑎0 = 37 Thus the formula is: 𝑎𝑛 = −5 ⋅ 𝑛 + 37. And 𝑎42 = −5 ⋅ 42 + 37 = −173


term, 𝑎7 = 15 and the 9𝑡ℎ term, 𝑎9 = 135.

Find the 10𝑡ℎ term 𝑎10.

First we determine the common ratio (𝑟), i.e. the base of the exponential function. Going from 𝑎7 = 15, to 𝑎9 = 135, we must multiply by 𝑟 two times, thus 15 ⋅ 𝑟2 = 135

from which 𝑟2 =135

15= 9 → 𝑟 = 3.

Now we can use either point to determine 𝑎0 in the formula 𝑎𝑛 = 𝑎0 ⋅ 3

𝑛. Take (𝑛 = 7, 𝑎7 =

15) to get: 15 = 𝑎0 ⋅ 37 → 𝑎0 =

5

36.

Thus 𝑎𝑛 =5

36⋅ 3𝑛 and 𝑎10 =

5

36⋅ 310 = 405.

Fibonacci Sequence and the Golden Ratio

The Fibonacci sequence is an interesting sequence that has applications to many areas. It is usually

defined in a recursively which means the value of the 𝑛𝑡ℎ element of the sequence is give in terms of

one or more of the elements previous to the 𝑛𝑡ℎ. A recursive formula for a sequence would thus look

like: 𝑎𝑛 =(some expression in the variables 𝑎𝑛−1, 𝑎𝑛−2, …

Fibonacci sequence recursively defined is given by: 𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2 and 𝑎1 = 1, 𝑎2 = 1 .

With this recursive formula it is easy to make a list

of the elements of the sequence. Thus:

𝑎1 = 1, 𝑎2 = 1,

𝑎3 = 1 + 1 = 2,

𝑎4 = 1 + 2 = 3,

𝑎5 = 2 + 3 = 5,

𝑎6 = 3 + 5 = 8…

Arithmetic and geometric sequences can also be defined recursively. For arithmetic, the 𝑛𝑡ℎ element is

obtained by adding 𝑑 to the (𝑛 − 1)𝑡ℎ element. Thus a recursive formula for an arithmetic sequence is

given by; 𝑎𝑛 = 𝑎𝑛−1 + 𝑑, and the starting value 𝑎0 or 𝑎1 must be given. Similarly, the 𝑛𝑡ℎ element of

a geometric sequence is obtained by multiplying the previous element by the common ratio (𝑟) and a

recursive formula is given by: 𝑎𝑛 = 𝑎𝑛−1 ⋅ 𝑟 where 𝑎0 or 𝑎1 must be given.

One question one might ask is whether it is possible to obtain the Fibonacci sequence as an explicit

function that just involves the index of the sequence 𝑛. I.e., can we get a formula for the 𝑛𝑡ℎ term

Fibonacci Sequence in Optimal Branching Pattern


𝑎𝑛 = (some expression involving only 𝑛)? Well, it turns out that it can be done and the formula is a bit

complicated, but comes down to the difference of two separate geometric sequences.

Explicit formula for Fibonacci Sequence 𝑎𝑛 =1

√5((

1+√5

2 )𝑛

− (1−√5

2)𝑛

)

It is easy to see that with 𝑛 = 1 the formula does produce 𝑎1 = 1. One can also compute 𝑎0 = 0 from

the formula. Use your calculator with 𝑛 = 6 to show that the above formula produces the correct value

of 𝑎6 = 8.

The base 1+√5

2 in the Fibonacci formula actually shows up in another context where it is called the

Golden Ratio. It happens to show up in “pleasing” dimensions for openings in architecture. It is the

length to width ratio 𝑙

𝑤 of a rectangle with the special property that when the rectangle is truncated by

slicing off a 𝑤 by 𝑤 square, the rectangle that remains has this same ratio of length to width.

Thus 𝑙

𝑤=

𝑤

𝑙−𝑤=

1𝑙

𝑤−1 . If we call

𝑙

𝑤= 𝑟, we have 𝑟 =

1

𝑟−1. We can solve this for 𝑟 to get: 𝑟(𝑟 − 1) =

1 → 𝑟2 − 𝑟 − 1 = 0. Solving with the quadratic formula, we have 𝑟 =1±√5

2 and these are the two

bases in the explicit formula for the Fibonacci sequence. The positive of these 𝑟 =1+√5

2≈ 1.618 is

called the Golden Ratio.

Golden Ratio is 𝑙

𝑤

Golden Spiral where the radius increases by the golden ratio for each quarter turn.


Section1.3b Worksheet Date:______________ Name:__________________________


What is a sequence?

What is an arithmetic sequence?

What is a geometric sequence?

Give an example of a sequence that is neither an arithmetic sequence nor a geometric sequence.

What is a Fibonacci sequence?

What is the golden ratio?

Give an example of an arithmetic sequence.

Give an example of an geometric sequence

Given an example of a sequence that is neither arithmetic nor geometric.



Exercises 1.3b

1. For the sequences below determine if they are arithmetic or geometric. Then find the formula for the 𝑎𝑛 and then fill in the missing terms column with value of that term.

Sequence Type 𝑓(𝑛) = 𝑎𝑛 = 𝑛𝑡ℎ term {22, 26, 30, 34,… . } Arithmetic

Geometric Neither

{6, 12, 24, . . . } Arithmetic Geometric Neither

{9, 13, 17,… . } Arithmetic Geometric Neither

{7, 11, 19,… . } Arithmetic Geometric Neither



First term

Second term

Third term

Fourth term

𝑎𝑛 = 4(1

3)2𝑛+1

, 𝑛 =

0,1,2,…


𝑎𝑛 = 3𝑛 + 5, 𝑛 =3,4,5,…


𝑎𝑛 =2𝑛

𝑛+3, 𝑛 = 3,4,5,… Arithmetic

Geometric Neither

𝑎𝑛 =(−1)𝑛

𝑛, 𝑛 = 1,2, … Arithmetic

Geometric Neither

3. For a given arithmetic sequence, the 82𝑛𝑑

element, 𝑎82 = −370 and the 6𝑡ℎ element, 𝑎6 = 10. Find the linear function

for 𝑎𝑛 and find the 33𝑟𝑑 element 𝑎33.


element 𝑎7 =23

25 and the 10𝑡ℎ element 𝑎10 =

115. Find a formula for this exponential

sequence and evaluate the14𝑡ℎ element 𝑎14.

5. Find the 20th and 21st elements of the Fibonacci sequence. Also compute the ratio 𝑎21

𝑎20 to show

that this ratio is very near in value to the Golden Ratio.

6. Consider the sequence of continued fractions: 𝑎1 = 1, 𝑎2 = 1 +1

1 , 𝑎3 = 1 +

1

1+1

1

,

𝑎4 = 1 +1

1+1

1+11

, …𝑎𝑛+1 = 1 +1

𝑎𝑛. Simplify the first 5 elements and express how the results

relate to the Fibonacci sequence. What number does this sequence tend to?


7. Create your own sequence where 𝑎5 = 32. Is your sequence unique? Explain your answer.

8. Create a sequence of each type listed below a. Geometric b. Arithmetic c. Constant d. Neither

Even and Odd Functions

A function is called even if and only if 𝑓(−𝑥) = 𝑓(𝑥).

This says that the function output 𝑦 is the same at any positive number 𝑥 = 𝑎 and the negative number

𝑥 = −𝑎 . On the graph of an even function the points (𝑎, 𝑏) and (−𝑎, 𝑏) always show up directly

across the 𝑦-axis from each other. Geometrically speaking it means that the graph of the function 𝑦 =

𝑓(𝑥) is symmetric with respect to the 𝑦-axis.

Examples

1. All polynomials functions of the type 𝑓(𝑥) = 𝑥𝑛 ,where 𝑛 is an even number, and 𝑥 is any real

number are even functions.

The reason for why the above function is even is that (−𝑥)𝑛 = 𝑥𝑛, when 𝑛 is an even since

(−𝑥)𝑛 = (−1)𝑛(𝑥)𝑛 and (−1) multiplied by itself an even number of times equals one.

2. The function 𝑦 = |𝑥|, where 𝑥 is any real number is also an even function.

The reason 𝑓(𝑥) = |𝑥| is an even function is the fact that |−𝑥| = |𝑥|, for all real numbers 𝑥.

3. All the functions below are even functions.

a.

b.

c.

A function is called odd if and only if 𝑓(−𝑥) = −𝑓(𝑥).

This says that the function output 𝑦 at any negative number 𝑥 = −𝑎 and is the negative of the 𝑦-value

at the positive number 𝑥 = 𝑎 . On the graph of an odd function the points (𝑎, 𝑏) and (−𝑎,−𝑏) always

show up opposite each other across the point (0, 0) since (0, 0) is the midpoint of the line segment

joining these two points. Geometrically speaking it means that the graph of the function 𝑦 = 𝑓(𝑥) is


symmetric with respect to the origin. This symmetry also means that the graph of an odd function

remains the same when rotated by 180 degrees.

Examples

1. All polynomials functions of the type 𝑓(𝑥) = 𝑥𝑛, where 𝑛 is an odd number are odd functions.

The reason that 𝑓(𝑥) = 𝑥𝑜𝑑𝑑 is even is the fact that 𝑦 = (−𝑥)𝑛 = (−1)𝑛𝑥𝑛, and (−1)𝑜𝑑𝑑 =

−1 and hence 𝑦 = −(𝑥𝑛).

2. All the functions below are odd functions since the graphs are symmetric with respect to the

origin.

a.

b.

c.

Practice Problems

1. Determine if the functions below are odd, even, or neither.

a. 𝑓(𝑥) = 2𝑥2 + 4

b. 𝑓(𝑥) = 3𝑥2 − 2𝑥 + 1

c. 𝑓(𝑥) = 𝑥3 − 𝑥2

d. 𝑓(𝑥) = 𝑥3 − 𝑥

e. 𝑓(𝑥) = 𝑥3 + 3𝑥 + 1

f.

g.

h.


1. Determine if the functions below are odd, even, or neither.

a. 𝑓(𝑥) = 2𝑥2 + 4

𝑓(−𝑥) = 2(−𝑥)2 + 4 = 2𝑥2 + 4 = 𝑓(𝑥) so the function is even by the definition.

b. 𝑓(𝑥) = 3𝑥2 − 2𝑥 + 1


𝑓(−𝑥) = 3(−𝑥)2 − 2(−𝑥) + 1 = 3𝑥2 + 2𝑥 + 1. This is different from 𝑓(𝑥) = 3𝑥2 −

2𝑥 + 1 and also different from −𝑓(𝑥) = −3𝑥2 + 2𝑥 − 1. Therefore the function is

neither even or odd.

c. 𝑓(𝑥) = 𝑥3 − 𝑥2

𝑓(−𝑥) = (−𝑥)3 − (−𝑥)2 = −𝑥3 − 𝑥2. This is also different from 𝑓(𝑥) = 𝑥3 − 𝑥2 and

from −𝑓(𝑥) = −𝑥3 + 𝑥2 , therefore the function is neither odd or even.

d. 𝑓(𝑥) = 𝑥3 − 𝑥

𝑓(−𝑥) = (−𝑥)3 − (−𝑥) = −𝑥3 + 𝑥 = −(𝑥3 − 𝑥) = −𝑓(𝑥), therefore the function is

an odd function.

e. 𝑓(𝑥) = 𝑥3 + 3𝑥 + 1

𝑓(−𝑥) = (−𝑥)3 + 3(−𝑥) + 1 = −𝑥3 − 3𝑥 + 1 ≠ 𝑓(𝑥) 𝑜𝑟 − 𝑓(𝑥), therefore the

function is neither odd nor even.

f.

This would be an odd function since for (𝑥, 𝑦) on the graph so is (−𝑥,−𝑦) on the graph. The graph is symmetric with respect to the origin.

g.

Neither. Since the graph is not symmetric with respect to origin or the 𝑦-axis.

h.

This would be an even function since for (𝑥, 𝑦) on the graph so is (−𝑥, 𝑦) on the graph. The graph is symmetric with respect to the 𝑦-axis.

(𝑥, 𝑦)

(−𝑥,−𝑦)

(𝑥, 𝑦) (𝑥, 𝑦)


Section1.3c Worksheet Date:______________ Name:__________________________


What is an even function?

What is an odd function?

What kinds of functions are symmetric with respect to 𝒚-axis?

What kinds of functions are symmetric with respect to 𝒙-axis?

What kinds of functions are symmetric with respect to the origin?

Give graphical examples of an a) Even function b) Odd Function c) Neither

What does it mean algebraically when a functions an even function? Give examples and explain.

What does it mean algebraically when a functions an odd function? Give examples and explain.



Exercises 1.3c

1. Determine if the functions below are odd, even, neither.

a. 𝑦 = 𝑥2 − 𝑥 Odd Even Neither

b. 𝑦 = |𝑥| Odd Even Neither

c. 𝑦 = 𝑥3 − 3𝑥 Odd Even Neither

d.

Odd Even Neither

f.

Odd Even Neither

g.

Odd Even Neither

h.

Odd Even Neither

i. 𝑓(𝑥) = 5𝑥2 + 3 Odd Even Neither

j. 𝑔(𝑥) =3

𝑥

Odd Even Neither

2. Determine whether: a. An exponential function could be even or odd.

b. A Linear function could be even or odd.

c. Is there any function that could be both even and odd?

3. Consider the function 𝑓(𝑥) = 𝑥2 − 3𝑥 + 5 and try to write it as a sum of an even and an odd function, i.e., find formulas for 𝑔(𝑥) = , and ℎ(𝑥) = such that 𝑔(𝑥) + ℎ(𝑥) = (𝑔 + ℎ)(𝑥) = 𝑥2 − 3𝑥 + 5.

4. Show that for 𝑓(𝑥) = 𝑥2 − 3𝑥 + 5 then 𝑔(𝑥) =𝑓(𝑥)+𝑓(−𝑥)

2 is an even function.


5. Create a function of each type listed below. a. Odd b. Even c. Neither

Which of these functions are one-to-one? Do you think an even function can be one-to-one? Explain your answer

1.4 Arithmetic and Composition of Functions In the previous sections we’ve seen many kinds of functions. As a mathematician any time we get new

objects to play with (which in this case is functions), we want to see if we can do arithmetic with it.

Playing

Below are some functions and we will see what happens when we do arithmetic with them. Attempt the

problems using your instinct before looking at answers on the next page.

Practice Examples

1. Evaluate the following

A. 𝑓(𝑥) = 8, 𝑓𝑜𝑟 − 3 ≤ 𝑥 ≤ 11,

𝑔(𝑥) = 2, 𝑓𝑜𝑟 − 8 ≤ 𝑥 ≤ 3

I. 𝑓(𝑥) + 𝑔(𝑥) =

II. 𝑓(𝑥) − 𝑔(𝑥) =

III. 𝑓(𝑥) × 𝑔(𝑥) =

IV. 𝑓(𝑥) ÷ 𝑔(𝑥) or 𝑓(𝑥)

𝑔(𝑥)=

B. 𝑓(𝑥) = 𝑥 − 1, for all real numbers 𝑥 𝑔(𝑥) = 2, for all real numbers 𝑥

I. 𝑓(𝑥) + 𝑔(𝑥) = II. 𝑓(𝑥) − 𝑔(𝑥) = III. 𝑓(𝑥) × 𝑔(𝑥) =

IV. 𝑔(𝑥) ÷ 𝑓(𝑥) or 𝑔(𝑥)

𝑓(𝑥)=

V. 𝑓(4) + 𝑔(4) =

Please attempt before looking at the solutions ahead.


Practice Examples Solutions

1. Evaluate the following

A. 𝑓(𝑥) = 8, 𝑓𝑜𝑟 − 3 ≤ 𝑥 ≤ 11, 𝑔(𝑥) = 2, 𝑓𝑜𝑟 − 8 ≤ 𝑥 ≤ 3

Picture for parts I. and II.

Picture for parts III. and IV.

1. You can see from the pictures above that adding, subtracting, multiplying or dividing two functions

together is not even possible unless we restrict our domains for both functions 𝑓(𝑥) and 𝑔(𝑥) to the

interval −3 ≤ 𝑥 ≤ 3 (which is the intersection of the two domains). This is the interval where both

functions are defined and we can do arithmetic with the output-values of the two given function. So

domain of all the functions below is [−3,3].

I. 𝑓(𝑥) + 𝑔(𝑥) = 8 + 2 = 10, for −3 ≤ 𝑥 ≤ 3

II. 𝑓(𝑥) − 𝑔(𝑥) = 8 − 2 = 6, for −3 ≤ 𝑥 ≤ 3

III. 𝑓(𝑥) × 𝑔(𝑥) = 8 × 2 = 16, for −3 ≤ 𝑥 ≤ 3

IV. 𝑓(𝑥) ÷ 𝑔(𝑥) or 𝑓(𝑥)

𝑔(𝑥)=

8

2= 4, for −3 ≤ 𝑥 ≤ 3


B. 𝑓(𝑥) = 𝑥 − 1, for all real numbers 𝑥

𝑔(𝑥) = 2, for all real numbers 𝑥

I. 𝑓(𝑥) + 𝑔(𝑥) = 𝑥 − 1 + 2 = 𝑥 + 1 for all real

numbers (note that, the effect of adding a 2

to the function 𝑥 − 1 is all the points on the

line 𝑦 = 𝑥 − 1 shifted up 2 units)

Domain here is all real numbers

II. 𝑓(𝑥) − 𝑔(𝑥) = 𝑥 − 1 − 2 = 𝑥 − 3 for all real

numbers (note that, the effect of subtracting

2 from all the outputs of the function 𝑥 − 1 is

all that all points on the line 𝑦 = 𝑥 − 1 move

vertically down 2 units)


III. 𝑓(𝑥) × 𝑔(𝑥) = (𝑥 − 1)2 = 2𝑥 − 2


IV. 𝑔(𝑥) ÷ 𝑓(𝑥) or 𝑔(𝑥)

𝑓(𝑥)=

2

𝑥−1, domain is all real numbers not equal to 1. Since when 𝑥 = 1 we get zero

in the denominator. So we can see that division of functions is only possible as long as the

denominator remains non-zero.

Domain is (−∞, 1) ∪ (1,∞)

V. 𝑓(4) + 𝑔(4) = 4 − 1 + 2 = 5

As you can see adding, subtracting, multiplying or dividing the outputs of two functions creates new

functions. We use the following notation for them

𝑓(𝑥) + 𝑔(𝑥) = (𝑓 + 𝑔)(𝑥)

𝑓(𝑥) − 𝑔(𝑥) = (𝑓 − 𝑔)(𝑥)

𝑓(𝑥) × 𝑔(𝑥) = (𝑓 ∙ 𝑔)(𝑥) 𝑓(𝑥)

𝑔(𝑥)= (

𝑓

𝑔) (𝑥)

The notation of 𝑓 + 𝑔, 𝑓 − 𝑔, 𝑓 ∙ 𝑔,𝑓

𝑔 are the names of the function that we get by doing the arithmetic

operations respectively. Remember the domain of these new functions is the intersection of the

domains of the individual functions involved in the operations. In the case of division we need to also be

sure to exclude any inputs (𝑥) that make the denominator function = 0.

In applications, when adding or subtracting functions, the outputs of the two functions must have the

same units in order that addition is meaningful. For products of two functions, the product of the two

outputs must be meaningful too. Thus we would not multiply the Wisconsin and Minnesota population

Picture for part I. and II.


functions, be we might add them to get the total population function for both states. We could

multiply the Wisconsin population function by the Average annual income of Wisconsin residents to

obtain the total Wisconsin income as a function of the year (𝑡). I.e. 𝑇𝑜𝑡𝑎𝑙 𝑊𝐼 𝐼𝑛𝑐𝑜𝑚𝑒 (𝑡) =

[𝑊𝐼 𝑃𝑜𝑝(𝑡)] ⋅ [𝑊𝐼 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝐼𝑛𝑐𝑜𝑚𝑒(𝑡)].


Section1.4a Worksheet Date:______________ Name:________________________


How is the sum function 𝒇 ± 𝒈 defined? I.e. how do you compute outputs of this function?

How do you determine the domain of the sum function 𝒇 ± 𝒈 ?

For the sum function 𝒇 ± 𝒈, what is necessary about the units of the output for 𝒇 and for 𝒈 so that 𝒇 ± 𝒈 is meaningful?

How is the product function 𝒇 × 𝒈 defined? I.e. how do you compute outputs of this function?

How do you determine the domain of the product function 𝒇 × 𝒈 ?

How is the quotient function 𝒇

𝒈 defined? I.e.

how do you compute outputs of this function?

How do you determine the domain of the

quotient function 𝒇

𝒈 ?



Exercises 1.4a

1 𝑓(𝑥) = 3𝑥 − 1 and 𝑔(𝑥) = 𝑥2 + 2 i. (𝑓 + 𝑔)(𝑥) = ____________________

ii. Domain of (𝑓 + 𝑔)

iii. (𝑓 + 𝑔)(3) = ____________________

iv. (𝑓𝑔)(𝑥) = ____________________

v. Domain of (𝑓𝑔) = ________________

vi. (𝑓𝑔)(0) = ____________________

2 𝑓(𝑥) = 5𝑥2 + 1 and 𝑔(𝑥) = 𝑥 − 2

i. (𝑓

𝑔) (𝑥) = ____________________

ii. Domain of (𝑓

𝑔)

iii. (𝑓

𝑔) (3) = ____________________

iv. (𝑓 − 𝑔)(𝑥) = ____________________

v. Domain of (𝑓 − 𝑔) = ________________

vi. (𝑓 − 𝑔)(0) = ____________________

3. If 𝑓(𝑥) = (2 + 𝑥)(−4 + 𝑥) and 𝑔(𝑥) = (1 − 𝑥)(1 + 𝑥). Find all values that are NOT in the domain

of 𝑓

𝑔. If there are more than one value separate them with commas.

4. EXTRA CREDIT: If 𝑓(𝑥) = √𝑥 − 1 and 𝑔(𝑥) = ln (𝑥 − 3). Evaluate the following A. Domain of 𝑓 + 𝑔

B. Domain of 𝑓

𝑔

5. Let the number of bushels of corn produced in each county of Wisconsin at year 𝑡 be denoted by 𝐵𝑐𝑜𝑢𝑛𝑡𝑦(𝑡) and the number of acres planted in corn in the county by 𝐴𝑐𝑜𝑢𝑛𝑡𝑦(𝑡).

a. Describe what the function 𝐵𝑐𝑜𝑢𝑛𝑡𝑦(𝑡)

𝐴𝑐𝑜𝑢𝑛𝑡𝑦(𝑡) represents.

b. Describe what the sum over all the counties 𝐵𝑎𝑑𝑎𝑚𝑠 + 𝐵𝑑𝑎𝑛𝑒 +𝐵𝑚𝑎𝑟𝑎𝑡ℎ𝑜𝑛 +⋯ over all the counties in the state function represents.


Composition of Functions

Composition of functions Part 1 https://www.youtube.com/watch?v=q3KcPmQfACo

Composition of functions Part 2 https://www.youtube.com/watch?v=HnKUOLC_PKs

Another important way to combine two or more functions is through composition. What this means is

that we start with an input to one function, then take the output of that function an use it as the input

of another function. This could go on for several levels of composition.

Playing

If we let the function 𝑓(𝑥) = 𝑥 − 1 and 𝑔(𝑥) = √𝑥 then if we start with 𝑥 as the input to 𝑔 and then

take the output √𝑥 and use that as the input to 𝑓 we call this the composition of the functions 𝑓 and 𝑔

with 𝑔 acting first. The notation for this would look like:

𝑓(𝑔(𝑥)) = 𝑓(√𝑥) = √𝑥 − 1

When evaluating a composition of functions as 𝑓(𝑔(𝑥)), we start the evaluation from the inside

working our way out. We might say here that 𝑔 is the inner function and 𝑓 the outer function. This

two-step process is denoted as above or by the notation 𝑓 ∘ 𝑔(𝑥) which means the same as 𝑓(𝑔(𝑥)).

The domain of 𝑓 ∘ 𝑔 is all those inputs of 𝑔 that will produce outputs that are in the domain of 𝑓.

Composition of Functions : Given two functions 𝑓(𝑥) and 𝑔(𝑥), a third function called the

composite function with notation (𝑓 ∘ 𝑔)(𝑥) is a function defined by evaluating 𝑓(𝑔(𝑥)) for

appropriate values of 𝑥, or you can think of it as 𝑥𝑔→𝑔(𝑥)

𝑓→𝑓(𝑔(𝑥)).

The domain of (𝑓 ∘ 𝑔)(𝑥) : {𝑥|𝑥 is in the domain of 𝑔 and also 𝑔(𝑥) is in the domain of 𝑓}

So you can think about (𝑓 ∘ 𝑔)(𝑥) as evaluating function of a function starting with the input of 𝑥,

getting an output from computing 𝑔(𝑥), this output then becomes the input in the function 𝑓 and the

output 𝑓(𝑔(𝑥)) is the answer to the composite function (𝑓 ∘ 𝑔)(𝑥). The domain of (𝑓 ∘ 𝑔)(𝑥) function

contains only those values of domain of 𝑔 for which the output values 𝑔(𝑥) are in domain of 𝑓.

Practice Examples

1. Evaluate the composite functions below. Find the domain of the composite functions.

a. Let 𝑓(𝑥) = 2𝑥 − 1 and 𝑔(𝑥) = √𝑥. Find i. (𝑓 ∘ 𝑔)(𝑥)

𝐷𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 (𝑓 ∘ 𝑔)(𝑥)

ii. (𝑔 ∘ 𝑓)(𝑥)

iii. 𝐷𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 (𝑔 ∘ 𝑓)(𝑥)

b. Let 𝑓(𝑥) =2

𝑥−1 and 𝑔(𝑥) = 𝑥 + 5. Find

i. (𝑓 ∘ 𝑔)(𝑥)

ii. (𝑔 ∘ 𝑓)(5)

iii. 𝐷𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 (𝑓 ∘ 𝑔)(𝑥)


c. Let 𝑓(𝑥) = 2𝑥 − 1 and 𝑔(𝑥) =𝑥+1

2. Find

i. (𝑓 ∘ 𝑔)(3)

ii. (𝑔 ∘ 𝑓)(5)

iii. (𝑓 ∘ 𝑔)(𝑥)

iv. (𝑔 ∘ 𝑓)(𝑥)

v. 𝑓−1(𝑥)

vi. 𝑔−1(𝑥)

d. Let 𝑓(𝑥) =1

𝑥−1 and 𝑔(𝑥) =

1

𝑥+ 1. Find

i. (𝑓 ∘ 𝑔)(6)

ii. (𝑔 ∘ 𝑓)(2)

iii. (𝑓 ∘ 𝑔)(𝑥)

iv. (𝑔 ∘ 𝑓)(𝑥)

v. 𝑓−1(𝑥)

vi. 𝑔−1(𝑥)

e. After solving problems 1c and 1d: i. What can you say when you have two functions 𝑓 and 𝑔 and 𝑓(𝑔(𝑥)) = 𝑥 ?

ii. If 𝑔 is the same function as 𝑓−1 then what is the function 𝑔−1 = (𝑓−1)−1, i.e. what is the inverse of the inverse of a function?


f. Use the graph to evaluate: i. 𝑓(1) =

ii. 𝑓(𝑓(0)) =

iii. 𝑓−1(−1) = iv. 𝑓−1(0) = v. Plot the graph of 𝑦 = 𝑓−1(𝑥)

vi. Find a formula for 𝑦 = 𝑓(𝑥) vii. Find a formula for 𝑦 = 𝑓−1(𝑥)

Practice Examples Solutions

1. Evaluate the composite functions below. Find the domain of the composite functions.

a. Let 𝑓(𝑥) = 2𝑥 − 1 and 𝑔(𝑥) = √𝑥. Find

i. (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) = 𝑓(√𝑥) =

2√𝑥 − 1 (see below for alternate explanation)

𝑥𝑔→√𝑥

𝑓→2√𝑥 − 1

Domain:[0,∞)

ii. (𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔(2𝑥 − 1) =

√2𝑥 − 1

𝑥𝑓→2𝑥 − 1

𝑔→√2𝑥 − 1

Domain:[1

2, ∞) since otherwise the end

result would not make sense.

b. Let 𝑓(𝑥) =2

𝑥−1 and 𝑔(𝑥) = 𝑥 + 5. Find

i. (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) = 𝑓(𝑥 + 5) =2

𝑥+5−1=

2

𝑥+4 (see below for alternate

explanation) Domain of 𝑓 ∘ 𝑔 requires 𝑥 ≠ −4.

𝑥𝑔→𝑥 + 5

𝑓→

2

𝑥 + 5 − 1=

2

𝑥 + 4

ii. (𝑔 ∘ 𝑓)(5) = 𝑔(𝑓(5)) = 𝑔 (2

5−1) = 𝑔 (

2

4) =

𝑔 (1

2) = 5 +

1

2= 5

1

2

𝑥𝑓→

2

5 − 1=2

4=1

2

𝑔→

1

2+ 5

c. Let 𝑓(𝑥) = 2𝑥 − 1 and 𝑔(𝑥) =𝑥+1

2.

Find

i. (𝑓 ∘ 𝑔)(3) = 𝑓(𝑔(3)) = 𝑓 (3+1

2) =

𝑓(2) = 2(2) − 1 = 4 − 1 = 3

3𝑔→3 + 1

2=4

2= 2

𝑓→2(2) − 1 = 4 − 1 = 3

Notice input was 3 and output is 3 also.

ii. (𝑔 ∘ 𝑓)(5) = 𝑔(𝑓(5)) = 𝑔(10 − 1) =

d. Let 𝑓(𝑥) =1

𝑥−1 and 𝑔(𝑥) =

1

𝑥+ 1. Find

i. (𝑓 ∘ 𝑔)(6) = 𝑓(𝑔(6)) = 𝑓 (1

6+ 1) =

11

6+1−1

=11

6

= 6

You can see that 𝑓 is undoing what 𝑔 did to 6.

ii. (𝑔 ∘ 𝑓)(2) = 𝑔(𝑓(2)) = 𝑔 (1

2−1 ) = 𝑔(1) =

1

1+ 1 = 2

You can see that 𝑔 is undoing what 𝑓 did to

https://www.youtube.com/watch?v=q3KcPmQfACo

https://www.youtube.com/watch?v=HnKUOLC_PKs


𝑔(9) =9+1

2= 5

5𝑓→2(5) − 1 = 10 − 1 = 9

𝑔→9 + 1

2=10

2= 5

Again notice input was 5 and output was 5.

iii. (𝑓 ∘ 𝑔)(𝑥) = 𝑓 (𝑥+1

2) = 2 (

𝑥+1

2) − 1 =

𝑥

iv. (𝑔 ∘ 𝑓)(𝑥) = 𝑔(2𝑥 − 1) =2𝑥−1+1

2= 𝑥

v. 𝑓−1(𝑥) 𝑦 = 2𝑥 − 1 𝑥 = 2𝑦 − 1 𝑥 + 1

2= 𝑦

𝑓−1(𝑥) =𝑥 + 1

2

vi. 𝑔−1(𝑥)

𝑦 =𝑥 + 1

2

𝑥 =𝑦 + 1

2

2𝑥 = 𝑦 + 1 𝑜𝑟 2𝑥 − 1 = 𝑦 𝑔−1(𝑥) = 2𝑥 − 1

2.

iii. (𝑓 ∘ 𝑔)(𝑥) = 𝑓 (1

𝑥+ 1 ) =

11

𝑥+1−1

=1

1𝑥

= 𝑥

iv. (𝑔 ∘ 𝑓)(𝑥) = 𝑔 (1

𝑥−1) =

11

𝑥−1

+ 1

= 𝑥 − 1 + 1 = 𝑥

v. 𝑓−1(𝑥)

𝑦 =1

𝑥−1 or 𝑥 =

1

𝑦−1

𝑦 − 1 =1

𝑥 or 𝑦 =

1

𝑥+ 1

𝑓−1(𝑥) =1

𝑥+ 1

vi. 𝑔−1(𝑥)

𝑦 =1

𝑥+ 1 or 𝑥 =

1

𝑦+ 1

𝑥 − 1 =1

𝑦 or 𝑦 =

1

𝑥−1

𝑔−1(𝑥) =1

𝑥 − 1

e. After solving problems 1c and 1d: i. What can you say when you have two functions 𝑓 and 𝑔 and 𝑓(𝑔(𝑥)) = 𝑥 ?

It seems when this happens that 𝑓 and g are inverse functions of each other!

ii. If 𝑔 is the same function as 𝑓−1 then what is the function 𝑔−1 = (𝑓−1)−1, i.e. what is the inverse of the inverse of a function?

The inverse of the inverse of a function is the original function!


f. Use the graph to evaluate:

i. 𝑓(1) = −1

2 , i.e., 𝑥 = 1 → 𝑦 = −1/2

ii. 𝑓(𝑓(0)) = 𝑓(−1) = −3/2

iii. 𝑓−1(−1) = 0 , since 𝑦 = −1 → 𝑥 = 0. iv. 𝑓−1(0) = 2, since 𝑦 = 0 → 𝑥 = 2 v. Plot the graph of 𝑦 = 𝑓−1(𝑥)

vi. Find a formula for 𝑦 = 𝑓(𝑥)

𝑦 = 𝑓(𝑥) = {1

2𝑥 − 1 𝑓𝑜𝑟 − 2 ≤ 𝑥 ≤ 2

3𝑥 − 6 𝑓𝑜𝑟 2 < 𝑥 ≤ 4}

vii. Find a formula for 𝑦 = 𝑓−1(𝑥)

𝑦 = 𝑓−1(𝑥) = {

2𝑥 + 2 𝑓𝑜𝑟 − 2 ≤ 𝑥 ≤ 01

3𝑥 + 2 𝑓𝑜𝑟 0 < 𝑥 ≤ 6

}

For any one-to-one function 𝑦 = 𝑓(𝑥), 𝑓−1 ∘ 𝑓 (𝑥) = 𝑥 and also 𝑓 ∘ 𝑓−1(𝑥) = 𝑥.

This makes sense in that if we start with 𝑥 in the domain of 𝑓 and compute 𝑦 = 𝑓(𝑥), then acting on this

𝑦 with 𝑓−1 will take us back to 𝑥 in the domain of 𝑓. Also for 𝑓 ∘ 𝑓−1(𝑥) = 𝑥, we start with some value

(𝑥 = 𝑏) in the range of 𝑓 and 𝑓−1 takes us back to the value 𝑎 = 𝑓−1(𝑏) in the domain of 𝑓 that

produces 𝑏 . Finally we act on this that 𝑎 with 𝑓 to get back to 𝑥 = 𝑏 in the range of 𝑓. These

compositions just go back and forth between a domain and range value of a one-to-one function 𝑓.

𝒚 = 𝒇−𝟏(𝒙)


Section1.4b Worksheet Date:______________ Name:________________________


What is the composite of two functions?

How do you find domain and range of a composite function? Explain.

What is the result of the composition of a one-to-one function and its inverse?

Give example of a composite function that is its own inverse.



Exercises 1.4b

1. Two functions 𝑔 and 𝑓 are defined in the figure below. Find the domain and range of the compositions 𝑓o𝑔(𝑥) = 𝑓(𝑔(𝑥)), and 𝑔o𝑓(𝑥) = 𝑔(𝑓(𝑥)). Then evaluate the function values below.

Domain of 𝑓

Domain of 𝑔

Domain of (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥))

Domain of (𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥))

Range of 𝑓

Range of 𝑔

Range of (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥))

Range of (𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥))

a. 𝑓 ∘ 𝑔(3)

b. 𝑔 ∘ 𝑓(9)

c. 𝑓 ∘ 𝑔(7)

d. 𝑔 ∘ 𝑓(0)


2. Consider the functions 𝑦 = 𝑓(𝑥) and 𝑦 = 𝑔(𝑥) which are given by their graphs below. a. Evaluate 𝑓(4) = b. Evaluate 𝑔(4) = c. Evaluate (𝑓 + 𝑔)(4) = d. Evaluate (𝑓 + 𝑔)(2) = e. Evaluate (𝑓 + 𝑔)(1) = f. Plot the graph of 𝑦 = (𝑓 + 𝑔)(𝑥) g. Evaluate (𝑓𝑔)(3) =

h. Evaluate 𝑔

𝑓(4) =

i. Evaluate 𝑓 ∘ 𝑔(4) = j. Evaluate 𝑔 ∘ 𝑓(2) = k. Evaluate 𝑓−1(4) = (Approximately)


4. For the real valued functions𝑔(𝑥) = √𝑥 + 8, and 𝑓(𝑥) = 𝑥2 + 7 find the compositions listed below and specify domains of these functions using interval notation. Then evaluate the values of the function listed below.

(𝑓o𝑔)(𝑥) = 𝑓(𝑔(𝑥)) =

(𝑔o𝑓)(𝑥) = 𝑔(𝑓(𝑥)) =

Domain of 𝑓 Domain of 𝑔

Range of 𝑓 Range of 𝑔

Domain of (𝑓o𝑔)(𝑥) = 𝑓(𝑔(𝑥)) Domain of (𝑔o𝑓)(𝑥) = 𝑔(𝑓(𝑥))

Range of (𝑓o𝑔)(𝑥) = 𝑓(𝑔(𝑥)) Range of (𝑔o𝑓)(𝑥) = 𝑔(𝑓(𝑥))

e. 𝑓o𝑔(𝑥)

f. 𝑔o𝑓(1)

g. 𝑔o𝑓(𝑥)

h. 𝑓o𝑔(1)

3. For the real valued functions 𝑔(𝑥) =𝑥+6

𝑥−5, and 𝑓(𝑥) = 2𝑥 − 7 find the compositions listed below

and specify domains of these functions using interval notation. Then evaluate the values of the function listed below.

(𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) =

(𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥)) =

Domain of 𝑓 Domain of 𝑔

Range of 𝑓 Range of 𝑔

Domain of (𝑓o𝑔)(𝑥) = 𝑓(𝑔(𝑥)) Domain of (𝑔o𝑓)(𝑥) = 𝑔(𝑓(𝑥))

Range of (𝑓o𝑔)(𝑥) = 𝑓(𝑔(𝑥)) Range of (𝑔o𝑓)(𝑥) = 𝑔(𝑓(𝑥))

a. (𝑓o𝑔)(𝑥)

b. (𝑔o𝑓)(6)

c. (𝑓o𝑔)(6)

d. (𝑔o𝑓)(0)


5. For each of the pairs of functions below find 𝑓(𝑔(𝑥)) and𝑔(𝑓(𝑥)). Then determine whether 𝑓 and 𝑔 are inverses of each other. Simplify your answers as much as possible. (Assume that your expressions are defined for all 𝑥 in the domain of the composition. It is a good idea to write the domain of each of the functions to make sure you know what 𝑥 values make sense in the functions.

a. 𝑓(𝑥) = 6𝑥 + 3 and 𝑔(𝑥) = 6𝑥 − 3 𝑓(𝑔(𝑥)) 𝑔(𝑓(𝑥))

o 𝑓 and 𝑔 are inverses of each other o 𝑓 and 𝑔 are not inverses of each other

b. 𝑓(𝑥) =2

𝑥 and 𝑔(𝑥) =

2

𝑥

𝑓(𝑔(𝑥)) 𝑔(𝑓(𝑥))


c. 𝑓(𝑥) = √𝑥 and 𝑔(𝑥) = 𝑥2, for 𝑥 ≥ 0 𝑓(𝑔(𝑥)) 𝑔(𝑓(𝑥))


d. 𝑓(𝑥) = √𝑥 and 𝑔(𝑥) = 𝑥2 𝑓(𝑔(𝑥)) 𝑔(𝑓(𝑥))



6. Given the formula for the composite function

𝑓 ∘ 𝑔(𝑥) = 𝑓(𝑔(𝑥)) = √2𝑥2 − 4 , state possible

formulas for what each of the functions 𝑔 and 𝑓. 𝑔(𝑥) =

𝑓(𝑥) =

7. Come up with formulas for two functions 𝑓 and 𝑔 so that when you

compose them, then 𝑓(𝑔(𝑥)) = 𝑥.

𝑓(𝑥) = 𝑔(𝑥) =

8. Evaluate the following for each of the one-to-one functions given by their graphs below.

a. 𝑓(1)

b. 𝑓−1(1)

c. 𝑓(2)

d. 𝑓−1(−4)

e. 𝑓(𝑓−1(−3))

f. 𝑓−1(𝑓(−4))

a. 𝑓(2)

b. 𝑓−1(0)

c. 𝑓(−5)

d. 𝑓−1(−2)

e. 𝑓(𝑓−1(−2))

f. 𝑓−1(𝑓(−5))


9. Find the inverses of the following one-to-one functions. Then find the domains and ranges of the functions and their inverses.

a) 𝑓(𝑥) =7𝑥+1

2𝑥−1

b) 𝑔(𝑥) = √2𝑥 − 1 for 𝑥 ≥1

2

Domain of 𝑓

Range of 𝑓−1

Domain of 𝑔

Range of 𝑔−1

Domain of 𝑓−1

Range of 𝑓

Domain of 𝑔−1

Range of 𝑔

(𝑓 ∘ 𝑓−1)(𝑥) =

(𝑓−1 ∘ 𝑓)(𝑥) =

(𝑔 ∘ 𝑔−1)(𝑥) = (𝑔−1 ∘ 𝑔)(𝑥) =


c) ℎ(𝑥) = 2𝑥

d) 𝑝(𝑥) = 𝑙𝑛𝑥

Domain of ℎ

Range of ℎ−1

Domain of 𝑝

Range of 𝑝−1

Domain of ℎ−1

Range of ℎ

Domain of 𝑝−1

Range of 𝑝

(ℎ ∘ ℎ−1)(𝑥) = (ℎ−1 ∘ ℎ)(𝑥) = (𝑝 ∘ 𝑝−1)(𝑥) = (𝑝−1 ∘ 𝑝)(𝑥) =


10. For the function 𝑦 = 𝑓(𝑥) graphed below, Use the graph to evaluate: a. 𝑓(1) =

b. 𝑓(𝑓(0)) =

c. 𝑓−1(−1) =

d. 𝑓−1(0) =

e. Plot the graph of 𝑦 =

𝑓−1(𝑥)

f. Find a formula for 𝑦 =𝑓(𝑥) (Piecewise defined!)

g. Find a formula for 𝑦 =𝑓−1(𝑥)

11. Show that 𝐴 = 𝑓(𝑡) = 200𝑒0.055𝑡 and 𝑔(𝐴) =1

0.055ln

𝐴

200 are inverses of each other by

showing that both 𝑓 ∘ 𝑔(𝐴) = 𝐴 and also that 𝑔 ∘ 𝑓(𝑡) = 𝑡.

Graphing Functions and Relations Page 101

Chapter 2: Graphing Functions and Relations

2.1 Library of Functions Lecture

Graphing Part 1 (14 min) https://www.youtube.com/watch?v=oxILQDhM3yM

Graphing Part 2 (15min) https://www.youtube.com/watch?v=zdO_bttea_E

Graphing Part 3 (15 min) https://www.youtube.com/watch?v=4xwIHBko9RA

In this chapter we will focus on the graphs of relations and functions. We already have some basic

understanding of specific functions from last chapter. You graphed versions of the basic polynomial,

rational, square root, absolute value, exponential and logarithmic functions. One way to graph functions

and relations is to plot enough points to give you a good sense of the main features of the graph. On the

other hand as a mathematician we can gain deeper understanding of functions and relations by

exploring their generic forms and observe how certain modifications of these forms or coefficients give

rise to predictable changes in the graphs.

Before we begin, let us review a basic library of functions that we have already encountered along with

their graphs. This is necessary so we can recognize how modifications of the basic formulas account for

changes in graphs and the other way too. This will allow us to recognize from a graph the basic type of

function formula for that graph and guide us in what the modifications of the basic equation form are.

Library of graphs of functions

1. 𝑓(𝑥) = 𝑥

𝑥 𝑦 = 𝑥

−2 −2

−1 −1

0 0

1 1

2 2

This is an odd function with the graph being symmetric with respect to the origin. And (0,0) is the 𝑥-intercept and 𝑦-intercept.

https://www.youtube.com/watch?v=oxILQDhM3yM

https://www.youtube.com/watch?v=zdO_bttea_E

https://www.youtube.com/watch?v=4xwIHBko9RA


2. 𝑓(𝑥) = |𝑥|

𝑥 𝑦 = |𝑥|

−2 2

−1 1

0 0

1 1

2 2

Note that the graph of 𝑦 =|𝑥| is symmetric with respect to the 𝑦-axis, i.e., 𝑓(𝑎) = 𝑓(−𝑎) and 𝑓(𝑥) =|𝑥| is an Even Function. Also, (0,0) is the 𝑥-intercept and 𝑦-intercept.

Notice how this graph differs from the graph of 𝑦 = 𝑥 on the left. You can see that all the absolute value changed in the graph of 𝑦 = 𝑥 is to invert bottom left part and make it positive.

3. 𝑓(𝑥) = 𝑥2

𝑥 𝑦 = 𝑥2 −2 4

−1 1

0 0

1 1

2 4

This is an even function with the graph being symmetric with respect to the 𝑦-axis. Also, (0,0) is the 𝑥-intercept and 𝑦-intercept.


4. 𝑓(𝑥) = 𝑥3

𝑥 𝑦 = 𝑥3 −2 −8 −1 −1 0 0 1 1 2 8

This is an odd function with its graph symmetric with respect to the origin. And (0,0) is the 𝑥-intercept and the 𝑦-intercept.

5. 𝑓(𝑥) = √𝑥

𝑥 𝑦 = √𝑥

0 0

1 1

4 2

9 3

This function is neither odd nor even. And (0,0) is the 𝑥-intercept and the 𝑦-intercept.


6. 𝑓(𝑥) =1

𝑥

𝑥 𝑦 =

1

𝑥

𝑥 𝑦 =

1

𝑥

−0.1 1

−0.1= −10

−10 1

−10= −0.1

−0.01 1

−0.01= −100

−100 1

−100= −0.01

−0.001 1

−0.001= −1000

−1000 1

−1000= −0. .001

0.1 1

0.1= 10

10 1

10= 0.1

0.01 1

0.01= 100

100 1

100= 0.01

0.001 1

0.001= 1000

1000 1

1000= 0. .001

This is an odd function as the graph is symmetric with respect to the origin. There are no 𝑥 or 𝑦 intercepts. The lines 𝑦 = 0 and 𝑥 = 0 are horizontal and vertical asymptotes respectively.


7. 𝑓(𝑥) = 𝑎𝑥

𝑥 𝑓(𝑥) = 𝑎𝑥

−1 1

𝑎

0 1

1 𝑎

Domain (−∞, ∞) Range (0, ∞)

8. 𝑓(𝑥) = 𝑙𝑜𝑔𝑎𝑥

𝑥 𝑓(𝑥) = 𝑙𝑜𝑔𝑎𝑥

1

𝑎

−1

1 0

𝑎 1

Domain (0, ∞) Range (−∞, ∞)

Playing

Now equipped with the library of graphs of functions above we try to figure out how these graphs

change when we change a few parameters or coefficients in the formulas for 𝑓. In other words, we are

trying to figure out how the graph of the function 𝑦 = 𝑎𝑓(𝑥 − ℎ) + 𝑘 relates to that of 𝑦 = 𝑓(𝑥). So

let’s play with functions below to figure out how the numbers 𝑎, ℎ and 𝑘 actually lead to predictable

modifications of the graph of 𝑦 = 𝑓(𝑥).


Horizontal and Vertical Shift of Function Graphs

Examples

Complete the table of values for each transformation of the functions 1-4 below and plot the respective

graph of each transformation, and please make sure you label each graph as shown for 1a below. Please

use colored pencils with a different color for each parts a.-e.

1. 𝑓(𝑥) = |𝑥|

𝑥 𝑦 = |𝑥|

−2 2

−1 1

0 0

1 1

2 2

a. 𝑦 = 𝑓(𝑥) + 1 = |𝑥| + 1

𝑥 𝑦 = |𝑥| + 1

−2

−1

0

1

2

b. 𝑦 = 𝑓(𝑥) − 1 = |𝑥| − 1

𝑥 𝑦 = |𝑥| − 1

−2

−1

0

1

2

c. 𝑦 = 𝑓(𝑥 − 2) = |𝑥 − 2|

𝑥 𝑦 = |𝑥 − 2|

0

1

2

3

4

d. 𝑦 = 𝑓(𝑥 + 2) = |𝑥 + 2|

𝑥 𝑦 = |𝑥 + 2|

−4

−3

−2

−1

0

e. 𝑦 = 𝑓(𝑥 − 2) + 1 = |𝑥 − 2| + 1

𝑥 𝑦 = |𝑥 − 2| + 1

0

1

2

3

4

Observations The graph of 𝑦 = 𝑓(𝑥 − ℎ) + 𝑘 = |𝑥 − ℎ| + 𝑘 is the same shape as the graph of

𝑦 = |𝑥| but is shifted ______ units up/down, and shifted ________ units left/right.


2. 𝑓(𝑥) = 𝑥2

𝑥 𝑦 = 𝑥2 −2 2

−1 1

0 0

1 1

2 2

a. 𝑦 = 𝑓(𝑥) + 1 = 𝑥2 + 1

𝑥 𝑦 = 𝑥2 + 1 −2

−1

0

1

2

b. 𝑦 = 𝑓(𝑥) − 1 = 𝑥2 − 1

𝑥 𝑦 = 𝑥2 − 1 −2

−1

0

1

2

c. 𝑦 = 𝑓(𝑥 − 2) = (𝑥 − 2)2

𝑥 d. 𝑦 = (𝑥 − 2)2

0

1

2

3

4

e. 𝑦 = 𝑓(𝑥 + 2) = (𝑥 + 2)2

𝑥 𝑦 = (𝑥 + 2)2 −4

−3

−2

−1

0

f. 𝑦 = 𝑓(𝑥 + 2) + 1 = (𝑥 + 2)2 + 1

𝑥 𝑦 = (𝑥 + 2)2 + 1 −4

−3

−2

−1

0

Observations

The graph of 𝑓(𝑥) = (𝑥 − ℎ)2 + 𝑘 is the same shape as the graph of 𝑦 = 𝑥2 but is shifted ______ units

up/down, and shifted ________ units left/right.


3. 𝑓(𝑥) = 3𝑥 𝑥 𝑦 = 3𝑥

−2

−1

0

1

2

Horizontal Asymptote _______

a. 𝑦 = 𝑓(𝑥) + 1 = 3𝑥 + 1 𝑥 𝑦 = 3𝑥 + 1

−2

−1

0

1

2


b. 𝑦 = 𝑓(𝑥) − 1 = 3𝑥 − 1 𝑥 𝑦 = 3𝑥 − 1

−2

−1

0

1

2


a. 𝑦 = −𝑓(𝑥) = −3𝑥 𝑥 𝑦 = −3𝑥

−2

−1

0

1

2


b. 𝑦 = 𝑓(𝑥 − 2) = 3𝑥−2 𝑥 𝑦 = 3𝑥−2 0

1

2

3

4


c. 𝑦 = 𝑓(𝑥 + 2) = 3𝑥+2 𝑥 𝑦 = 3𝑥+2

−4

−3

−2

−1

0


Observations: The graph of 𝑦 = 𝑓(𝑥 − ℎ) + 𝑘 = 𝑎𝑥−ℎ + 𝑘 is the same shape as the graph of

𝑦 = 𝑎𝑥 but is shifted ______ units up/down, and shifted ________ units left/right. The

horizontal asymptote of the graph of 𝑓(𝑥) = 𝑎𝑥−ℎ + 𝑘 is given by 𝑦 =_____________.


4. 𝑓(𝑥) = 𝑙𝑜𝑔3𝑥 𝑥 𝑦 = 𝑙𝑜𝑔3𝑥 1

9

1

3

1

3

9

Vertical Asymptote 𝑥 =___

a. 𝑦 = 𝑓(𝑥) + 2 = 𝑙𝑜𝑔3𝑥 + 2 𝑥 𝑦 = 𝑙𝑜𝑔3𝑥 + 2 1

9

1

3

1

3

9

Vertical Asymptote 𝑥 =_____

b. 𝑦 = 𝑓(𝑥) − 2 = 𝑙𝑜𝑔3𝑥 − 2 𝑥 𝑦 = 𝑙𝑜𝑔3𝑥 − 2 1

9

1

3

1

3

9

Vertical Asymptote 𝑥 =____

c. 𝑦 = −𝑓(𝑥) = −𝑙𝑜𝑔3𝑥 𝑥 𝑦 = −𝑙𝑜𝑔3𝑥 1

9

1

3

1

3

9


d. 𝑦 = 𝑓(𝑥 − 2) = 𝑙𝑜𝑔3(𝑥 − 2) 𝑥 𝑦 = 𝑙𝑜𝑔3(𝑥 − 2)

1

9+ 2

1

3+ 2

1 + 2 = 3

3 + 2 = 5

9 + 2 = 11


e. 𝑦 = 𝑓(𝑥 + 2) = 𝑙𝑜𝑔3(𝑥 + 2) 𝑥 𝑦 = 𝑙𝑜𝑔3(𝑥 + 2)

1

9− 2

1

3− 2

1 − 2 = −1

3 − 2 = 1

9 − 2 = 7


Observations:

➢ The graph of 𝑦 = 𝑓(𝑥 − ℎ) + 𝑘 = 𝑙𝑜𝑔𝑎(𝑥 − ℎ) + 𝑘 is the same shape as the graph of 𝑦 = 𝑙𝑜𝑔𝑎𝑥 but is shifted ______ units up/down, and shifted ________units left/right.

➢ The vertical asymptote of the graph of 𝑦 =𝑙𝑜𝑔𝑎(𝑥 − ℎ) + 𝑘 is given by 𝑦 =_____.

➢ In general our observations are that the graph of 𝑦 = 𝑓(𝑥 − ℎ) + 𝑘 is the same shape as the graph of 𝑦 = 𝑓(𝑥) but is shifted ______ units up/down, and shifted ________units left/right.

➢ If the graph of 𝑦 = 𝑓(𝑥) has vertical or horizontal asymptotes of the type 𝑥 = 𝑎, or 𝑦 = 𝑎, then in the new graph of 𝑦 = 𝑓(𝑥 − ℎ) + 𝑘 the vertical asymptote of the graph is at 𝑥 =____________ and the horizontal asymptote of the graph is at 𝑦 =_____________.


Horizontal and Vertical Stretch/Compression and Reflections

Examples

Complete the table of values for each transformation of the functions 5-6 below and plot the respective

graphs of the transformations as in the previous problems, and please make sure you label each graph.

Please use colored pencils with a different color for each of the transformations b. through g.

Plot the graphs of the transformations of functions indicated below based on the original graphs

of 𝑓(𝑥) = √4 − 𝑥2 , and 𝑦 = 𝑔(𝑥) shown.

Function Transformation

𝑓(𝑥) = √4 − 𝑥2

𝑥 𝑦 = √4 − 𝑥2

−2 0

0 2

2 0

A. 𝑦 = 2𝑓(𝑥) = 2√4 − 𝑥2

𝑥 𝑦 = 2√4 − 𝑥2

−2

0

2

B. 𝑦 = 4𝑓(𝑥) = 4√4 − 𝑥2

𝑥 𝑦 = 4√4 − 𝑥2

−2

0

2

C. 𝑦 =1

2𝑓(𝑥) =

1

2√4 − 𝑥2

𝑥 𝑦 =

1

2√4 − 𝑥2

−2

0

2

D. 𝑦 =1

4𝑓(𝑥) =

1

4√4 − 𝑥2

𝑥 𝑦 =

1

4√4 − 𝑥2

−2

0

2

E. 𝑦 = 𝑓(2𝑥) = √4 − 4𝑥2

𝑥 𝑦 = √4 − 4𝑥2

−1

0

1

F. 𝑦 = 𝑓 (1

2𝑥) = √4 −

𝑥2

4

𝑥

𝑦 = √4 −𝑥2

4

−4

0

4

G. = 𝑓 (1

4𝑥) = √4 −

𝑥2

16

𝑥

𝑦 = √4 −𝑥2

16

−8

0

8


Observations: The graph of 𝑦 = 𝑎𝑓(𝑥) = 𝑎√4 − 𝑥2 has a similar shape as the graph of 𝑦 =

𝑓(𝑥) = √4 − 𝑥2 but is stretched by a factor of ______ in the _____ direction when 𝑎 > 0.

The graph of 𝑦 = 𝑎𝑓(𝑥) = 𝑎√4 − 𝑥2 is the same shape as the graph of 𝑦 = 𝑓(𝑥) = √4 − 𝑥2

but is stretched by a factor of ______ in the _____ direction and reflected across ___-axis,

when 𝑎 < 0.

5. 𝑦 = 𝑔(𝑥)

𝑥 𝑦 = 𝑔(𝑥)

0 0

1 1

2 0

3 −1

4 0

a. 𝑦 = 2𝑔(𝑥)

𝑥 𝑦 = 2𝑔(𝑥)

0

1

2

3

4

b. 𝑦 = 3𝑔(𝑥)

𝑥 𝑦 = 3𝑔(𝑥)

0

1

2

3

4

c. 𝑦 = −𝑔(𝑥)

𝑥 𝑦 = −𝑔(𝑥)

0

1

2

3

4

d. 𝑦 = −2𝑔(𝑥)

𝑥 𝑦 = −2𝑔(𝑥)

0

1

2

3

4

e. 𝑦 =1

2𝑔(𝑥)

𝑥 𝑦 =

1

2𝑔(𝑥)

0

1

2

3

4

f. 𝑦 = 𝑔(1

2𝑥)

𝑥 𝑦

= 𝑔(1

2𝑥)

0

1

2

3

4


Observations: The graph of 𝑦 = 𝑎𝑔(𝑥) has a similar shape as the graph of 𝑦 = 𝑔(𝑥) but is

stretched by a factor of ______ in the _____ direction when 𝑎 > 0.

The graph of 𝑦 = 𝑎𝑔(𝑥) is the same shape as the graph of 𝑦 = 𝑔(𝑥) but is stretched by a factor

of ______ in the _____ direction and reflected across the ___-axis, when 𝑎 < 0.

6. 𝑓(𝑥) = √𝑥

𝑥 𝑦 = √𝑥

0 0

1 1

4 4

a. 𝑦 =1

2𝑓(𝑥) =

1

2√𝑥

𝑥 𝑦 =

1

2√𝑥

0

1

4

b. 𝑦 =1

4𝑓(𝑥) =

1

4√𝑥

𝑥 𝑦 =

1

4√𝑥

0

1

4

c. 𝑓(𝑥) = −√𝑥

𝑥 𝑦 = √𝑥

0

1

4

d. 𝑦 = −2𝑓(𝑥) = −2√𝑥

𝑥 𝑦 = −2√𝑥 0

1

4

e. 𝑦 = −1

2𝑓(𝑥) = −

1

2√𝑥

𝑥 𝑦 = −

1

2√𝑥

0

1

4


Observations: The graph of 𝑦 = 𝑎𝑓(𝑥) = 𝑎√𝑥 is the same shape as the graph of 𝑦 = √𝑥 but is

streched______ units _______________ when 0 < 𝑎 < 1.

The graph of 𝑦 = −𝑓(𝑥) = −√𝑥 is the same shape as the graph of 𝑦 = √𝑥 but is reflected

across the ___-axis.

The graph of 𝑦 = 𝑎𝑓(𝑥) = 𝑎√𝑥 is the same shape as the graph of 𝑦 = √𝑥 but is reflected

across the ___-axis and streched______ units _______________ when 𝑎 < −1.

The graph of 𝑦 = 𝑎𝑓(𝑥) = 𝑎√𝑥 is the same shape as the graph of 𝑦 = √𝑥 but is reflected

across the ___-axis and streched______ units _______________ when −1 < 𝑎 < 0.

7. 𝑦 = 𝑔(𝑥) 𝑥 𝑦 = 𝑔(𝑥)

0 0

1 1

2 0

3 −1

4 0

a. 𝑦 =1

2𝑔(𝑥)

𝑥 𝑦 = 2𝑔(𝑥)

0

1

2

3

4

b. 𝑦 = 𝑔(−𝑥) 𝑥 𝑦 = −𝑔(𝑥)

0

−1

−2

−3

−4

c. 𝑦 = 𝑔(2𝑥)

𝑥 𝑦 = 𝑔(2𝑥)

0

1

2

1

3

2

2

d. 𝑦 = 𝑔 (1

2𝑥)

𝑥 𝑦 = 𝑔 (

1

2𝑥)

0

2

4

6

8

e. 𝑦 = 𝑔 (−1

2𝑥)

𝑥 𝑦 = 𝑔 (−

1

2𝑥)

0

−2

−4

−6

−8


Observations: The graph of 𝑦 = 𝑎𝑔(𝑥) is the same shape as the graph of 𝑦 = 𝑔(𝑥) but is

stretched______ units ______________ when 0 < 𝑎 < 1.

The graph of 𝑦 = 𝑔(𝑎𝑥) is the same shape as the graph of 𝑦 = 𝑔(𝑥) but is stretched______

units _______________ when 𝑎 > 1.

The graph of 𝑦 = 𝑔(𝑎𝑥) is the same shape as the graph of 𝑦 = 𝑔(𝑥) but is stretched____ units

____________ and reflected across ___-axis, when 0 < 𝑎 < 1.

The graph of 𝑦 = 𝑔(−𝑥) is the same shape as the graph of 𝑦 = 𝑔(𝑥) but is reflected across

___-axis.

You probably noticed by doing examples above that with the vertical and horizontal stretch

transformations, when the factor 𝑎 is negative, the stretch also induces a reflection in the direction of

the stretch. Thus 𝑦 = −1

2√𝑥 stretches all the 𝑦-coordinates from the graph of 𝑦 = √𝑥 by a factor of

1

2

and then also makes them negative. Thus 𝑦 = 𝑎𝑓(𝑥) involves a stretch and a reflection across the

horizontal axis. Likewise the graph of 𝑦 = 𝑓(𝑎𝑥) is obtained by horizontally stretching by the factor 1

𝑎

and when 𝑎 is negative, it also reflects the graph across the 𝑦-axis.

For the graph of 𝑦 = 𝑔(𝑥) that we plotted in the examples 6 and 8 above, the graphs of 𝒚 = 𝒈(𝟐𝒙) and

of 𝒚 = 𝒈 (−𝟏

𝟐𝒙) are given below in case you struggled with it. Note that the graph of 𝑔 (−

1

2𝑥)

stretches the graph of 𝑦 = 𝑔(𝑥) by a factor of 1

1/2= 2, and reflects across the 𝑦-axis.


Once you do enough examples and find the patterns. We must investigate why is the effect of the

constants in the different parts of 𝑦 = 𝑓(𝑥 + ℎ) + 𝑘 what they are.

Another way to write 𝑦 = 𝑓(𝑥 + ℎ) + 𝑘 is 𝑦 − 𝑘 = 𝑓(𝑥 + ℎ). If a coordinate (𝑐, 𝑑) is on the original

graph 𝑦 = 𝑓(𝑥), that means 𝑓(𝑐) = 𝑑 in the new graph then the 𝑦 − 𝑘 = 𝑓(𝑥 + ℎ), we must have

𝑐 − ℎ as our input and the output must be 𝑑 + 𝑘. That would give us 𝑑 + 𝑘 − 𝑘 = 𝑓(𝑐 − ℎ + ℎ) giving

us 𝑑 = 𝑓(𝑐). So on the new graph the coordinate must be (𝑐 − ℎ, 𝑑 + 𝑘)explaining why the graph

translates up/down 𝑘 units (up if 𝑘 > 0, and down if 𝑘 < 0), and left/right ℎ units (left if ℎ > 0, and

right if ℎ < 0).

A similar argument can tell us what happens to the functions 𝑦 = 𝑎𝑓(𝑥) and 𝑦 = 𝑓(𝑎𝑥) (for all 𝑎 ≠ 0).

If (𝑐, 𝑑) is a point on the original 𝑦 = 𝑓(𝑥) graph, again 𝑓(𝑐) = 𝑑. Then the coordinate (𝑐, 𝑎𝑑) would be

on the new graph. Making the transformed graph stretch vertically since the 𝑦-coordinate is getting

multiplied by a factor of 𝑎 (if 𝑎 < 0 the reflection across the 𝑥-axis is predicted along with the vertical

stretch). Now in 𝑦 = 𝑓(𝑎𝑥) if the 𝑥 =𝑐

𝑎 we get 𝑓 (𝑎 ∙

𝑐

𝑎) = 𝑓(𝑐) = 𝑑 so (

𝑐

𝑎, 𝑑) is a coordinate on the

graph of 𝑦 = 𝑓(𝑎𝑥) this explains the fact that it is a horizontal stretch (if 𝑎 < 0 the graph will also have

reflection across the 𝑦-axis).

Practice Problems

1. Identify the standard (original) functions that was transformed to the equations below. Explain what

kind of transformation results in each of the functions below compared to their standard functions.

Sketch the graphs of the original and the transformed function on the graph.

a. 𝑦 = 3|𝑥|

b. 𝑦 = (𝑥

3)

2= (

1

3𝑥)

2 or viewed another way 𝑦 =

1

9𝑥2

c. 𝑦 = −3√𝑥

d. 𝑦 = √−1

2𝑥

Solutions


a. 𝑦 = 3|𝑥| 𝒚 = 𝟑|𝒙|, This is a vertical stretch by a factor of 3 of the 𝑦 = |𝑥| graph. Its graph is the V-shaped graph to the right.

b. 𝑦 = (𝑥

3)

2= (

1

3𝑥)

2 or viewed another way

𝑦 =1

9𝑥2

As 𝒚 = (𝟏

𝟑 𝒙)

𝟐, this can be viewed as a

horizontal stretch by a factor of 1

1/3= 3 of the

standard graph 𝑦 = 𝑥2. Alternatively, viewed

as 𝑦 =1

9 𝑥2 this can be viewed as a vertical

stretch by a factor of 1/9 of the graph of 𝑦 =𝑥2. Its graph is the parabola graph to the right.

c. 𝑦 = −3√𝑥

𝒚 = −𝟑√𝒙, This is a vertical stretch by a factor of three and also reflection across the

𝑥-axis of the graph of 𝑦 = √𝑥.


d. 𝑦 = √−1

2𝑥

𝒚 = √−𝟏

𝟐𝒙, This is a

horizontal stretch by 1

1/2= 2

and reflection across the 𝑦-

axis of the graph of 𝑦 = √𝑥.

Combination of Translations, Stretches and Reflections

By using a combination of these transformation methods, we are able to move, stretch and reflect our

basic function shapes anywhere in the coordinate plane. Four parameters will determine the location

and stretch/reflection of the new graph relative to the standard graph of any function type as indicated

below.

Function Transformation Summary:

The transformation of a function 𝑦 = 𝑓(𝑥) to 𝑦 = 𝑎𝑓(𝑏(𝑥 − ℎ)) + 𝑘

corresponds to transforming the graph of 𝑦 = 𝑓(𝑥) by:

1. A vertical stretch by a factor of 𝑎 (if 𝑎 is negative, in addition we will have a vertical reflection

across the 𝑥-axis too)

2. A horizontal stretch by a factor of 1

𝑏 (if 𝑏 is negative, in addition we will have a reflection

across the 𝑦-axis too)

3. A vertical shift of 𝑘 units

4. A horizontal shift of ℎ units

Note that the order of operations requires we do parenthesis first, then followed with multiplication

before addition or subtraction. That means we start with start by doing the stretch/reflections when

transforming a standard function graph. We need to know our standard function graphs’ main features

in order that we can apply these transformations correctly. The standard functions we should be

familiar with were discussed in at beginning of this section when we listed the library of functions, which

include: 𝑦 = 𝑥, 𝑦 = 𝑥2, 𝑦 = 𝑥3, 𝑦 =1

𝑥, 𝑦 = √𝑥, 𝑦 = 𝑎𝑥 , 𝑦 = loga 𝑥 , 𝑦 = |𝑥|, 𝑦 =

1

𝑥. If you are

unsure of the general shape get some points to help solidify your mental imagery of the basic shapes.

Examples:

1. Describe how each function is a transformation of a basic function and sketch its graph.

a. 𝑦 =1

2(𝑥 + 4)2 − 5

b. 𝑦 = −3√𝑥 − 2 − 1

c. 𝑦 = −2 ⋅ 2𝑥−3 + 7


d. 𝑦 = log4(𝑥 − 1) + 3

e. 𝑦 = −21

𝑥+4+ 5

2. Obtain a formula for each of the functions a-d shown in the graphs below as a transformation of one

of the standard functions listed below. Explain your reasoning.

𝑦 = 𝑥, 𝑦 = 𝑥2, 𝑦 = 𝑥3, 𝑦 =1


1

𝑥.

a.

b.

c.

d.

Solutions

1. Describe how each function is a transformation of a basic function and sketch its graph.

a. 𝑦 =1

2(𝑥 + 4)2 − 5

The standard function is 𝑦 =𝑓(𝑥) = 𝑥2 and this function

is: 𝑦 =1

2𝑓(𝑥 + 4) − 5. It is a

➢ Vertical stretch by ½ ➢ Horizontal shift of 4 to the

left, ➢ Vertical shift of 5 down


b. 𝑦 = −3√𝑥 − 2 − 1 The standard function is 𝑦 =

𝑓(𝑥) = √𝑥 and this function is: 𝑦 = −3𝑓(𝑥 − 2) − 1. It is a

➢ Vertical stretch by 3 ➢ A reflection across the 𝑥-

axis ➢ Horizontal shift of 2 to the

right ➢ Vertical shift of 1 down

c. 𝑦 = −2 ⋅ 2𝑥−3 + 7 The standard function is 𝑦 =𝑓(𝑥) = 2𝑥 and this function is: 𝑦 = −2𝑓(𝑥 − 3) + 7.

➢ Vertical stretch by 2 ➢ Reflection across the 𝑥-axis ➢ Horizontal shift of 3 to the

right ➢ Vertical shift of 7 up ➢ Horizontal Asymptote is at

𝑦 = 7

d. 𝑦 = log4(𝑥 − 1) + 3 The standard function is

𝑦 = 𝑓(𝑥) = log4 𝑥 and this function is: 𝑦 = 𝑓(𝑥 − 1) + 3. ➢ Horizontal shift of 1 to the

right ➢ Vertical shift 3 up. ➢ Vertical asymptote is at

𝑥 = 1


e. 𝑦 = −21

𝑥+4+ 5

The standard function is 𝑦 =

𝑓(𝑥) =1

𝑥 and this function is:

𝑦 = −2𝑓(𝑥 + 4) + 5. ➢ Vertical stretch by 2 ➢ Reflection across the

𝑥-axis. ➢ Horizontal shift of 4

to the left ➢ Vertical shift of 5 up ➢ Asymptotes are: 𝑥 =

−4 & 𝑦 = 5.

2.

a. Original graph is 𝑦 = |𝑥|. The graph is shifted to the left 3, up 4 and stretched

vertically by a 1

2 . So the transformed graph

would be 𝑦 =1

2|𝑥 + 3| + 4

b. Original graph is 𝑦 = 𝑥3. The graph is shifted to the right 3, up 4 and a vertical

stretch by a 1

2. So the transformed graph

would be 𝑦 =1

2(𝑥 − 3)3 + 4

c. Original graph is 𝑦 = √𝑥. The graph is shifted right 2, down 3, reflected across the 𝑥-axis and a vertical stretch of 2. So the transformed graph would be 𝑦 = −2√𝑥 − 2 − 3

d. Original graph is 𝑦 =1

𝑥. The graph is

shifted left 2, down 3. The vertical asymptote is at 𝑥 = −2 and horizontal asymptote is 𝑦 = −3. So the

transformed graph would be 𝑦 =1

𝑥+2−

3




1. Vertical Shift

2. Horizontal Shift

3. Vertical Stretch

4. Horizontal Stretch

5. Reflection across 𝒙-axis

6. Reflection across y-axis

7. Transformation of a function



Exercises 2.1

1. Please fill in the blanks below.

a. The graph of the function 𝑦 = 𝑓(𝑥) + 𝑘 is shifted ______________________________ from the

graph of the original function 𝑦 = 𝑓(𝑥).

b. The graph of the function 𝑦 = 𝑓(𝑥) − 𝑘 is shifted ______________________________ from the


a. Statements 1 and 2 can also be written as statements 3 and 4 below.

c. The graph of the function 𝑦 − 𝑘 = 𝑓(𝑥) is shifted ______________________________ from the


d. The graph of the function 𝑦 + 𝑘 = 𝑓(𝑥) is shifted ______________________________ from the


e. The graph of the function 𝑦 = 𝑓(𝑥 − ℎ) is shifted ______________________________ from the


f. The graph of the function 𝑦 = 𝑓(𝑥 + ℎ) is shifted ______________________________ from the


g. The graph of the function 𝑦 = 𝑎𝑓(𝑥), 𝑎 > 1 is _________________________ from the graph of

the original function 𝑦 = 𝑓(𝑥).

h. The graph of the function 𝑦 = 𝑐𝑓(𝑥), 0 < 𝑐 < 1 is _________________________ from the


i. The graph of the function 𝑦 = −𝑓(𝑥) is _____________________________________________

from the graph of the original function.

j. The graph of the function 𝑦 = 𝑓(−𝑥) is _____________________________________________

from the graph of the original function.

k. For quadratic functions, completing the squares allows you to find the vertex of the parabola. So

if we had the function 𝑦 − 𝑘 = 𝑎(𝑥 − ℎ)2, the point (ℎ, 𝑘) is referred to as the ______________

of the parabola.

l. For quadratic functions, completing the squares allows you to find the vertex of the parabola. So

if we had the function 𝑦 + 𝑘 = 𝑎(𝑥 + ℎ)2, the point (−ℎ, −𝑘) is referred to as the

______________ of the parabola.

m. An even function is symmetric with respect to the ______________.

n. An odd function is symmetric with respect to the ______________________.


2. Please find the following for all functions below.

I. Sketch the graph of the functions and relations below. Explain clearly how you decided the

graph was the shape you drew. Show all relevant parts of the graph.

II. Find the domain and range of the functions and relations that you graphed based on your graph.

a. 𝑦 = |𝑥| + 5

b. 𝑦 = |𝑥| − 5

c. 𝑦 = |𝑥 + 3|

d. 𝑦 = |𝑥 − 3|

e. 𝑦 = |𝑥 − 3| + 5

f. 𝑦 = |𝑥 + 3| + 5

g. 𝑦 = |𝑥 − 3| − 5

h. 𝑦 = |𝑥 − 3| − 5

i. 𝑦 = 2|𝑥|

j. 𝑦 = −2|𝑥|

k. 𝑦 = 2|𝑥| + 5

l. 𝑦 = −2|𝑥| + 5

m. 𝑦 = 2|𝑥 − 3|

n. 𝑦 = −2|𝑥 + 3| − 5

o. 𝑦 = √𝑥 − 4 − 5

p. 𝑦 = √−𝑥

q. 𝑦 = 2√𝑥 − 4 − 5

r. 𝑦 = 2√−𝑥 − 4 − 5

s. 𝑦 = −2√−𝑥 + 4 − 5

t. 𝑦 = 𝑥2 + 3

u. 𝑦 = −2𝑥3

v. 𝑦 = −𝑥4+1

w. 𝑦 = 3(𝑥 − 1)2 − 4

x. 𝑦 = |2𝑥| − 1

y. 𝑦 = 2√𝑥 + 5 − 4

z. 𝑦 − 5 = −3(𝑥 − 1)2

aa. 𝑦 = −1

2(𝑥 − 4)2 + 5

bb. 𝑦 = √−1

2(𝑥 + 2) − 1


3. Please find the following for all functions below.

I. Sketch the graph of the functions and relations below. Explain clearly how you decided the

graph was the shape you drew. Show all relevant parts of the graph.

II. Find the domain and range of the functions and relations that you graphed based on your

graph.

III. List all the vertical and horizontal asymptotes if any.

a. 𝑦 = 2𝑥 + 3

b. 𝑦 = 2𝑥 − 3

c. 𝑦 = 2𝑥+3

d. 𝑦 = 2𝑥−3

e. 𝑦 = 𝑒𝑥 + 3

f. 𝑦 = 𝑒𝑥 − 3

g. 𝑦 = 𝑒𝑥+3

h. 𝑦 = 𝑒𝑥−3

i. 𝑦 = 2𝑥−1 + 3

j. 𝑦 = 𝑒𝑥−2 − 3

k. 𝑦 = (1

2)

𝑥+1− 3

l. 𝑦 = 2𝑒𝑥−1 + 3

m. 𝑦 = 2𝑒−𝑥 + 1

n. 𝑦 = 300 (1

2)

𝑡

4

o. 𝑦 =1

2𝑒−2𝑥 + 1

p. 𝑦 = 300 (1

2)

𝑡

4

q. 𝑦 = log(𝑥) + 3

r. 𝑦 = 𝑙𝑜𝑔2(𝑥)

s. 𝑦 = −3 log(𝑥)

t. 𝑦 = ln(𝑥 + 1)

u. 𝑦 = − log2(𝑥 + 3) + 4 v. 𝑦 = 2 ln(−𝑥) + 5 w. 𝑦 = −2 log(𝑥 − 1) + 4

x. 𝑦 =1

𝑥−2

y. 𝑦 =1

𝑥−2 + 5

z. 𝑦 = − 3

𝑥−2− 5


4. Write a formula for each of the functions a-d shown in the graphs below so that they are a

transformation of one of the standard functions listed below. Explain your reasoning.

𝑦 = 𝑥, 𝑦 = 𝑥2, 𝑦 = 𝑥3, 𝑦 =1


1

𝑥.

a.

b.

c.

d.

e.

f.


5. For each original graph 𝑦 = 𝑓(𝑥) please graph the transformations, and label all relevant parts of the new graph appropriately.

a. 𝑦 = 𝑓(−𝑥) b. 𝑦 = −𝑓(𝑥) c. 𝑦 = 𝑓(𝑥 + 3) + 2 d. 𝑦 = 𝑓(−𝑥 + 2) − 3

e. 𝑦 = −1

2𝑓(𝑥)

a. 𝑦 = 𝑔(𝑥 + 1) + 3 b. 𝑦 = −𝑔(−𝑥)

c. 𝑦 =1

2𝑔(𝑥)

d. 𝑦 − 2 = 𝑔(−𝑥 − 1)

a. 𝑦 = 𝑙(−𝑥 − 3) + 1 b. 𝑦 = −𝑙(𝑥) c. 𝑦 = −𝑙(𝑥) d. 𝑦 = 2(𝑙(𝑥) − 1)


a. 𝑦 = 𝑠(−𝑥) − 3 b. 𝑦 = −𝑠(𝑥) + 2 c. 𝑦 = 𝑠(𝑥 − 1) d. 𝑦 = 𝑠(𝑥) + 2

6. Sketch the graph of the function below. Please show all your work and clearly show relevant

points.

𝑦 = 𝑔(𝑥) has the graph below, use that to find the graph of 𝑦 =1

3𝑔(−𝑥).

7. Given the two graphs below, find a algebraic relationship between the two so that one function

is the result of a transformation of the other.

In other words, write a formula connecting 𝑓(𝑥) and 𝑔(𝑥).


8. The graph of a function 𝑔 is shown. Use it to sketch the graph of 𝑦 = −𝑔(𝑥 + 1) − 2 on the

same axes. Show all your intermediary transformations using colored pens and label all the

transformations.

9. Investigate the difference between the graphs of the following functions. What is the conclusion you draw from it? You can use a graphing utility to explore these relations.

a. 𝑦 = 𝑥2 + 2

b. 𝑦 = 𝑥2 + 2.5

c. 𝑦 = 𝑥2 + 2.7

d. 𝑦 = 𝑥2 + 3

e. 𝑦 = 𝑥2 + 5

f. 𝑦 = 𝑥2 − 2

g. 𝑦 = 𝑥2 − 2.5

h. 𝑦 = 𝑥2 − 2.7

i. 𝑦 = 𝑥2 − 3

j. 𝑦 = 𝑥2 − 5

k. 𝑦 = 𝑥2 + 𝑥


2.2 Graphing Conic Sections

Conic Sections Part 1 (17 min) https://www.youtube.com/watch?v=UZlB9Bs8hQg

Conic Section Part 2 (12 min) https://www.youtube.com/watch?v=ofZn7v1jJ0U

Part 3 Equations of Circles (10 mins) http://www.youtube.com/watch?v=fzNXmoCHRCk

Conic Sections: Conic sections are the curves formed by the intersection of a plane and a cone. A

mathematical cone can be imagined as two ice-cream cones attached at their tips as shown below. The

angle that the intersecting plane makes with the axis of the cone determines which of three basic types

of conic curve you get. These are among the earliest curves to be studied back to ~350 B.C. Their initial

definition in terms of a cone and a plane precedes the development of algebra by nearly 2000 years.

Since then, several equivalent ways to define conic curves have been discovered. The “Locus” definition

below is one, while we will mostly work with graphs of algebraic equations in 𝑥 and 𝑦 to define conic

curves.

A conic is the "Locus" (which is another word for "path“) of a point 𝑃 in the plane so that its

distance from a fixed point 𝐹 (called the “focus”) in the plane has a constant ratio “𝑒” (called

the “eccentricity”) to its distance from a fixed straight line 𝑑 (called the “directrix”) in the plane.

➢ 𝑒 = 1 the conic is called a parabola

➢ 𝑒 < 1 the conic is called an ellipse

➢ 𝑒 > 1 the conic is called a hyperbola.

Mathematical Cone

Intersections of a plane and a cone.

https://www.youtube.com/watch?v=UZlB9Bs8hQg

https://www.youtube.com/watch?v=ofZn7v1jJ0U

http://www.youtube.com/watch?v=fzNXmoCHRCk


2.2a Parabolas

Parabola: A parabola is collection of all points equidistant from a fixed point (focus) and a fixed

line (directrix).

Consider a fixed point 𝐹 and line 𝐿 as shown below. The locus or collection of all points equidistant from

the fixed point focus and the fixed line Directrix are shown below. You may notice this shape as you

have seen it before in form of quadratic function in one variable.

The distance 𝑃𝐿 and 𝑃𝐹 are the same in a parabola. You can see from the picture that the collection of

all points (shown in red here) form the parabola. In the picture above, the vertex is at some point (ℎ, 𝑘),

the focus is above this 𝑝 units at (ℎ, 𝑘 + 𝑝), and the directrix is 𝑝 units below the vertex at 𝑦 = 𝑘 − 𝑝.

The point 𝑃 at (𝑥, 𝑦) represents a general point on the parabola and thus the distance from 𝑃 to the

focus at (ℎ, 𝑝 + 𝑘) must be equal to the distance straight down to the directrix at 𝑦 = 𝑘 − 𝑝.

The distance 𝑃𝐿̅̅̅̅ is the difference in the 𝑦-coordinates from 𝑃 at (𝑥, 𝑦) to the line 𝑦 = 𝑘 − 𝑝, or

(𝑦 − (𝑘 − 𝑝)).

The distance 𝑃𝐹̅̅ ̅̅ from 𝑃 at (𝑥, 𝑦) to 𝐹 at (ℎ, 𝑘 + 𝑝) is obtained by the distance formula or Pythagorean

Theorem by (𝑃𝐹̅̅ ̅̅ )2 = (𝑥 − ℎ)2 + (𝑦 − (𝑘 + 𝑝))2

Squaring the distance (𝑃𝐿̅̅̅̅ )2 = (𝑦 − (𝑘 − 𝑝))2

and setting this equal to

(𝑃𝐹̅̅ ̅̅ )2 = (𝑥 − ℎ)2 + (𝑦 − (𝑘 + 𝑝))2

we get: (𝑦 − (𝑘 − 𝑝))2

=(𝑥 − ℎ)2 + (𝑦 − (𝑘 + 𝑝))2

.

Rearranging to (𝑦 − 𝑘 + 𝑝)2 − (𝑦 − 𝑘 − 𝑝)2 = (𝑥 − ℎ)2 and factoring the left side as a difference of

squares, we obtain:

[ (𝑦 − 𝑘 + 𝑝) − (𝑦 − 𝑘 − 𝑝) ] [ (𝑦 − 𝑘 + 𝑝) + (𝑦 − 𝑘 − 𝑝) ] = 2𝑝 ⋅ 2(𝑦 − 𝑘) = (𝑥 − ℎ)2

4𝑝(𝑦 − 𝑘) = (𝑥 − ℎ)2 𝒐𝒓 𝑦 =1

4𝑝(𝑥 − ℎ)2 + 𝑘

You may notice that this is the graph of 𝑦 = 𝑥2 stretched vertically by the factor 1

4𝑝 and translated to the

vertex of the parabola (ℎ, 𝑘). Thus the 𝑦 = 𝑥2 graph is a parabola!


Putting the focus below the directrix is accomplished by letting 𝑝 < 0 which causes the vertical stretch

factor to be negative and thus the parabola opens down.

Playing

As a mathematician we would logically ask the questions about what changes might occur if we make

the directrix vertical? As we might predict, this simply reverses the roles of 𝑥 and 𝑦 in the form of the

equation and we get 𝑥 =1

4𝑝(𝑦 − 𝑘)2 + ℎ. This implies that the parabolas facing up/down and left/right

with vertex at (ℎ, 𝑘) and distance from vertex to focus equal to 𝑝 can be expressed in Vertex Form as

one of the equations: 𝑦 =1

4𝑝(𝑥 − ℎ)2 + 𝑘 𝒐𝒓 𝑥 =

1

4𝑝(𝑦 − 𝑘)2 + ℎ respectively.

In general then,

𝑦 =1

4𝑝(𝑥 − ℎ)2 + 𝑘 , for 𝑝 > 0

𝑦 =1

4𝑝(𝑥 − ℎ)2 + 𝑘 , for 𝑝 < 0

𝑥 =

1

4𝑝(𝑦 − 𝑘)2 + ℎ, for 𝑝 > 0

𝑥 =1

4𝑝(𝑦 − 𝑘)2 + ℎ, for 𝑝 < 0


Example

1. Consider the parabola with equation 𝑦 =1

8(𝑥 − 2)2 − 3. Find the focus, directrix, and the vertex of

this parabola and sketch the graph.

Solution: We see the vertex is shifted 2 Right and 3 down and 1

4𝑝=

1

8 or 𝑝 = 2 > 0 which means the

parabola faces up. The directrix is 𝑝 units below the vertex at 𝑦 = −5. The focus is 𝑝 units above the

focus at (2, −1).

Reflective Property of Parabola: One of the properties of a parabola is that if an energy source

located far away transmits energy and the parabola is made of a reflective surface the energy

bounces of the parabola and merges onto the focus. Similarly if an energy source is located at

the focus it will bounce off the parabola along the lines perpendicular to the point where it hits

the parabola.

This reflective property use can be seen in satellite dishes, telescopic mirrors, car headlights, and search

beams, making parabolas useful in real life in many different domains.


All quadratic equations with only one of 𝑥 or 𝑦 being squared and the other to the first power can be

transformed into one of the standard Vertex-Forms of a parabola. Thus we recognize these curves from

their equations by the presence of either 𝑥 or 𝑦 being squared. When there are both quadratic and

linear terms present, we complete the square to put the equation in vertex form.

Equation of an up/down opening parabola in vertex form is given by: 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘

Where (ℎ, 𝑘) = Vertex of the parabola, 𝑎 determines the direction of the parabola (up, or down) and contributes the width of the parabola. Also from above,

𝑎 =1

4𝑝 where 𝑝 is the distance from the vertex to the

focus. ➢ If 𝑎 > 0, the parabola opens up. ➢ If 𝑎 < 0, the parabola opens down. ➢ If |𝑎| > 1, the parabola is vertically stretched or is

narrower. ➢ If |𝑎| < 1, the parabola is vertically compressed, or

is wider. ➢ The graph moves up/down 𝑎 units as 𝑥 moves one

unit right or left from the vertex.

Equation of a left/right opening parabola in vertex form is given by: 𝑥 = 𝑎(𝑦 − 𝑘)2 +h

Where (ℎ, 𝑘) = Vertex of the parabola and 𝑎 determines the direction of the parabola (left or right) and impacts the width of the parabola. The distance

from the vertex to focus is 𝑝 and 𝑎 =1

4𝑝

➢ If 𝑎 > 0, the parabola opens to the right. ➢ If 𝑎 < 0, the parabola opens to the left. ➢ If |𝑎| > 1, the parabola is narrower. ➢ If |𝑎| < 1, the parabola is wider. ➢ The graph moves left/right 𝑎 units as 𝑦 moves one

unit up or down from the vertex.


Example

1. Sketch the graph of 𝑦 = 5𝑥2 + 3𝑥 + 2

Solution: We will try to get the equation in the standard form which will help in the graphing and

finding the vertex. Use completing the square process.

𝑦 = 5 (𝑥2 +3

5𝑥) + 2

To complete the square, add half of the middle term squared and subtract the same so the net

result is no change on the right side.

𝑦 = 5 (𝑥2 +3

5𝑥 + (

3

2(5))

2

) + 2 − 5 (3

2(5))

2

𝑦 = 5 (𝑥 +3

10)

2

+ 111

20

Vertex is at (−3

10, 1

11

20)

𝒙 𝒚

−6

10

2

Vertex −3

10 31

20

0 2


In general, for any quadratic function in standard form y = a𝑥2 + 𝑏𝑥 + 𝑐 we can complete square

and get the equation into vertex-form. If we apply this to a quadratic in 𝑥 in standard form, we

obtain general results as to where the vertex and focus are located in terms of the coefficients 𝑎, 𝑏,

and 𝑐.

𝑦 = a𝑥2 + 𝑏𝑥 + 𝑐

𝑦 = a (𝑥2 +𝑏

𝑎𝑥) + 𝑐 add the square of half of the 𝑥-coefficient and also subtract this outside the ().

𝑦 = a (𝑥2 +𝑏

𝑎𝑥 + (

𝑏

2𝑎)

2

) + 𝑐 − 𝑎 (𝑏

2𝑎)

2

𝑦 = a (𝑥 +𝑏

2𝑎)

2

+ 𝑐 −𝑏2

4𝑎

Vertex is at (−𝑏

2𝑎, 𝑐 −

𝑏2

4𝑎). Recall that −

𝑏

2𝑎 is the first part of the quadratic formula. This is very useful

in that −𝑏

2𝑎 very quickly locates the 𝑥-coordinate of the vertex. It is perhaps easiest to obtain the 𝑦-

coefficient by plugging 𝑥 = −𝑏

2𝑎 into the original formula. Two additional points can be obtained by

moving one unit in either direction from the vertex and 𝑦 will move 𝑎 units. If one needs to locate the

focus, we have that 𝑎 =1

4𝑝 which solved for 𝑝 gives the focus-vertex distance as 𝑝 =

1

4𝑎 .

The graph of 𝑦 = 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐, is a parabola with vertex at 𝑥 = −𝑏

2𝑎 and focus a

distance of 𝑝 =1

4𝑎 from the vertex. The 𝑦-coordinate of the vertex is at 𝑓 (−

𝑏

2𝑎).

When 𝑎 > 0, the function output at the vertex, 𝑦 = 𝑓 (−𝑏

2𝑎), is the minimum value of the function.

If 𝑎 < 0, the parabola opens downward and 𝑦 = 𝑓 (−𝑏

2𝑎) is the maximum value of the function.

Quadratic functions are used to model many situations where either a maximum or minimum output of

the function is desired. Some examples include finding optimum revenue by controlling the number of

items that are to be sold and determining the maximum height of an object that is thrown in the air.

For parabolas opening left/right we can solve for 𝑥 = 𝑎𝑦2 + 𝑏𝑦 + 𝑐 with 𝑎 ≠ 0. The vertex is

located at = 𝑘 = −𝑏

2𝑎 . Plug this 𝑦 into the equation to get: = 𝑎 (−

𝑏

2𝑎)

2

+ 𝑏 (−𝑏

2𝑎) + c.

The focus is 𝑝 =1

4𝑎 units horizontally from the vertex. (Opens right if 𝑎 > 0, left if 𝑎 < 0.) Also,

the stretch factor 𝑎 is how far left or right we move from the vertex 𝑥-value, as 𝑦 moves up or

down one unit from the vertex value.


Practice Problems

1. For the problems below find the vertex, and the focus. Then sketch the graph of the function.

a. 𝑦 =1

3𝑥2 − 8𝑥 + 2

b. 𝑦 = −0.05𝑥2 + 𝑥 + 2

c. 𝑥 = −2𝑦2 − 12𝑦 − 14

d. 𝑥 =1

12𝑦2 − 𝑦 + 5

2. Find the equation of the parabola whose graph is given below.

3. Find an equation of the parabola that goes through the points (3, 3), (-1, 3), and (1, 5).

Solutions

1.

a. 𝑦 =1

3𝑥2 − 8𝑥 + 2

Vertex is at 𝑥 =8

2/3= 12, 𝑦 = −46

Focus is at 𝑝 =1

4/3=

3

4 up from vertex.

b. 𝑦 = −0.05𝑥2 + 𝑥 + 2

Vertex is at 𝑥 = −1

2(−0.05)= 10, 𝑦 = 7

Focus is at 𝑝 =1

4(−0.05)= −5; 5 down from

vertex. Note that 𝑦 moves 𝑎 units as 𝑥 goes one in either direction from the vertex. Here as 𝑥 moves from 10 to 11 or to 9, y moves from 7 down 0.05 units to 6.95.


c. 𝑥 = −2𝑦2 − 12𝑦 − 14

Vertex is at 𝑦 =12

−4= −3, 𝑥 = 4

Focus is at =1

−8 ;

1

8 left from vertex.

d. 𝑥 =1

12𝑦2 − 𝑦 + 5

Vertex is at 𝑦 =1

1/6 = 6, 𝑥 = 2

Focus is at 𝑝 =1

1/3 = 3; 3 right of

vertex.

2. Find the equation of the parabola whose graph is given below.

The vertex is at (8, −2) and 𝑦 goes up ¼ unit as 𝑥

moves one unit left/right from 𝑥 = 2, so 𝑎 = 1/4

and 𝑦 =1

4(𝑥 − 8)2 − 2

3. Find an equation of the parabola that goes through the points (3, 3), (-1, 3), and (1, 5).

First plot the points.

From the points and symmetry, the point ( 1,5) must be the vertex. Also we move

down 1

2 units when 𝑥 moves 1 left/right, so

𝑎 = −1

2 and 𝑦 = −

1

2(𝑥 − 1)2 + 5.


Section2.2a Worksheet Date:______________ Name:__________________________


1. What is a parabola?

2. What are all the different forms in which we can write equation of a parabola?

3. What are the different kinds of parabolas?

4. What equation forms are possible for a parabola with vertex at (𝟑, −𝟓). There is one parameter 𝒂 that is not given.

𝑥 = Or 𝑦 =

5. What is the maximum value of a function that is quadratic?

6. What is the minimum value of a quadratic function?



Exercises 2.2a

1. Sketch the graphs of the following relations. Find the vertex of the parabola and other relevant

information if asked.

a. 𝑦 = 2(𝑥 − 3)2 − 5 b. 𝑦 = −2(𝑥 − 3)2 − 5 c. 𝑥 = 2(𝑦 − 3)2 − 5 d. 𝑥 = −2(𝑦 − 3)2 − 5 e. 𝑦 = 3(𝑥 + 1)2 + 2 f. 𝑥 = −3(𝑦 + 1)2 + 2 g. 𝑥2 − 4𝑥 + 1 = 𝑦

(Hint: Use completing the squares)

h. −𝑥2 − 4𝑥 + 1 = 𝑦 (Hint: Use completing the squares)

i. 2

3𝑥2 − 4𝑥 +

1

3= 𝑦

j. −2

3𝑦2 − 4𝑦 −

1

3= 𝑥

k. 𝑦 = 4𝑥2 l. 𝑦 = 4(𝑥 − 1)2 − 4 m. 𝑦 = 4𝑥2 − 8𝑥 n. 𝑥2 + 8𝑥 + 3𝑦 + 22 = 0 o. 2𝑦2 − 16𝑦 − 𝑥 + 29 = 0 p. 3𝑥2 − 5𝑥 + 4 − 2𝑦 = 10

2. Graph and find an equation of the parabola with vertex (4,3) and focus (4,6).

3. Graph and find an equation of the parabola passing through (−1,3), (3,3) and (1,2). What is

the vertex of this parabola?

4. Graph and find an equation of the parabola with vertex (−4,3) and focus (4,3).

5. An old satellite dish is has a cross-section that is in the shape of a parabola. The dish is 48 inches across and 8 inches deep at its center. The vertex is at this lowest point in the parabola. Put the vertex at (0, 0) and find an equation for the parabola cross section. Also determine the value of 𝑝 and state how high above the vertex the focus is.

6. Create a parabola for a parabolic trough solar cooker that can be contained inside of a copier paper box. These boxes are 17 inches long and 10 inches deep. Make it so your parabola has vertex within one inch of the bottom of the box and so that the focus is within 3 inches of the bottom center of the box. Find an equation for this parabola when the origin is at the center line of the bottom of the box.


2.2b Circles and Ellipses

A locus definition for a circle would be all points that lie a fixed distance called the radius from a

fixed point called the center of the circle. We saw earlier that all circles lead to equations of the

form (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑅2 when the center and radius are known to be (ℎ, 𝑘) & 𝑅.

An ellipse can be obtained by stretching the graph of a circle to make it oblong in shape. The locus

definition for an ellipse requires two fixed points each called a focus point and a length (2𝑎) called the

major diameter of the ellipse.

A locus definition for an ellipse is all points 𝑃 such that the sum of the two distances from 𝑃 to

each focus is equal to the major diameter of the ellipse.

This provides an easy way to draw an ellipse when the two foci and the major diameter are given. You can place pins at each focus and connect them with a string that is the major diameter long. Then take your pencil and use it to keep the string taught as you move it along the ellipse.

The locus definition leads to the standard form of the equation of an ellipse as:

(𝑥 − ℎ)2

𝑎2+

(𝑦 − 𝑘)2

𝑏2= 1

Here (ℎ, 𝑘) is the center and 𝑎 and 𝑏 are each half of the major and minor diameters which are

horizontal and vertical depending on whether 𝑎 > 𝑏, or 𝑎 < 𝑏. The larger of 2𝑎 or 2𝑏 is

referred to as major diameter. The two focus points are located on the major diameter at a

distance 𝑐 = √|𝑎2 − 𝑏2| from the center. The endpoints of the major diameter are called the

vertices of an ellipse.

➢ If 𝑎 < 𝑏, the ellipse major diameter is vertical, and the Foci are given by (ℎ, 𝑘 ± 𝑐) and

vertices by (ℎ, 𝑘 ± 𝑏).

➢ If 𝑎 > 𝑏 the major diameter is horizontal and the Foci are given by (ℎ ± 𝑐, 𝑘) and vertices

by (ℎ ± 𝑎, 𝑘).

➢ If 𝑎 = 𝑏 this ellipse equation reduces to that of a circle of radius 𝑎 and center(ℎ, 𝑘).


Reflective Property of Ellipse: A light or sound wave leaving in any direction from one focus of

an ellipse will reflect off of the ellipse and arrive at the other focus of the ellipse.

This property is used in whispering galleries where the walls of a room are in the shape of an ellipse and if a person speaks softly at one focus, a listener at the other focus point will receive all of the sound from the whisperer from all parts of the wall. Elliptical reflectors are also used in medicine to treat kidney stones without surgery. Shockwaves are generated at one focus outside of the body and an elliptical shell is adjusted so that the kidney is positioned at the other focus of the elliptical shell where it receives the concentrated energy of the waves and gets pulverized.

Examples:

1. Sketch the graph of (𝑥+2)2

64+

(𝑦−3)2

25= 1. Find the center, major and minor axis, and the foci.

2. Sketch the graph of the ellipse and find the center of the ellipse.

4𝑥2 + 16𝑥 + 9𝑦2 + 18𝑦 = 119

3. Sketch the graph of the ellipse with equation 25𝑥2 − 150𝑥 + 9𝑦2 + 36𝑦 = −36 and locate its

center, vertices and foci.

4. Find equations and the location of the foci for the ellipses which are graphed below.

a.

b.


Solutions

1. Sketch the graph of (𝑥+2)2

64+

(𝑦−3)2

25= 1. Find the center, major and minor axis, and the foci.

Solution: Center is (−2,3). 𝑎 = 8,

𝑏 = 5, 𝑐 = √|64 − 25| = √39 and major axis is of length 16 and minor axis of length 10. Foci are

located at (−2 + √39, 3), (−2 −

√39, 3). The vertices are located

at (−2 + 8, 3) = (6, 3) and at (−2 − 8, 3) = (−10, 3)

2. Sketch the graph of the ellipse and find the center of the ellipse.

4𝑥2 + 16𝑥 + 9𝑦2 + 18𝑦 = 119

Solution: 4(𝑥2 + 4𝑥 + 𝟒) + 9(𝑦2 + 2𝑦 + 𝟏) =119+𝟏𝟔 + 𝟗 4(𝑥 + 2)2 + 9(𝑦 + 1)2 = 144 (𝑥 + 2)2

36+

(𝑦 + 1)2

16= 1

Center is (−2, −1), 𝑎 = 6, 𝑏 = 4. Vertices are (−2 + 6,1) = (4,1) and (−2 − 6,1) = (−8,1)


3. Sketch the graph of the ellipse with equation 25𝑥2 − 150𝑥 + 9𝑦2 + 36𝑦 = −36 and locate its

center, vertices and foci.

Solution: Complete the squares: 25(𝑥2 − 6𝑥 + 𝟗) + 9(𝑦2 + 4𝑦 + 𝟒) = −36 + 𝟐𝟐𝟓 + 𝟑𝟔 25(𝑥 − 3)2 + 9(𝑦 + 2)2 = 225 (divide both sides by 225)

(𝒙 − 𝟑)𝟐

𝟗+

(𝒚 + 𝟐)𝟐

𝟐𝟓= 𝟏

Center is at (3, −2) major diameter is vertical and 2𝑏 =10 units long, while the minor diameter is 2𝑎 = 6 units

long. The foci are vertically 𝑐 = √|9 − 25| = √16 = 4 units up and down from the center since 𝑎 = 3 < 𝑏 = 5 and the vertices are 5 units up and down from the center.

4. Find equations and the location of the foci for the ellipses which are graphed below.

a.

Center is at (−2, 3), 𝑎 = 5, 𝑏 = 2,

𝑐 = √|52 − 22| = √21

Foci at (−2 ± √21, 3),

Vertices at (−2 ± 5, 3 )

Equation is: (𝑥+2)2

25+

(𝑦−3)2

4= 1 .

b.

Center is at (3, −4), 𝑎 = 4, 𝑏 = 3,

𝑐 = √42 − 32 = √7

Foci at (3, −4 ± √7),

Vertices at (3, ±4)

Equation is: (𝑥−3)2

9+

(𝑦+4)2

16= 1


Circles

Equations of Circles

http://www.youtube.com/watch?v=fzNXmoCHRCk (10 mins)

We derive the equation of a circle with a given the center and radius using the distance formula.

Hint: The distance between two points; the center at (𝑥1, 𝑦1) = (𝒉, 𝒌), and a point on the circle (𝑥2, 𝑦2) = (𝑥, 𝑦) is given by:

𝑑 = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2 𝑑2 = (𝒙 − 𝒉)2 + (𝑦 − 𝒌)2

With this distance set to 𝑑 = 𝑅, the equation of the circle centered at (ℎ, 𝑘) = (0, 0) with radius of 𝑅 = 4 becomes: 𝑥2 + 𝑦2 = 16.

Equation of a Circle in Standard Form

The equation (𝒙 − 𝒉)𝟐 + (𝒚 − 𝒌)𝟐 = 𝑹𝟐 represents the circle with center (ℎ, 𝑘) and radius 𝑅.

Examples

1. Sketch the graph of (𝑥 − 2)2 + (𝑦 + 1)2 = 9

2. Find equation of a circle with

a. Center at (3,5) and radius of 2.

b. Center (−3,7) and radius of 5.

3. Find the center and radius of the circle given by (𝑥 − 6)2 + (𝑦 + 3)2 = 49

4. Complete the squares to establish that the equation 𝑥2 − 12𝑥 + 𝑦2 + 8𝑦 = 12 represents a

circle and determine its center and radius.

5. Find the equation of the circle whose graph is given below.

http://www.youtube.com/watch?v=fzNXmoCHRCk


Solutions

1. Sketch the graph of (𝑥 − 2)2 + (𝑦 + 1)2 = 9 So the graph of the equation (𝑥 − 2)2 + (𝑦 + 1)2 = 32 would be a circle with radius of 3 and center of (2, −1) since (𝑥 − 2)2 + (𝑦 + 1)2 = 32 can be written as (𝑥 − 2)2 + (𝑦 − (−1))2 = 32

2. Find equation of a circle with

a. Center at (3,5) and radius of 2.

Solution: An equation for the circle centered at (3, 5) with radius 2 is

(𝑥 − 3)2 + (𝑦 − 5)2 = 22

b. Center (−3,7) and radius of 5.

Solution: The circle of radius 5 centered at (−3, 7) is represented by

(𝑥 − (−3))2

+ (𝑦 − 7)2 = 52 or (𝑥 + 3)2 + (𝑦 − 7)2 = 25.

3. Find the center and radius of the circle given by (𝑥 − 6)2 + (𝑦 + 3)2 = 49

Solution: The equation (𝑥 − 6)2 + (𝑦 + 3)2 = 49 represents a circle centered at (6, −3) of

radius 7.

4. Complete the squares to establish that the equation 𝑥2 − 12𝑥 + 𝑦2 + 8𝑦 = 12 represents a

circle and determine its center and radius.

𝑥2 − 12𝑥 + 𝟑𝟔 + 𝑦2 + 8𝑦 + 𝟏𝟔 = 12 + 𝟑𝟔 + 𝟏𝟔

(𝑥 − 6)2 + (𝑦 + 4)2 = 64.

Thus the graph of 𝑥2 − 12𝑥 + 𝑦2 + 8𝑦 = 12 is the circle of radius 8 centered at (6, -4).

5. Find the equation of the circle whose graph is given below.

The equation for the circle graphed to the right is obtained by reading off the center as (3, −2) and radius as 𝑟 = 3 from the graph. Thus the equation is (𝑥 − 3)2 + (𝑦 + 2)2 = 9.


Section2.2b Worksheet Date:______________ Name:__________________________


1. What is a circle?

2. What is an ellipse? How is it different than a circle?

3. What are the major axis and minor axis of an ellipse?

4. How do we find the center and the foci of an ellipse if we are given its equation?

5. How do we find the center and radius of a circle from its equation?



Exercises 2.2b

1. Sketch the graph of each circle and state its center and radius.

a. 𝑥2 + 𝑦2 = 36 b. (𝑥 − 3)2 + (𝑦 − 4)2 = 25 c. (𝑥 + 2)2 + (𝑦 − 1)2 = 9 d. 4𝑥2 + 8𝑥 + 4𝑦2 − 24𝑦 = 24 e. Plot the graph of(𝑥 + 5)2 + (𝑦 − 3)2 = 9.

f. Plot the graph of(𝑥 − 2)2 + (𝑦 + 1)2 =25

4.

g. 𝑥2 + 4𝑥 + 𝑦2 − 6𝑦 = 12

h. 2𝑥2 + 8𝑥 + 2𝑦2 − 14𝑦 = −29

2

2. Sketch graphs and write equations that represent each circle described below.

a. The center is at (2, 7) and the radius is 𝑟 = 9.

b. Two endpoints of a diameter are the points (1, 2) and (5,2).

c. Find equation of the circle with center (−3, 5) and radius 5.

d. Find the equation of a circle with center (2,5), and passing through (−2,2).

e. Find equation of the circle whose diameter has endpoints (−1,4), and (4, −2).

f. Find an equation of the circle with center (-3, 5) that goes through the point (5, -1).

3. Write the equation in standard form for the graph of the circle shown below.

4. Write equation of the circle in standard form for the graph of the circle shown below.

5. Plot the ellipses below and locate their center, vertices, foci, and the values of 𝑎 and 𝑏.

a. (𝑥+3)

16

2+

(𝑦−1)

25

2= 1

b. 16𝑥2 + 𝑦2 − 16 = 0

c. 4𝑥2 − 16𝑥 + 9𝑦2 − 18𝑦 = 11

d. (𝑥−3)

9

2+ (𝑦 + 2)2 = 1


6. Find an equation of each ellipse described or plotted below and find the location of their foci.

a. The ellipse is centered at (3, 6) with a horizontal major diameter of 10 and minor diameter of 6.

b. The ellipse has vertices at (−3, 6) and (−3, −2) and (−1, 2) is the endpoint of a minor diameter.

c. The ellipse plotted below:

d. The ellipse plotted below:

2.2c Hyperbolas One final type of conic section call a hyperbola can be constructed when given two fixed points called

foci and either a single point on the hyperbola or a distance 2𝑎 which will again be the distance between

two vertices of a hyperbola.

A locus definition for a hyperbola is all points 𝑃 such that the difference between the distance

from 𝑃 to one focus is 2𝑎 more than or less than the distance from 𝑃 to the other focus.

General form of the equation of the hyperbola is given by:

(𝒙 − 𝒉)𝟐

𝒂𝟐 −

(𝒚 − 𝒌)𝟐

𝒃𝟐 = 𝟏, 𝒐𝒓 −

(𝒙 − 𝒉)𝟐

𝒃𝟐 +

(𝒚 − 𝒌)𝟐

𝒂𝟐 = 𝟏

Here (ℎ, 𝑘) is the center, the two vertices are at distance 𝑎 units from the center in the

direction of the positive variable, asymptotes are diagonal lines through a box with sides 𝑎 and

𝑏 units away from the center in the direction of the positive and negative variables respectively.

The foci are located a distance of 𝑐 = √𝑎2 + 𝑏2 from the center and lie beyond the vertices.

Vertices for (𝑥−ℎ)2

𝑎2−

(𝑦−𝑘)2

𝑏2= 1 are located at (ℎ ± 𝑎, 𝑘).

Vertices for −(𝑥−ℎ)2

𝑎2 +(𝑦−𝑘)2

𝑏2 = 1 are located at or (ℎ, 𝑘 ± 𝑏) respectively.

For extra credit, find out about the reflective property of hyperbolas and an application of the reflective

property.


Examples

1. Sketch the graph of (𝑥−2)2

9−

(𝑦−3)2

4= 1. Find the center, the foci and the asymptotes.

2. Plot the graph of hyperbola and find the center, asymptotes, and foci.

−4𝑥2 − 16𝑥 + 9𝑦2 + 18𝑦 = 43

Solutions

1. Sketch the graph of (𝑥−2)2

9−

(𝑦−3)2

4= 1. Find the center, the foci and the asymptotes.

Solution: Center is (2, 3), 𝑎 = 3, 𝑏 = 2, 𝑐 = √9 + 4 = √13, foci are located at (2 ± √13, 3).

Asymptotes are given by 𝑦 = 3 ±2

3(𝑥 − 2). Vertices are at (2 ± 3, 3) = (5, 3)𝑜𝑟 (−1, 3).

2. Plot the graph of hyperbola and find the center, asymptotes, and foci.

−4𝑥2 − 16𝑥 + 9𝑦2 + 18𝑦 = 43

−4(𝑥2 + 4𝑥) + 9(𝑦2 + 2𝑦) =43 (Complete the squares → (𝑥 + 2)2 & (𝑦 + 1)2)

−4(𝑥 + 2)2 + 9(𝑦 + 1)2 = 43 − 16 + 9

−4(𝑥 + 2)2 + 9(𝑦 + 1)2 = 36

−(𝑥 + 2)2

9+

(𝑦 + 1)2

4= 1

Center is (−2, −1), 𝑎 = 2, 𝑏 = 3, asymptote box is 3 R/L and 2 U/D from center and asymptote

equations are 𝑦 = −1 ±2

3(𝑥 + 2). Vertices are located at (−2, −1 ± 2) = (−2,1)𝑜𝑟 (−2, −3).

The foci are at 𝑐 = ±√𝑎2 + 𝑏2 Up/Down from the center at (−2, −1 ± √5).


In general then, an equation of the type 𝑎𝑥2 + 𝑏𝑦2 + 𝑐𝑥𝑦 + 𝑑𝑥 + 𝑓𝑦 + 𝑔 = 0 will represent a

conic section.

➢ If 𝑐 = 0 and either 𝑎 or 𝑏 is zero but not both, we will get a parabola.

➢ If 𝑐 = 0 and both 𝑎 and 𝑏 are both nonzero real numbers with the same sign, we will get an

ellipse with its diameters parallel to the 𝑥 and 𝑦 axes.

➢ If 𝑐 = 0 and both 𝑎 and 𝑏 are both nonzero real numbers with the opposite sign, we will get

a hyperbola aligned with the coordinate axes.

In the case ≠ 0 , we still end up with one of these three curves, but they will be rotated through some

angle from the cases we studied here with 𝑐 = 0. Some examples are shown below. Finding the

center, foci, and even what type of conic an equation represents requires a little more advanced

knowledge of trigonometry and is a motivation for you take more math if you like what you see below.

1. 𝑥2 − 𝑥𝑦 + 𝑦 − 2 + 2𝑦2 = 0

2. 𝑥2 − 𝑥𝑦 + 𝑦 − 2 − 2𝑦2 = 0

3. 2𝑥2 − 𝑥𝑦 + 𝑦 − 2 + 𝑦2 = 0

4. −2𝑥2 − 𝑥𝑦 + 𝑦 − 2 + 𝑦2 = 0


Summary and tips on Identifying Conic Sections

Equations of the type 𝑎𝑥2 + 𝑏𝑦2 + 𝑐𝑥 + 𝑑𝑦 + 𝑓 = 0 have graphs that are one of the three conic section

types. (There are a few pathological cases that can happen where the equation has no solutions, e.g.,

𝑥2 + 𝑦2 = −5.) The type of conic can be determined by the presence and sign of the coefficients 𝑎 and

𝑏. The table below summarizes what we’ve done in this chapter to identify type and plot these conic

section graphs.

Conic Section Tips to Identify

Parabola When only one variable has a squared term, it is a parabola.

If the equation contains an 𝑥2 term this means it is an up/down parabola. Put the 𝑦 term on one side and move the rest to the other side.

If equation contains 𝑦2 term this means it is a left/right parabola. Put the 𝑥 term on one side and move the rest to the other side.

If the coefficient of the 𝑥2 term is > 0, then the parabola faces up. Example 3𝑥2 − 4𝑦 + 𝑥 − 1 = 0

or 3

4𝑥2 +

1

4𝑥 −

1

4= 𝑦

If the coefficient of the 𝑥2 term is < 0, then the parabola faces down. Example 3𝑥2 + 4𝑦 + 𝑥 − 1 = 0

or 𝑦 = −3

4𝑥2 −

1

4𝑥 +

1

4

If the coefficient of the 𝑦2 term is > 0, then the parabola faces right. Example

𝑥 =3

4𝑦2 −

1

4𝑦 +

1

4

If the coefficient of the 𝑥2 term is < 0, then the parabola faces down. Example

𝑥 = −3

4𝑦2 −

1

4𝑦 +

1

4

Focus is 𝑝 units from vertex where 𝑝 =1

4𝑎

Ellipse Get all terms on the same side. Both 𝑥2, and 𝑦2 terms are of the same sign.

Circle Ellipse that is not a circle

Both 𝑥2, and 𝑦2 terms are of the same sign and the same coefficient. Example 4𝑥2 + 4𝑦2 − 8𝑥 + 4𝑦 = 10

Both 𝑥2, and 𝑦2 terms are of the same sign and different coefficients. Example 8𝑥2 + 3𝑦2 − 8𝑥 + 4𝑦 = 10


Hyperbola Get all terms on the same side, complete the squares if there are 𝑥 or 𝑦-terms and finally make the other side equal to one. To be a hyperbola, the (𝑥 − ℎ)2, and (𝑦 − 𝑘)2 terms have coefficents of opposite sign.

When both 𝑥2, and 𝑦2 terms are on the same side the (𝑥 − ℎ)2 is positive, the hyperbola opens left/right. Example 𝑥2 − 4𝑦2 + 6𝑥 + 8𝑦 = 11 →

(𝒙 + 𝟑)𝟐

𝟏𝟔−

(𝒚 − 𝟏)𝟐

𝟒= 𝟏

When both 𝑥2, and 𝑦2 terms are on the same side the (𝑦 − 𝑘)2 is positive, the hyperbola opens up/down. Example − 𝑥2 + 4𝑦2 − 6𝑥 − 8𝑦 = 21 →

−(𝑥 + 3)2

16+

(𝑦 − 1)2

4= 1

Examples

Determine the type of conic, locate vertex(s) and focus(s) and asymptotes (if any) and graph.

1. 𝑥2 + 5𝑦2 − 4𝑥 + 30𝑦 = −29 Solution: Complete Squares

𝑥2 − 4𝑥 + 5(𝑦2 + 6𝑦 ) = −29 (𝑥−2)2

20+

(𝑦+3)2

4= 1

This is the equation of an ellipse with a horizontal

major axis, center at (2, −3), 𝑎 = √20, 𝑏 = 2

𝑐 = √20 − 4 = √16 = 4

Vertices at (2 ± √20, −3) Foci at (2 ± 4, −3).

2. 𝑦2 − 6𝑦 − 8𝑥 − 15 = 0 Solution: Complete Squares 𝑦2 − 6𝑦 = 8𝑥 + 15 (𝑦 − 3)2 = 8𝑥 + 24

𝑥 =1

8(𝑦 − 3)2 − 3

The equation is a parabola with vertex at (−3, 3).

𝑝 =1

4⋅1

8

=11

2

= 2 > 0 means it opens to the right.

Focus is at (-1, 3).


3. 9𝑥2 − 16𝑦2 + 36𝑥 + 96𝑦 = −36 Solution: Complete Squares

9(𝑥2 + 4𝑥 + 4) − 16(𝑦2 − 6𝑦 + 9) = −36 + 36 − 144 9(𝑥 + 2)2 − 16(𝑦 − 3)2 = −144 Divide both sides by −144

−(𝑥 + 2)2

16+

(𝑦 − 3)2

9= 1

Hyperbola! Center at (−2, 3), opens up/down since the negative sign is with the 𝑥 term, 𝑎 = 3, 𝑏 = 4. Vertices at (−2, 3 ± 3).

𝑐 = √9 + 16 = 5 Foci are at (−2, 3 ± 5) = (−2, −2) and (−2,8).

Asymptotes are given by 𝑦 = ±3

4(𝑥 + 2) + 3 are shown in

the dotted red lines. The purple dotted box is used a guide to plot the asymptotes.


Section2.2c Worksheet Date:______________ Name:__________________________


1. What is a hyperbola?

2. What are the different kinds of hyperbolas?

3. How do you find center and foci of a hyperbola?

4. How do we find the asymptotes of a hyperbola?



Exercises 2.2c

1. In problems a through f, locate the center, vertices, foci, and the asymptotes. Then sketch the graph

and show all the information you found in your graphs.

a. (𝑦+3)

9

2−

(𝑥−1)

16

2= 1

b. (𝑥+3)

16

2−

(𝑦−1)

9

2= 1

c. 16𝑥2 − 𝑦2 − 16 = 0

d. 4𝑥2 − 16𝑥 − 9𝑦2 + 18𝑦 = 29

e. 4𝑦2 − 16𝑦 − 25𝑥2 − 150𝑥 = 309

f. −4𝑥2 − 16𝑥 + 9𝑦2 + 18𝑦 = 43

2. For problems a through h, complete the square as needed to identify what conic section the

equation represents, and locate the vertex(center), vertices, and focus(foci) and asymptotes as

appropriate.

a. 𝑦2 + 6𝑦 − 2𝑥 + 5 = 0 Ellipse Hyperbola

Circle Parabola

b. 𝑥2 + 6𝑥 − 8𝑦 − 31 = 0

Ellipse Hyperbola

Circle Parabola

c. 16𝑥2 − 64𝑥 + 9𝑦2 + 108𝑦 = −244 Ellipse Hyperbola

Circle Parabola

d. 16𝑦2 − 96𝑦 − 9𝑥2 − 18𝑥 = 9 Ellipse Hyperbola

Circle Parabola


e. 𝑦2 = 2𝑥 − 𝑥2 Ellipse Hyperbola

Circle Parabola

f. 𝑥2 + 𝑦2 − 2𝑥 + 4𝑦 = 4 Ellipse Hyperbola

Circle Parabola

g. 3𝑦2 − 4𝑥 + 6𝑦 = 4

Ellipse Hyperbola

Circle Parabola

h. 3𝑦2 + 4𝑥 + 6𝑦 = 4

Ellipse Hyperbola

Circle Parabola

3. In problems a through d find an equation of the conic section for the given graphs

a. b.


c. d.

4. Two friends live 10 miles from away from each other along an east-west highway. One day as

they are chatting on their cell-phones, the eastern friend hears a loud clap of thunder over the

cell phone and 40 seconds later hears the sound of the same thunder reaching her house

directly. Since the sound of thunder travels a mile in 5 seconds, this means that the thunder

traveled an additional 40 seconds to reach the eastern friend. Thus the distance from the

lightning strike to the eastern friend was 8 miles longer than the distance to the western friend.

Use the locus definition of a hyperbola to locate the possible locations of the lightning strike.

(Hint: Draw a graph with the friends located on the 𝑥-axis at ± 5 and then draw an appropriate

hyperbola. )

Also, if the western friend notices that the flash of lightning that produced this thunder was

directly north of her, determine exactly where the strike was.


5. Match each relation to its appropriate graph. If there is no match, please state so.

Match all the quantities in Column B that are equivalent to quantities in Column A. Some of the

column B quantities may not have any corresponding items in column A, but all items in column

A have at least one or more corresponding items in column B. Explain your reasoning for the

choices you made. (3 pt each)

Column A Answer Column B

i. (𝑥 + 2)²

16+

(𝑦 − 2)²

4= 1

Reasoning:

a.

ii.

(𝑥 − 2)²

4−

(𝑦 − 3)²

16= 1

Reasoning:

b.

iii.

−(𝑥 − 2)2

4+

(𝑦 − 3)²

16= 1

Reasoning:

c.

iv. 𝑥2 + 𝑦2 = 4 d.


2.3 Graphing Polynomial Functions We concluded the last chapter with graphs of second degree polynomials in 𝑥 and 𝑦 which graphically

are represented by parabolas, ellipses and hyperbolas. While these curves have been studied for over

2000 years, about 400 years ago with the development of algebra came infinitely many new curves to

study by looking at graphs of equations. In this section we look at general polynomial functions in one

variable. Back in chapter 1 we saw examples of some simple polynomial functions which we review

below.

1. Polynomial function: A function defined as 𝑃(𝑥) = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + 𝑎3𝑥3 + ⋯ + 𝑎𝑛𝑥𝑛, where 𝑎0, 𝑎1, 𝑎2, 𝑎3, … 𝑎𝑛 are real numbers and n ≥ 0 is a whole number. The domain of these functions is all real numbers.

Examples:

i) Constant Function: A function defined as 𝑓(𝑥) = 𝑎, where 𝑎 is any real number. This is a polynomial function of degree zero.

Generic Graph

Domain: All real numbers Range: {𝑎} One-to-One: NO

Example If 𝑓(𝑥) = 5, then find

a. 𝑓(3) b. 𝑓(−2) c. 𝑓(−3456) d. 𝑓(𝑎 + ℎ) e. Sketch the graph

Solutions a. 𝑓(3) = 5

b. 𝑓(−2) = 5 c. 𝑓(−3456) = 5 d. 𝑓(𝑎 + ℎ) = 5

e. As you can see from the parts a,b,and c, no matter what 𝑥-coordinate you plot the 𝑦-coordinate is always 5 so its graph is a horizontal line as shown above.


ii) Linear Function: A function defined as (𝑥) = 𝑚𝑥 + 𝑏 . With 𝑚 ≠ 0, this is a first degree polynomial function. Recall that 𝑚 =slope of the line, 𝑏 = 𝑦-intercept. This is a polynomial function of degree one.

Example 1. If 𝑓(𝑥) = 2𝑥 − 3, then

a. Find 𝑓(−1) b. Find 𝑓(0) c. Find 𝑓(4) d. Find the inverse function e. Sketch the graph of 𝑦 = 𝑓(𝑥)

Solution a. 𝑓(−1) = 2(−1) − 3 = −2 −

3 = −5 b. 𝑓(0) = −3 c. 𝑓(4) = 2(4) − 3 = 8 − 3 =

5 d. Inverse function

𝑦 = 2𝑥 − 3 𝑥 = 2𝑦 − 3

𝑥 + 3 = 2𝑦 𝑥 + 3

2= 𝑦

Inverse function is

𝑓−1(𝑥) =𝑥+3

2

e. Graph of the function is to the right with slope of 2 and 𝑦-intercept of −3.

2. An Olympic size swimming pool holds 2,500,000 liters of water. If the pool currently holds 100,000 liters of water, and water is being pumped at 400,000 liters/hour into the tank.

a. Write a function that represents the amount of water in the tank after 𝑡 hours. b. Find the domain and range of this function. c. Sketch the graph of this function.

Solution: a. 𝐴(𝑡) = 100000 + 400000𝑡

Liters. The domain restriction 0 ≤ 𝑡 ≤ 6 for this application reflects the fact that the water inflow starts at 𝑡 = 0 and at 𝑡 = 6 the poo is filled with 2,500,000 liters of water.

b. Domain of 𝐴 = [0,6] and Range of 𝐴 = [100000,2500000]

c. Graph is to the right. The scale is each tick mark represents 1 million liters on the 𝑦-axis and 1 hour on the 𝑡-axis.


iii) Square Function: A function defined as (𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 . This is a general degree-two polynomial function (𝑎 ≠ 0) and is also called a quadratic function.

Example If 𝑓(𝑥) = 𝑥2, then

a. Find 𝑓 (2

3)

b. Find𝑓(−2) c. Find 𝑓(2) d. Find 𝑓(𝑎 + ℎ) e. Sketch the graph the function 𝑦 = 𝑥2 f. Is the function one-to-one?

Solution:

a. 𝑓 (2

3) = (

2

3)

2=

4

9

b. 𝑓(−2) = (−2)2 = 4 c. 𝑓(2) = (2)2 = 4 d. 𝑓(𝑎 + ℎ) = (𝑎 + ℎ)2 = 𝑎2 + 2𝑎ℎ + ℎ2 e. Plot a few points to sketch the graph of

the function. See to the left f. No. The function is not one-to-one as it

does not pass the horizontal line test.

𝑥 𝑦 = 𝑥2 −2 (−2)2 = 4 −1 (−1)2 = 1 0 (0)2 = 0 1 (1)2 = 1 2 (2)2 = 4

iv) Cubic Function: A function defined as 𝑓(𝑥) = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑

A special case of a polynomial function of degree three (𝑎 ≠ 0).

Example If 𝑓(𝑥) = 𝑥3, then

a. Find 𝑓 (2

3)

b. Find𝑓(−2) c. Find 𝑓(2) d. Sketch the graph the function 𝑦 = 𝑥3 e. Is the function one-to-one?

Solution:

a. 𝑓 (2

3) = (

2

3)

3=

8

27

b. 𝑓(−2) = (−2)3 = −8 c. 𝑓(2) = (2)3 = 8 d. Plot a few points to sketch the graph of

the function. See to the left e. Yes. The function is one-to-one as it does

pass the horizontal line test. f. Finding inverse function write 𝑦 = 𝑥3,

then we get 𝑥 = 𝑦3 solving for 𝑦 we get

that inverse function is 𝑓−1(𝑥) = √𝑥3

𝑥 𝑦 = 𝑥3 −2 (−2)3 = −8 −1 (−1)3 = −1 0 (0)3 = 0 1 (1)3 = 1 2 (2)3 = 8


So we now have a fairly complete understanding of the graphs of constant functions, linear functions,

quadratic functions, and the basic cubic function 𝑓(𝑥) = 𝑥3 . I.e., constant functions and linear

polynomial graphs are lines. In conic sections, we saw that quadratic functions of the type 𝑓(𝑥) =

𝑎𝑥2 + 𝑏𝑥 + 𝑐 have parabola graphs with vertex (−𝑏

2𝑎, 𝑓 (

−𝑏

2𝑎)) and open up when 𝑎 > 0 and down when

𝑎 < 0.

We will use this information and the information we learned about transformation of functions to

explore the behavior of more complicated polynomial function graphs.

Let us start with exploring functions of the type 𝑓(𝑥) = 𝑥𝑛, where 𝑛 is a whole number.

1 When 𝑛 = 0 we have a constant function 𝑓(𝑥) = 1 which is a horizontal straight line.

2 When 𝑛 = 1 we have 𝑓(𝑥) = 𝑥 which is also a straight line passing through (0,0)

3 When 𝑛 = 2 we have a quadratic 𝑓(𝑥) = 𝑥2 which is a parabola with vertex (0,0) facing up.


4 In general then when 𝑛 > 2 and an even number we know it will behave similar to the graph of 𝑓(𝑥) = 𝑥2 in its generic shape. Plotting a few points we notice that a) When −1 < 𝑥 < 1, 𝑥𝑛 < 𝑥2. For

example, 0.14 = 0.0001 < 0.12 =0.01.

b) When −1 > 𝑥, 𝑜𝑟 𝑥 > 1, 𝑥𝑛 > 𝑥2. For example, 104 = 10000 > 102 =100 or (−10)4 = 10000 >(−10)2 = 100.

See graphs of some of the functions of the type 𝑓(𝑥) = 𝑥𝑛 in comparison to each other on the right.

5 When 𝑛 = 3 we have a cubic function 𝑓(𝑥) = 𝑥3 which is passes through (0,0) and is symmetric with respect to origin.

6 In general then when 𝑛 > 3 and an odd number we know it will behave similar to the graph of 𝑓(𝑥) = 𝑥3 in its generic shape. Plotting a few points we notice that a) When 0 < 𝑥 < 1, 𝑥𝑛 < 𝑥3 . For

example, 0.15 = 0.00001 < 0.13 =0.001

b) When −1 < 𝑥 < 0, 𝑥𝑛 > 𝑥3. For

example, (−0.1)5 = −0.00001 >(−0.1)3 = −0.001

c) When < −1, 𝑥𝑛 < 𝑥3. For example,

(−10)5 = −100000 < 103 = 1000. d) When 𝑥 > 1, 𝑥𝑛 > 𝑥3. For example,

105 = 100000 > 103 = 1000

See graphs of some of the functions of the type 𝑓(𝑥) = 𝑥𝑛 in comparison to each other below.


All polynomial functions are continuous (~no breaks or holes in the graph) on their

domain (−∞, ∞).

Intermediate Value Theorem: For a polynomial function 𝑓(𝑥), if 𝑓(𝑎) < 𝑓(𝑏), with

𝑎 < 𝑏, then for every number 𝐶 for which 𝑓(𝑎) ≤ 𝐶 ≤ 𝑓(𝑏), there exists a value of 𝑥

with 𝑎 ≤ 𝑥 ≤ 𝑏 so that 𝑓(𝑥) = 𝐶. That means 𝑓 attains all 𝑦-values between 𝑓(𝑎), and

𝑓(𝑏) at least at one value 𝑥 in the interval [𝑎, 𝑏].

How can we use this fact to our benefit? The Intermediate Value Theorem graphically means that every

horizontal line between the 𝑦-values at 𝑥 = 𝑎 and 𝑥 = 𝑏 intersects the polynomial graph at least one

time.

In scenario at least once the function attains the

value of 𝐶.

In scenario the function attains the value of 𝐶

several times.

As you can see, even from the graphs of the basic polynomial functions if you travel on the graph from

left to right, sometimes the graph of the function does not rise or fall as in the case of the constant

function, and other times it rises or falls like riding a roller coaster.

The definitions below are for any function and not just polynomial functions.

A function 𝑓(𝑥) is constant on an interval (𝑎, 𝑏) iff for all 𝑥1 < 𝑥2 in the interval (𝑎, 𝑏),

𝑓(𝑥1) = 𝑓(𝑥2).

A function 𝑓(𝑥) is strictly increasing on an interval (𝑎, 𝑏) iff for all 𝑥1 < 𝑥2 in the interval

(𝑎, 𝑏), 𝑓(𝑥1) < 𝑓(𝑥2). Graphically, 𝑦-coordinates get higher as 𝑥 moves to the right.

A function 𝑓(𝑥) is strictly decreasing on an interval (𝑎, 𝑏) iff for all 𝑥1 < 𝑥2 in the interval

(𝑎, 𝑏), 𝑓(𝑥1) > 𝑓(𝑥2). Graphically, 𝑦-coordinates go down as 𝑥 moves to the right.


Example

1. Determine all the intervals where the function below is increasing, decreasing and or constant.

The function is increasing on the interval (−∞, −1) ∪ (1,1.7)

The function is decreasing on the interval (−1,1)

The function is constant on the interval (1.7,7)

Practice problems:

Try to come up with rough sketches of the graphs the following functions. Try using the knowledge you

have accumulated so far (i.e., transformation of functions, arithmetic of functions). Do not peek after

this page to see what the answers are. Cultivating your intuition will aid in your deeper understanding of

the material. You can always resort to plotting points that might be of interest to you.

1. 𝑓(𝑥) = (𝑥 − 1)2(𝑥 + 1)3

= 𝑥5 + 𝑥4 − 2𝑥3 − 2𝑥2 + 𝑥 + 1

2. 𝑓(𝑥) = (𝑥 − 1)2(𝑥 + 1)3 + 2

= 𝑥5 + 𝑥4 − 2𝑥3 − 2𝑥2 + 𝑥 + 3

3. 𝑓(𝑥) =1

10(𝑥 − 4)(𝑥 + 4)(𝑥 − 2)3(𝑥 + 2)2 4. 𝑓(𝑥) = −

1

10(𝑥 − 4)(𝑥 + 4)(𝑥 − 2)3(𝑥 + 2)2


Solutions to practice problems:

1. 𝑓(𝑥) = (𝑥 − 1)2(𝑥 + 1)3 = 𝑥5 + 𝑥4 − 2𝑥3 − 2𝑥2 + 𝑥 + 1 We know that the function 𝑦 = (𝑥 − 1)2 is a parabola shifted one to the right, and 𝑦 = (𝑥 + 1)3 is a cubic function shifted one to the left. These individual graphs are shown below. Plotting them on the same coordinate axis will give us insight into the graph of their product. Note 𝑦-intercept is 𝑓(0) =(0 − 1)2(0 + 1)3 = 1

To plot the graph of 𝑓(𝑥) we need to see how the product of the two functions we know behaves.

𝑥 < −1 −1 < 𝑥 < 1 𝑥 > 1 Test Point −3 0 1.5

𝑓(𝑥) = (𝑥 − 1)2(𝑥 + 1)3

(−3 − 1)2(−3 + 1)3 = −128 < 0

(0 − 1)2(0 + 1)3 = 1 > 0

(1.5 − 1)2(1.5 + 1)3 3.90625 > 0

Graph of 𝑓(𝑥) is below the 𝑥-axis

Graph of 𝑓(𝑥) is above the 𝑥-axis

Graph of 𝑓(𝑥) is above the 𝑥-axis

𝑓(1) = (1 − 1)2(1 + 1)3 = 0 and 𝑓(−1) = (−1 − 1)2(−1 + 1)3 = 0 that means 𝑥 = 1, −1 are 𝑥-intercepts. So just knowing the sign of the function we can sketch a rough sketch. The domain of the function 𝑓(𝑥) is all real numbers. We can see the two functions = (𝑥 − 1)2 , and 𝑦 = (𝑥 + 1)3 have no breaks or holes in them and so that is what we would expect in the product function. You can plot several points to see what happens to convince yourself but each of them will preserve their shapes near to where each crosses the 𝑥-axis. Our rough sketch would look as shown below. We don’t know exactly know how high it will rise before turning back down when −1 < 𝑥 < 1 but we know it goes through the 𝑦-axis at (0, 1).

𝒚 = (𝒙 − 𝟏)𝟐

𝒚 = (𝒙 + 𝟏)𝟑


From the table below we can see that when 𝑥 approaches ∞ or −∞, the value of just the highest degree

term is a good approximation to the value of the full polynomial output. The lower degree terms are

much smaller at these 𝑥-values because they have fewer factors of the large number 𝑥 in them. This

means that terms other than the highest degree term will not contribute much make a dent in the

magnitude of the overall function value. In other words the highest degree terms will take over the

behavior of the function when 𝑥 → ±∞. This is called the end behavior of the function. See table below.

𝑥 𝑦 = 𝑥5 𝑦 = 𝑓(𝑥) = 𝑥5 + 𝑥4 − 2𝑥3 − 2𝑥2 + 𝑥 + 1

1000 1,000,000 ,000,000,000 1,000,997 ,998,001,001

10000 100 ,000,000,000 ,000,000,000 100, 009,997,999 ,800,010,001

100000 10000000 000000000 000000000 10000099 997999980 000100001

−1000 −1,000,000 ,000,000,000 −998,998 ,002,000,999

−10000 −100 ,000,000,000 ,000,000,000 −99 ,989,998,000, 200,009,999

−100000 −10,000,000,000,000,000,000,000,000 −9,999,899 ,998,000,020 ,000,099,999

The end behavior of 𝑓(𝑥) is that when 𝑥 → ±∞, the graph of 𝑦 = 𝑓(𝑥) is very much like 𝑦 = 𝑥5.

This example illustrates the importance

of using an appropriate viewing window

when using technology to create the

graph of a function. The details in the

top-left plot are totally lost in the lower

graph. We need to know where to look

to see the interesting details. In this

course, we’ll develop tools to see how

polynomials behave near their 𝑥-

intercepts. Other details, such as where

the local maximum is near 𝑥 = 0.2

require tools from calculus.


When 𝑓(𝑥) is a polynomial function with (𝑥 − 𝑎)𝑛 as a factor, we can lump the other factors together as

𝑔(𝑥) so 𝑓(𝑥) = 𝑔(𝑥)(𝑥 − 𝑎)𝑛 . We can do it so that 𝑔(𝑎) ≠ 0. Note that when 𝑥 = 𝑎 we get

𝑓(𝑎) = 𝑔(𝑎)(𝑎 − 𝑎)𝑛 = 0, and this is a reason why then 𝑥 = 𝑎 is said to be a zero of 𝑓 of multiplicity 𝑛.

The multiplicity refers to how many times 𝑥 = 𝑎 occurs as a zero in 𝑓(𝑥) (since(𝑥 − 𝑎)𝑛 = (𝑥 − 𝑎)(𝑥 −

𝑎) … (𝑥 − 𝑎) (a product of 𝑥 − 𝑎, 𝑛 times)).

Lets look at the behavior of the function 𝑓(𝑥) = 𝑔(𝑥)(𝑥 − 𝑎)𝑛 for 𝑥 = 𝑏, where 𝑏 very close to 𝑎.

Then at this value 𝑔(𝑏) will be near to 𝑔(𝑎) and we can think of 𝑓(𝑏) ≈ 𝑔(𝑎)(𝑏 − 𝑎)𝑛. In other words

near each zero 𝑥 = 𝑎 the graph will behave like a stretched version of 𝑦 = (𝑥 − 𝑎)𝑛 where the stretch

factor is equal to 𝑔(𝑎) , i.e., the number when 𝑥 = 𝑎 is plugged into all the 𝑥 locations except (𝑥 − 𝑎)𝑛.

Thus multiplicity 𝑛 = 1 means the graph crosses at 𝑥 = 𝑎 like a line with slope 𝑔(𝑎). With multiplicity

𝑛 = 2, the graph behaves like a parabola with vertex at 𝑥 = 𝑎 and stretch factor of 𝑔(𝑎). For

multiplicity 𝑛 = 3 the graph crossses at 𝑥 = 𝑎, but in a manner that is flat to the 𝑥-axis the same as how

𝑦 = 𝑥3 looks likes near 𝑥 = 0.

2. 𝑓(𝑥) = (𝑥 − 1)2(𝑥 + 1)3 + 2 = 𝑥5 + 𝑥4 − 2𝑥3 − 2𝑥2 + 𝑥 + 3 We know from before that the graph here will be similar to the one before just moved up 2 units.


Summary

End behavior of a polynomial function is governed by the its degree and the leading

coefficient.

When the leading coefficient is positive, then as 𝑥 → +∞ the 𝑦 = 𝑓(𝑥) → +∞.

If the degree is even, then 𝑦 = 𝑓(𝑥) → +∞ when 𝑥 → −∞ .

If the degree is odd, then 𝑦 = 𝑓(𝑥) → −∞ when 𝑥 → −∞.

When the leading coefficient is negative, then as 𝑥 → +∞ the 𝑦 → −∞.

If the degree is even, then 𝑦 = 𝑓(𝑥) → −∞ when 𝑥 → −∞.

If the degree is odd, then 𝑦 = 𝑓(𝑥) → +∞ when 𝑥 → −∞.

A pictorial summary of some of the points made above is below.

Degree of the Polynomial

Leading Coefficient

Positive

Negative

Even

Odd


3. 𝑓(𝑥) =1

10(𝑥 − 4)(𝑥 + 4)(𝑥 − 2)3(𝑥 + 2)2

Step 1: 𝑥-intercepts Setting 𝑓(𝑥) = 0 we get we get 𝑥 = 4, −4,2, −2 will be the 𝑥-intercepts.

Near 𝑥 = −2, 𝑦 ≈1

10(−6)(2)(−4)3(𝑥 + 2)2 = 76.8(𝑥 + 2)2 So the graph is like a steep parabola

opening up near 𝑥 = −2.

Likewise near 𝑥 = −4, 𝑦 ≈1

10(−8)(𝑥 + 4)(−6)3(−2)2 = 691.2(𝑥 + 4); a line with slope ~700.

And near 𝑥 = 2, 𝑦 ≈ 96(𝑥 − 2)3, and near 𝑥 = 4, 𝑦 ≈ 230.4(𝑥 − 4)

Step 2: 𝑦-intercept Setting 𝑥 = 0 in the function we get 𝑓 (0) =1

10(0 − 4)(0 + 4)(0 − 2)3(0 +

2)2 =256

5= 51.2 will be the 𝑦-intercept.

Step 3: End behavior

Note that if we were to expand the polynomial the degree of the polynomial would be 7 since we

have two linear terms, a cubic term and a square term. So our polynomial will behave like 1

10𝑥7 on

the ends (left hand side falls, right hand side rises).

Step 4: Graph the function using steps 1-3.

We use our knowledge of the end behavior and near 𝑥-intercepts to finish the graph making use of our

observation of where the function is linear, quadratic, or cubic. See if you can come with the graph

before looking on the next page using the information summarized from steps 1-3 below.

Putting all pieces together we have our rough sketch below.


Another way to sketch is to investigate signs of the function at different 𝑥 values.

We know that an odd number of negative numbers multiply to give us a negative number, and an

even number of negative numbers multiply to give us a positive number. An odd number of positive

numbers and an even number of positive numbers multiply to give a positive number.

We also know that a linear polynomial 𝑎𝑥 + 𝑏, we have 𝑎𝑥 + 𝑏 > 0 for all real numbers 𝑥 > −𝑏

𝑎,

and 𝑎𝑥 + 𝑏 < 0 for all real numbers 𝑥 < −𝑏

𝑎. Using these two facts we can determine whether the

function 𝑓(𝑥) > 0 or 𝑓(𝑥) < 0 for certain values of 𝑥 by doing a sign chart as shown on the next

page.


See details below.

𝒙 < −𝟒 −𝟒 < 𝒙 < −𝟐 −𝟐 < 𝒙 < 𝟐 𝟐 < 𝒙 < 𝟒 𝒙 > 𝟒

Test Pts −5 −3 0 3 5

𝒙 + 𝟒 − + + + +

(𝒙 + 𝟐)𝟐 + + + + +

(𝒙 − 𝟐)𝟑 − − − + +

𝒙 − 𝟒 − − − − +

𝒇(𝒙) − + + − +

𝑓(𝑥) =1

10(𝑥 − 4)(𝑥 + 4)(𝑥 − 2)3(𝑥 + 2)2

Looking at the sign of the function as shown to the left, and our knowledge of the 7th degree polynomial end behavior we can now draw a rough sketch of the graph. We can see that at 𝑥 = −2 we will still retain the parabola shape, and at 𝑥 = 2 we would retain the cubic polynomial shape. See below for the rough sketch of the graph. We still do not know how far the graph rises or falls at the local maximum and minimum points.


To really make sense of the chart and graph together, see below. The graph of the function was drawn

using a graphing utility since without it we cannot know how high or low the function rises or falls. As

you can see, our rough sketch is pretty accurate for the main features of the graph.


4. 𝑓(𝑥) = −1

10(𝑥 − 4)(𝑥 + 4)(𝑥 − 2)3(𝑥 + 2)2

We know that putting a negative in front of the graph will just reflect the graph in previous example

about the 𝑥-axis. So the graph will look like

The peaks and valleys we see have names. They are called local extrema or local maximum and

minimum values of the function output.

Local Maximum: In an interval (𝑎, 𝑏) a point (𝑐, 𝑓(𝑐)) where 𝑎 < 𝑐 < 𝑏 is called a local

maximum point if 𝑓(𝑐) ≥ 𝑓(𝑥) for all 𝑥 in the interval (𝑎, 𝑏). We’d say 𝑓(𝑐) is a local max.

Local Minimum: In an interval (𝑎, 𝑏) a point (𝑐, 𝑓(𝑐)) where 𝑎 < 𝑐 < 𝑏 is called a local

minimum if 𝑓(𝑐) ≤ 𝑓(𝑥) for all 𝑥 in the interval (𝑎, 𝑏). We’d say 𝑓(𝑐) is a local min.

Now attempt to graph the following functions.

5. 𝑔(𝑥) = (2𝑥 − 9)(𝑥 + 3)2(𝑥 − 1)2

𝑥-intercepts are ______________

𝑦-intercept is ____________

End Behavior like _______________

Number of local extremum _______

Graph of the polynomial

6. ℎ(𝑥) = 𝑥(𝑥 + 5)(𝑥 + 2)(𝑥 − 1)(𝑥 − 4)







Solution:

5. 𝑔(𝑥) = (2𝑥 − 9)(𝑥 + 3)2(𝑥 − 1)2

𝑥-intercepts are 𝑥 =9

2= 4.5, −3, and 1. The behavior is like a parabola at 𝑥 = 1 𝑎𝑛𝑑 𝑥 = −3 and the

graph crosses the 𝑥-axis at 𝑥 = 4.5 like a line. The 𝑦-intercept is 𝑦 = 𝑔(0) = (−9)(3)2(−1)2 = −81

End Behavior like 𝑦 = 2𝑥(𝑥2)(𝑥2) = 2𝑥5 (left hand side falls, right hand side rises).

The number of local maximums/minimums total 4 (as seen from the graph).

The computer generated graph shows how low the function dips between 𝑥 = 1, 𝑥 = 4. To locate the

exact location of these local max/min points requires tools from calculus. The knowledge of the

behavior at the intercepts, however, is enough to provide the basic shape and alert us to there being

local max and minimum points. We also recommend that you plot a few points that that lie in-between

the intercepts as a check that your intercept and end-behavior analysis is correct.


6. ℎ(𝑥) = 𝑥(𝑥 + 5)(𝑥 + 2)(𝑥 − 1)(𝑥 − 4)

𝑥-intercepts are 𝑥 = 0, −5, −2, 1 and 4 . 𝑦-intercept is ℎ(0) = 0(5)(−2)(1)(4) = 0.

End Behavior is 𝑦 ≈ 𝑥(𝑥)(𝑥)(𝑥)(𝑥) = 𝑥5 (left hand falls, right hand rises)

Number of local extrema is 4 (as seen from the graph).

Graph of the polynomial:


Observation:

You can see from the examples we have done so far that a factor of the type (𝑎𝑥 + 𝑏)𝑛

where 𝑎 ≠ 0, 𝑛 is a counting number has the graph either cross or touch (without crossing) the

𝑥-axis at 𝑥 = −𝑏

𝑎 .

When 𝑛 is an even number the graph of the polynomial function will touch the 𝑥-axis at

𝑥 = −𝑏

𝑎.

When 𝑛 is a odd number the graph of the polynomial function will cross the 𝑥-axis at

𝑥 = −𝑏

𝑎.

Maximum number of local extrema possible is one less than the degree of the polynomial

function.

As you can see the factors of a polynomial function provide a lot of information about its graph. This

makes the factored form very useful. We can also use this in reverse when a graph is given to

determine the form of an equation of a polynomial function..

7. For each graph below, find the factored form of a polynomial that would produce the graph of that

shape. You can label the axis hash marks as you please but the basic shape should match. You can

then use a graphing utility like a graphing calculator, or websites like www.wolframalpha.com or

https://www.desmos.com/calculator to graph your polynomial to check your answers.

a.

b.

Zoomed

In closer

http://www.wolframalpha.com/

https://www.desmos.com/calculator


Solutions:

a.

Since the two ends are facing up, the polynomial has even degree and a positive leading coefficient. The x-intercepts can be thought of as 𝑥 = −2, 0, 2, 4 and the powers of the factors that give you those intercepts would be odd, even, odd, even respectively. Note that the graph looks similar to 𝑥3 shape at 2 and -2, thus 𝑛 should be at least 3 for those factors. Thus, the simplest form for 𝑓 would be:

𝑓(𝑥) = 𝑥2(𝑥 + 2)3(𝑥 − 2)3(𝑥 − 4)2 Graph is below and this seems a very good match.

If we had used a 4th order factor at 𝑥 = 0, we get the graph below which has a similar shape but poorer fit.

𝑓(𝑥) =1

100𝑥4(𝑥 + 2)3(𝑥 − 2)3(𝑥 − 4)2

b.

We can set the right-most 𝑥-intercept at 𝑥 = 3 and this puts the next one at

𝑥 =1

2 (factor (2𝑥 − 1)) with even

multiplicity. Also 𝑥 = 0 is another intercept with odd multiplicity. The left-most intercept seems to be at

𝑥 ≈ −5

3 (factor (𝑥 +

5

3) 𝑜𝑟 𝑢𝑠𝑒 (3𝑥 +

5) with multiplicity of 𝑛 = 1. Also, the leading coefficient is negative. The polynomial below is the simplest to do this.

𝑔(𝑥) = −𝑥3(2𝑥 − 1)2(3𝑥 + 5)(𝑥 − 3)2 The graph of 𝑔 below confirms it works.

Zoomed

In closer


8. Match the functions on the right with their graphs on the left. Explain your answers.

I. 𝑓(𝑥) = (𝑥 − 1)2(𝑥 + 1)3 II. 𝑔(𝑥) = (𝑥 − 1)2(𝑥 + 1)2

III. ℎ(𝑥) = −10𝑥2(𝑥 − 1)2(𝑥 + 1)2 IV. 𝑅(𝑥) = 10𝑥3(𝑥 − 1)2(𝑥 + 1)2

A.

B.

C.

D.


Solution:

I. 𝑓(𝑥) = (𝑥 − 1)2(𝑥 + 1)3 This graph matches B. Since the end

behavior is like 𝑥5, and at 𝑥 = 1 we have a parabola shape, and a cubic shape at 𝑥 = −1.

II. 𝑔(𝑥) = (𝑥 − 1)2(𝑥 + 1)2

End behavior is like 𝑥4, and parabola shapes at 𝑥 = 1, −1, so must match graph C.

III. ℎ(𝑥) = −10𝑥2(𝑥 − 1)2(𝑥 + 1)2

Both ends down since leading coefficient is −10 and degree is 6. Parabola shapes at 𝑥 = 0, −1, and 1 so must match graph A.

IV. 𝑅(𝑥) = 10𝑥3(𝑥 − 1)2(𝑥 + 1)2

There is only one graph left D. It touches at 𝑥 = 1 and 𝑥 = −1 and behaves like 𝑥3 at 𝑥 = 0.

A.

B.

C.

D.


Playing

In all the examples we have done so far the polynomial function given to you was written in the factored

form which makes it easier for analyzing its graph. So one might wonder can all polynomial functions be

written in the factored form.

To explore this question please consider the polynomials below and answer the following questions.

a) Sketch the graph of the polynomials functions below.

b) Can any of them be written in factored form? If yes, please factor them, and then show what

information is more visible when the polynomial is written in the factored form.

c) Can any of them be written in a way that is a transformation of a polynomial that can be written in

factored form? If yes, please explain.

1. 𝑥2 + 2𝑥 − 2 2. 𝑥2 − 4𝑥 − 12 3. 𝑥4 + 16

4. 𝑥4 − 16 5. 𝑥3 − 2𝑥2 − 𝑥 + 2 6. 3𝑥3 − 2𝑥2 + 4

We will answer some of the questions posed here about whether or not all polynomials can be factored

or not in Chapter 3.




1. Degree of a polynomial

2. Leading coefficient of a polynomial

3. Local extrema

4. 𝒙-intercept

5. 𝒚-intercept

6. Difference between solving an equation in one variable, and sketching the graph of a function in one variable

7. Difference between solving an inequality in one variable, and sketching the graph of a function in one variable



Exercises 2.3

1. Answer the questions below

a. How does the degree of the polynomial affect the end behavior?

b. What does the leading coefficient of a polynomial control in its graph ?

c. What is the maximum number of local extrema you can expect in a polynomial of degree 𝑛?

d. How does the exponent 𝑛 in factors of the type (𝑎𝑥 + 𝑏)𝑛 influence the shape of the graph near

the 𝑥-intercept associated with this factor?

e. How do you find all the 𝑥-intercepts of a polynomial function?

f. How do you find all the 𝑦-intercepts of a polynomial function?

g. How would you solve a an equation of the type

𝑎(𝑥 − 𝑎1)𝑛1(𝑥 − 𝑎2)𝑛2(𝑥 − 𝑎3)𝑛3 … . (𝑥 − 𝑎𝑘)𝑛𝑘 = 0?

h. What is the significance of the solutions to the equation in question 7 to the graph of the

polynomial function 𝑓(𝑥) = 𝑎(𝑥 − 𝑎1)𝑛1(𝑥 − 𝑎2)𝑛2(𝑥 − 𝑎3)𝑛3 … . (𝑥 − 𝑎𝑘)𝑛𝑘?

i. How would you solve an inequality of the type

𝑎(𝑥 − 𝑎1)𝑛1(𝑥 − 𝑎2)𝑛2(𝑥 − 𝑎3)𝑛3 … . (𝑥 − 𝑎𝑘)𝑛𝑘 > 0

Or

𝑎(𝑥 − 𝑎1)𝑛1(𝑥 − 𝑎2)𝑛2(𝑥 − 𝑎3)𝑛3 … . (𝑥 − 𝑎𝑘)𝑛𝑘 < 0

j. How are the solutions to the inequalities in question 9 related to the graph of the polynomial

function 𝑓(𝑥) = 𝑎(𝑥 − 𝑎1)𝑛1(𝑥 − 𝑎2)𝑛2(𝑥 − 𝑎3)𝑛3 … . (𝑥 − 𝑎𝑘)𝑛𝑘?

k. What is a local extrema of a polynomial function?

l. Does every polynomial function have local extreme points? Explain your answer.

2. Sketch the graph of the functions below label all 𝑥-intercepts, 𝑦-intercepts, label all local extrema

points, show end behavior.

a. 𝑔(𝑥) = 𝑥2(𝑥 − 1)(𝑥 + 2)3






b. ℎ(𝑥) = 𝑥(𝑥 + 1)(𝑥 + 4)(𝑥 − 2)(𝑥 − 5)






c. 𝑔(𝑥) = (𝑥 + 1)3(𝑥 − 1)(𝑥 − 2) d. ℎ(𝑥) = (𝑥 − 1)(𝑥 + 1)2(𝑥 + 2)2(𝑥 − 3)












3. Determine the interval(s) on which the function is (strictly) increasing, or decreasing, or constant.

Write your answer in interval notation.

A.

Increasing: ___________ Decreasing: ___________ Constant: ___________

B.



C.


D.


4. Use the graph of the function 𝑓 below to find (If there is more than one answer, separate them

with commas)

I. All values of 𝑥 at which 𝑓 has local minimum

II. All values of 𝑥 at which 𝑓 has local maximums

III. All local minimum values of 𝑓

IV. All local maximum values of 𝑓.

a.

b.


c.

d.

5. Find all the 𝑥-intercepts and 𝑦-intercepts of the functions below. If there is more than one

answer, separate them with commas.

A. 𝑥-intercepts _________ 𝑦-intercepts _________

B. 𝑥-intercepts _________ 𝑦-intercepts _________

C. 𝑓(𝑥) = (𝑥 + 3)(𝑥 − 1)2(𝑥 + 5) 𝑥-intercepts _________ 𝑦-intercepts _________


D. (factor by grouping) 𝑓(𝑥) = 2𝑥3 + 2𝑥2 − 18𝑥 − 18 𝑥-intercepts _________ 𝑦-intercepts _________

E. 𝑓(𝑥) = 𝑥4 − 4𝑥2 − 𝑥3 + 4𝑥 𝑥-intercepts _________ 𝑦-intercepts _________

6. Mark the end behavior of the graph of each polynomial function below.

A. 𝑓(𝑥) = (𝑥 + 3)(𝑥 − 1)2(𝑥 + 5)

Left Right

Falls

Rises

Falls Rises

B. 𝑓(𝑥) = −2𝑥3 + 2𝑥2 − 18𝑥 − 18

Left Right

Falls

Rises

Falls Rises

E. 𝑓(𝑥) = 5𝑥11 − 4𝑥2 − 𝑥3 − 2𝑥

Left Right

Falls

Rises

Falls Rises

F. 𝑓(𝑥) = −4𝑥32 − 5𝑥20 + 4𝑥 − 1

Left Right

Falls

Rises

Falls Rises


7. Match the graphs below with the functions listed here.

I. 𝑓(𝑥) = −3(𝑥 + 1)2(𝑥 + 3)2

II. ℎ(𝑥) = 𝑥2(𝑥 − 2)3(𝑥 + 1)

III. 𝑔(𝑥) =1

2(𝑥3 − 𝑥2 − 6𝑥)

IV. 𝑟(𝑥) = (𝑥 − 1)(𝑥 + 1)(𝑥 − 2)

A.

B.

C.

D.


8. For the graphs below determine the following

I. What is the sign of the leading coefficient of the polynomial function (positive, negative

or not enough information?

II. Which of the following is a possibility of the degree of the function? Choose all that

apply 4, 5, 6, 7, 8, 9.

A.

B.

C.

9. Find a possible equation that represents the graphs below.

a.

b.

c.

d.


2.4 Graphing Quotients of Functions

Rational Functions We introduced the rational functions briefly in chapter 1.

Quotient of Function: A quotient of function is defined as the ratio of two functions denoted as


𝑄(𝑥), forl functions 𝑃(𝑥), and 𝑄(𝑥) with domain all real numbers 𝑥 for which 𝑄(𝑥) ≠ 0.

Rational Function: A rational function is defined as the ratio of two polynomial functions denoted as


𝑄(𝑥), for polynomial functions 𝑃(𝑥), and 𝑄(𝑥) with domain all real numbers 𝑥 for which

𝑄(𝑥) ≠ 0.

Linear Asymptotes of Rational Functions

Rational function graphs can have asymptotic behavior as 𝑥 → ±∞ and also as 𝑥 → 𝑎 for where the

denominator function is zero at 𝑥 = 𝑎. The curves that the graph of 𝑅(𝑥) gets close to as 𝑥 or 𝑦

approach ±∞ are called asymptotes of the function. We’ll look primarily at asymptotes that are lines

which can be vertical, horizontal, or oblique (slanted).

Vertical Asymptote: A vertical line of the form 𝑥 = 𝑎 is called a vertical asymptote to a function

𝑦 = 𝑓(𝑥) if and only if the function 𝑓(𝑥) approaches either ∞ or −∞ as 𝑥 approaches 𝑎 either

from the left or right. These are always caused by the denominator polynomial 𝑄(𝑥) being zero

at 𝑥 = 𝑎.

Horizontal or Oblique Asymptotes: A line 𝑦 = 𝑚𝑥 + 𝑏 is called a horizontal (when 𝑚 = 0) or

oblique (𝑚 ≠ 0) asymptote to a function 𝑦 = 𝑓(𝑥) if and only if the distance between the

function 𝑓(𝑥) and 𝑦 = 𝑚𝑥 + 𝑏 approaches zero as 𝑥 approaches either ∞ or −∞.

Example: Let us review the graphing of the simplest example of a rational function 𝑅(𝑥) =1

𝑥 .

a. Sketch the graph of 𝑅(𝑥).

b. What is domain of this function?

c. What linear asymptotes does this function have?

Play with it to see if you can recall what we did before looking on the next page.


We start by plotting a bunch of points and this will help us understand where and why the function has

asymptotes. As we can see the denominator cannot be zero, so the graph won’t have a point on the 𝑦-

axis. Thus the graph will be in two pieces, one on either side of the 𝑦-axis.

𝑥 𝑦 =

1

𝑥

𝑥 𝑦 =

1

𝑥

−0.1 1

−0.1= −10

−10 1

−10= −0.1

−0.01 1

−0.01= −100

−100 1

−100= −0.01

−0.001 1

−0.001= −1000

−1000 1

−1000= −0. .001

0.1 1

0.1= 10

10 1

10= 0.1

0.01 1

0.01= 100

100 1

100= 0.01

0.001 1

0.001= 1000

1000 1

1000= 0. .001

Domain: All nonzero real numbers

The left table above shows how, as 𝑥 gets close to where the denominator is zero, the 𝑦-values blow up. That means 𝑦 approaches either +∞ when 𝑥 is small but positive, or 𝑦 → −∞ when 𝑥 is small but

negative. The 𝑦-axis is a vertical asymptote of the function 𝑓(𝑥) =1

𝑥. (There will be a vertical

asymptote for any rational function at every 𝑥-value where the denominator is zero, but the numerator polynomial ≠ 0.) The right table above shows how as 𝑥 goes to positive infinity, then the 𝑦-values are small but positive and as to 𝑥 → −∞ then 𝑦 is small but negative. The 𝑥-axis a horizontal asymptote of the function

𝑓(𝑥) =1

𝑥. Not every rational function has a horizontal asymptote. Also, some rational functions have a

horizontal asymptote not at the 𝑥-axis, but at some other non-zero 𝑦-value.


Playing

We can generalize our observations from plotting the function 𝑅(𝑥) =1

𝑥. Note that for (𝑥) =

1

𝑥𝑛 ,

𝑦 =1

𝑥𝑛 → 0 when 𝑥 → ±∞ for any positive integer 𝑛. If you continue your study of functions in a

calculus course, limit notation makes these statements much more succinct as indicated below.

lim𝑥→∞

1

𝑥𝑛= 0 𝑎𝑛𝑑 lim

𝑥→ −∞

1

𝑥𝑛= 0

Also note that

a) When 𝑛 is even, 𝑓(𝑥) =1

𝑥𝑛 → ∞ when 𝑥 → 0 from both sides of 0.

b) When 𝑛 is odd, 𝑓(𝑥) =1

𝑥𝑛 → ∞ when 𝑥 → 0 from the right and 𝑓(𝑥) =1

𝑥𝑛 → −∞ when 𝑥 → 0 from

the left side .

Observation: 𝑓(𝑥) =1

𝑥𝑛, where 𝑛 is a counting number has a vertical asymptote is at 𝑥 = 0 and

horizontal asymptote at 𝑦 = 0.

Using our knowledge from transformation of functions we can see that for a counting number 𝑛, the

following statements can be made-

1. For (𝑥) =1

(𝑥−𝑎)𝑛 , the vertical asymptote is at 𝑥 = 𝑎, and the horizontal asymptote is at 𝑦 = 0.

2. For (𝑥) =1

𝑥𝑛 + 𝑏 , the vertical asymptote is at 𝑥 = 0, and the horizontal asymptote is at 𝑦 = 𝑏.

3. For (𝑥) =1

(𝑥−𝑎)𝑛 + 𝑏 , the vertical asymptote is at 𝑥 = 𝑎, and the horizontal asymptote is at 𝑦 = 𝑏.

4. For 𝑔(𝑥) =1

(𝑥−𝑎)𝑛 + 𝑚𝑥 + 𝑏, the vertical asymptote is at 𝑥 = 𝑎, and the oblique asymptote is at

𝑦 = 𝑚𝑥 + 𝑏. The reason the oblique asymptote is 𝑦 = 𝑚𝑥 + 𝑏 is the fact that 1

(𝑥−𝑎)𝑛 → 0 when

𝑥 → ±∞ making the vertical distance between 𝑔(𝑥) =1

(𝑥−𝑎)𝑛 + 𝑚𝑥 + 𝑏 and 𝑦 = 𝑚𝑥 + 𝑏 go to zero

when 𝑥 → ±∞.

5. For 𝑔(𝑥) =1

(𝑥−𝑎)𝑛 + 𝑄(𝑥), where 𝑄(𝑥) is a polynomial function, the vertical asymptote is at 𝑥 = 𝑎,

and 𝑦 = 𝑄(𝑥) is a polynomial function asymptote when 𝑥 → ∞ or 𝑥 → −∞. This means that the

graph of 𝑦 = 𝑔(𝑥) gets very close to the polynomial graph of 𝑦 = 𝑄(𝑥) as 𝑥 → ±∞.


Let’s take a look at all the scenarios of what a rational function graph may look like when you only have

linear asymptotes. We want you to explore all possibilities of what the graph of a rational function may

do if you knew the asymptotes.

Sketch the possibilities below for how a rational function could fit the three different asymptote

scenarios shown below.

a) 𝑦 = 𝑏 is a horizontal asymptote, what are our choices?

b) 𝑥 = 𝑎 is a vertical asymptote, what are our choices?

c) 𝑦 = 𝑚𝑥 + 𝑏 is an oblique asymptote, what are our choices?

Do you think you exhausted all possibilities?


a) For the horizontal asymptote 𝑦 = 𝑏 the graph of the function eventually must approach the line either from the above or the below. Those are the only choices as shown below.

b) For the vertical asymptote 𝑥 = 𝑎 the graph of the function must approach this line from the left or right and shoots up or down. Those are the only choices as shown below.

c) For the oblique asymptote 𝑦 = 𝑚𝑥 + 𝑏 the graph must get closer and closer from above or below as Those are the only choices as shown below as 𝑥 → ±∞.

So now lets us look at how we can put these observations together to imagine the different possible

graphs we can have with multiple given linear asymptotes.


Below are some possible graphs we can imagine if we had a vertical asymptote of 𝑥 = 𝑎, and a horizontal asymptote of 𝑦 = 𝑏.

a)

b)

c)

d)

e)

f)

As you can see there are countless possibilities. The last two may seem not possible but there can be

finite number of intersections of the graph of the rational function and the horizontal asymptotes. In

other words, 𝑦 = 𝑏 is a horizontal asymptote as long as the graph eventually stays closer and closer to

the line 𝑦 = 𝑏. You can also see that the left side of the graph in the last two examples gets to the

vertical asymptote a little slower than the right side. So it is actually a lot of fun to graph these. In

graphing polynomial functions we find the asymptotes, the intercepts, the points where the graph

intersects the horizontal asymptotes, and then make a sign chart to see what some of the values of the

function are in the domain, we will get a pretty good rough sketch.


Below are some possible graphs where 𝑥 = 𝑎 is the vertical asymptote, and 𝑦 = 𝑚𝑥 + 𝑏 is an oblique asymptote.

a)

b)

c)

d)

e)

f)

Again, the number of possible shapes is large. The complexity of the numerator and denominator functions lead to more asymptotes and potentially more local max/min points. The asymptotes are very important and provide a framework for the graph and suggest what 𝑥-values we would want to include in a small 𝑡-table of points to plot.


Reciprocal of a Function

Playing

How should one even begin to understand the graphs of rational functions? Mathematician’s start with

known facts and work their way up. We know that a rational function is a ratio of two polynomials. We

have already seen (in the previous section) how to get a rough sketch of a polynomial function that is in

the factored form.

We now know how a graph of the function 𝑦 =1

𝑥𝑛 , 𝑛 ≥ 1 looks like. So let’s investigate what happens to

the graph of a polynomial function when we invert it.


Practice Examples

1. Compare the graphs 𝑓(𝑥) = (𝑥 − 1)(𝑥 + 2) and 𝑦 =1

𝑓(𝑥)=

1

(𝑥−1)(𝑥+2)

The graph of 𝑦 = 𝑓(𝑥) = (𝑥 − 1)(𝑥 + 2) is a parabola with 𝑥-intercepts at 𝑥 = 1, −2. To graph 𝑦 =1

𝑓(𝑥)

we can see right away the domain will change since the denominator cannot be zero. The domain of the

function 𝑦 =1

(𝑥−1)(𝑥+2) is in three pieces: (−∞, −2) ∪ (−2,1) ∪ (1, ∞). The sign of 𝑦 = 𝑓(𝑥) and

𝑦 =1

𝑓(𝑥) is the same since the numerator is always positive. That means

1

𝑓(𝑥)> 0 for all 𝑥 < −2 and

𝑥 > 1. At values of 𝑥 closer to 1 but above 1 the graph will approach +∞, and the same behavior will

be noticed when values of 𝑥 are closer to −2 but 𝑥 < −2. Also 1

𝑓(𝑥)< 0 for all −2 < 𝑥 < 1. So the

graph will approach −∞ when values of 𝑥 are closer to 𝑥 = 1 and −2 but −2 < 𝑥 < 1.

We also know that the end behavior of the parabola 𝑓(𝑥) = (𝑥 − 1)(𝑥 + 2) is similar to 𝑦 = 𝑥2 which

means the end behavior of the function 𝑦 =1

𝑓(𝑥)=

1

(𝑥−1)(𝑥+2) is going to be like 𝑦 =

1

𝑥2 . The graph of

𝑦 =1

(𝑥−1)(𝑋+2) is similar to 𝑦 =

1

𝑥2 as 𝑥 → ±∞ making the line 𝑦 = 0 a horizontal asymptote.


So you can see that for the graph of 𝑦 =1

𝑓(𝑥) , the places where the graph of 𝑦 = 𝑓(𝑥) shoots to infinity,

the graph of 𝑦 =1

𝑓(𝑥) goes to zero, and where the graph of 𝑦 = 𝑓(𝑥) is zero then 𝑦 =

1

𝑓(𝑥) will have

vertical asymptotes and the signs are preserved. Thus if 𝑓(𝑥) → +∞, then 𝑦 =1

𝑓(𝑥) will be a small (+)

number.

In general let us investigate what happens to the graph of a function and its reciprocal.

2. Sketch the graphs of 𝑦 =1

𝑓(𝑥) where you are given the graphs of 𝑦 = 𝑓(𝑥). The principles we saw

above are valid here even when the function 𝑓(𝑥) is not a polynomial function.

a.

b.

c.

d.

This function is a periodic function with domain (−∞, ∞) and is not a polynomial function.

Try to work these out before peeking ahead.


Solutions to problem 2

2. Sketch the graphs of 𝑦 =1

𝑓(𝑥) where you are given the graphs of 𝑦 = 𝑓(𝑥). The principles we

observed below are preserved even of the function 𝑓(𝑥) is not a polynomial function.

a.

b.


c.

Since the original function 𝑦 = 𝑓(𝑥) has no 𝑥-intercepts,

the graph of 𝑦 =1

𝑓(𝑥) has no vertical asymptotes.

d.

Zeros of the Numerator and Denominator

We now need to see how the numerator polynomial influences the graph of a rational function. We

would predict that the basic features of the numerator polynomial function for 𝑥-values that are close to

the 𝑥-intercepts of the numerator will be preserved. A mathematician should always venture an

intuitive guess based on previous experience and check out a few examples, and then hypothesize a

theory that can be proven using mathematical principles. Let us use graphing utilities to see a few

examples.

Below is the process a mathematician might employ to get a grip on how the graph of

𝑓(𝑥) = (𝑥−1)2

(𝑥+3)3

𝑥 might compare to the graph of the function 𝑓(𝑥) = (𝑥 − 1)2(𝑥 + 3)3.

In the previous section we saw how to graph 𝑓(𝑥) = (𝑥 − 1)2(𝑥 + 3)3 , we also know from the

previous section that if we wanted to graph 𝑔(𝑥) = 2 (𝑥 − 1)2(𝑥 + 3)3 or ℎ(𝑥) =1

2 (𝑥 − 1)2(𝑥 + 3)3,


the basic graph would not change except for a scaling factor of 2 or 1

2 respectively as shown below. If we

wanted negative scaling factors , the original graph would reflect across the 𝑦-axis along with stretching.

So a natural question to ask is what is the effect of a variable scaling factor of say 1

𝑥 . If we can answer

that, we would know how the graph of 𝑓(𝑥) =1

𝑥( (𝑥 − 1)2(𝑥 + 3)3 ) looks like compared to

graph of the function 𝑔(𝑥) = (𝑥 − 1)2(𝑥 + 3)3 . Our intuitive guess would be that the basic

numerator features should be the same except for a variable stretching near the 𝑥-intercepts, and we

would also have to account for a negative scaling factor when variable scaling factor 1

𝑥< 0 which would

then reflect that part of the graph across the 𝑦-axis. So in the vicinity of the 𝑥 = 1, and 𝑥 = −3 we

expect the graph of the rational function 𝑓 to be much like the numerator graph 𝑔, but with stretch

factors at 1

1= 1 near 𝑥 = 1 and a stretch factor of −

1

3 near 𝑥 = −3. Also, given our work with

functions 𝑦 =1

𝑥𝑛 we know that near 𝑥 = 0 the graph of 𝑓 must blow with 𝑦 → ∞ or 𝑦 → −∞.

So let’s now look at our graph using a graphing utility to see if our analysis and intuitive guess is

consistent with actual graphs. After this example you might want to play on your own and explore more

functions to strengthen your intuition.


𝒈(𝒙) = (𝒙 − 𝟏)𝟐(𝒙 + 𝟑)𝟑

𝒇(𝒙) =(𝒙 − 𝟏)𝟐(𝒙 + 𝟑)𝟑

𝒙

The two functions 𝒇(𝒙) =(𝒙−𝟏)𝟐(𝒙+𝟑)𝟑

𝒙 and 𝒈(𝒙) = (𝒙 − 𝟏)𝟐(𝒙 + 𝟑)𝟑 together so you can

compare.

As you can see the basic features of the cubic and squares parts of the numerator got preserved. So the

denominator of 𝑥 in the rational function can be thought of as providing a local stretch factor to just

stretch or compress or reflect the shape of the graph of the numerator polynomial


We will see that for all rational functions, near the zeros of the numerator function the graph behaves

much like the graph of the numerator polynomial. Also at each zero of the denominator function the

graph has a vertical asymptote and behaves like 1

(𝑥−𝑎)𝑛 with a possible vertical stretch or reflection.

Let us now consider the same numerator polynomial but with the denominator being 𝑥2 i.e., apply the

variable stretch factor of 1

𝑥2 . Note that at 𝑥 = 0 the graph should have the shape of 1

𝑥2 and at

𝑥 = 1 𝑎𝑛𝑑 − 3 the graph should behave like the numerator 𝑦 = (𝑥 − 1)3(𝑥 + 3)2, but with stretch

factors 1

12 = 1 and 1

(−3)2 =1

9.

𝒈(𝒙) = (𝒙 − 𝟏)𝟐(𝒙 + 𝟑)𝟑

𝒇(𝒙) =(𝒙 − 𝟏)𝟐(𝒙 + 𝟑)𝟑

𝒙𝟐

The two functions 𝒇(𝒙) =(𝒙−𝟏)𝟐(𝒙+𝟑)𝟑

𝒙𝟐 and 𝒈(𝒙) = (𝒙 − 𝟏)𝟐(𝒙 + 𝟑)𝟑 together so you can

compare.


We can of course play with more examples to really get a good feel and so we will let you play with it.

Try to be creative when using the graphing utility to explore.

One more example:

𝒈(𝒙) = (𝒙 − 𝟏)𝟐(𝒙 + 𝟑)𝟑

𝒇(𝒙) =(𝒙 − 𝟏)𝟐(𝒙 + 𝟑)𝟑

(𝒙 − 𝟐)(𝒙 + 𝟏)

We plot functions 𝒇(𝒙) and 𝒈(𝒙) together so to see the similarities near the zeros of 𝒈(𝒙).

𝒙 = −𝟏, and 𝒙 = 𝟐 are shown as dotted lines as a reference only. This way you can see the

blow up of the graph near those values clearly.

Zoomed in


End Behavior of a Graph of a Rational Function

The top and bottom polynomials also determine how the graph behaves as 𝑥 goes off to ±∞. Long

division allows us to rewrite 𝑓(𝑥) in the form (𝑥) =𝑃(𝑥)

𝑆(𝑥)= 𝑄(𝑥) +

𝑅(𝑥)

𝑆(𝑥) , where 𝑄(𝑥) =Quotient, and

𝑅(𝑥) = remainder with degree less than that of 𝑆(𝑥). The term 𝑅(𝑥)

𝑆(𝑥) will always tend to zero as

𝑥 → ±∞ because the degree of 𝑅(𝑥) < degree of 𝑆(𝑥) and each of these polynomials behave like their

highest degree terms as 𝑥 → ±∞.

See below for the examples we just finished looking at.

𝒇(𝑥) =(𝑥 − 1)2(𝑥 + 3)3

𝑥

= 𝑥4 + 7𝑥3 + 10𝑥2 − 18𝑥 − 27 +27

𝑥

A crude end-behavior is to use just the highest degree

terms of the top and bottom so that 𝑦 ≈𝑥5

𝑥= 𝑥4

as 𝑥 → ±∞.

𝒇(𝑥) =(𝑥 − 1)2(𝑥 + 3)3

𝑥2

= 𝑥3 + 7𝑥2 + 10𝑥 − 18

+−27𝑥 + 27

𝑥

Crude end-behavior is 𝑦 ≈𝑥5

𝑥2= 𝑥3 as

𝑥 → ±∞.

𝑓(𝑥) =(𝑥 − 1)2(𝑥 + 3)3

(𝑥 + 1)(𝑥 − 2)

= 𝑥3 + 8𝑥2 + 20𝑥 + 18

+31𝑥 + 63

(𝑥 + 1)(𝑥 − 2)

Crude end-behavior is 𝑦 ≈𝑥5

𝑥2= 𝑥3 as

𝑥 → ±∞.


Practice examples

Use the graphs of the numerators and denominators provided to sketch a rough graph of the function

𝑦 =𝑓(𝑥)

𝑔(𝑥)

See example below

Let 𝑓(𝑥) = (𝑥 + 1)2(𝑥 − 3), and 𝑔(𝑥) = (𝑥 − 1)(𝑥 + 3). To sketch the graph of 𝑦 =(𝑥+1)2(𝑥−3)

(𝑥−1)(𝑥+3)

we can use what we have learnt above and put it together.

Note that expanding and doing long division gives: 𝑦 =(𝑥+1)2(𝑥−3)

(𝑥−1)(𝑥+3)= 𝑥 − 3 +

4𝑥−12

(𝑥−1)(𝑥+3).

Thus the end-behavior is 𝑦 = 𝑥 − 3 is the oblique asymptote. Note the crude estimate of the end-

behavior is 𝑦 =𝑥3

𝑥2 = 𝑥

In these examples we have accumulated some basic tools for connecting information on the zeros and

degrees of the numerator and denominator polynomials of a rational function to features of its graph.

You might ask why do we need to learn how to graph a rational function without a graphing utility? This

is a good question and one answer is that the effective use of a graphing utility requires selecting an

appropriate scale or viewing window to capture the main features. Understanding what equation


features cause certain graphical behaviors also allows one to come up with formulas for 𝑓(𝑥) when

given its graph.

We can summarize our observations as following

The Graph of a rational function is influenced by three factors:

The graph behaves like that of the Numerator polynomial near the zeros of the numerator.

The end behavior as 𝑥 → ±∞ is determined by the quotient function from long division of

the rational function. A crude version of this is the power function that is the ratio of the

highest degree term of the numerator over the highest degree term of the denominator.

The zeros of the denominator polynomial can cause vertical asymptotes at these 𝑥-values

with the behavior like 𝑦 = 𝐶 ⋅1

(𝑥−𝑎)𝑛 near each zero 𝑥 = 𝑎 of multiplicity 𝑛.

Steps to Graph a Rational Function

Let 𝑓(𝑥) =𝑃(𝑥)

𝑆(𝑥) be a rational function in lowest terms. We basically look at the three items above to

plot asymptotes, behavior near 𝑥-intercepts and near vertical asymptotes and the end-behavior. We

then sketch the graph to accommodate these local features..

Step 1 Use long division to write (𝑥) =𝑃(𝑥)

𝑆(𝑥)= 𝑄(𝑥) +

𝑅(𝑥)

𝑆(𝑥) , where 𝑄(𝑥) =Quotient, and

𝑅(𝑥) =remainder.

This tells us that as 𝑥 → ±∞ the rational function 𝑓(𝑥) will behave very much like 𝑄(𝑥). The equation

𝑦 = 𝑄(𝑥) will is the horizontal or oblique asymptote, or possibly an asymptotic polynomial function.

Note that whenever the degree of numerator 𝑃(𝑥) is less than the degree of denominator 𝑆(𝑥),

then the quotient 𝑄(𝑥) = 0 and there is a horizontal asymptote at the 𝑥-axis.

When the degree of the numerator and denominator is the same

𝑄(𝑥) =𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑃(𝑥)

𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑆(𝑥) so there is a horizontal asymptote at this value of 𝑦.

Step 2 Set denominator S(𝑥) = 0 to find all the vertical asymptotes (places where the rational function

blows up).

Step 3 Setting the numerator 𝑃(𝑥) = 0 and factoring it will give you all of the 𝑥-intercepts. This will

help you get the basic shape near the 𝑥-axis intercepts.

Step 4 Setting 𝑥 = 0 in the function 𝑓(𝑥) will give you the 𝑦-intercept.

Step 5 Setting the remainder 𝑅(𝑥) = 0 will give you points where the function intersects the horizontal

or oblique asymptotes if any. This provides fine detail that may not be needed for a rough sketch.

Step 6 Vertical asymptotes break the graph into separate pieces. It is useful to plot a table of points

with a couple of points from each piece of the graph.

Step 7 Use the previous steps to make a rough sketch and label all important parts of the graph.


Practice Problems

The first one is done for you, please attempt the others on your own.

1. For all the rational functions below please find the requested information below and then graph the

function. In your graph show all the relevant information and label all parts of the graph. Draw the

asymptotes if any as dotted lines.

a) End behavior by long division.

b) Vertical asymptotes where bottom is zero.

c) All the 𝑥-intercepts where top is zero.

d) 𝑦-intercept by plugging in 𝑥 = 0

e) Solve for 𝑅(𝑥) = 0 from a) to locate points (if any) where the graph intersects the horizontal or

oblique asymptotes.

f) Plot two points on either side of the vertical asymptotes and any additional points as needed to

finally sketch your graph.

I. 𝑓(𝑥) =1

𝑥+2

a) End behavior by long division. Nothing to do here as the numerator is already of degree less than that of the denominator.

Function rewriten as 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟

𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟+ 𝑄𝑢𝑜𝑡𝑖𝑒𝑛𝑡

𝑓(𝑥) =1

𝑥+2+ 0 that means the 𝑥-axis is

horizontal asymptote.

b) Denominator is zero at 𝑥 = −2. Vertical Asymptotes: 𝑥 = −2

c) 𝑥-intercepts: None since 1 ≠ 0 for all 𝑥.

d) 𝑦-intercept is at 𝑓(0). 𝑦-intercept 𝑦 = 𝑓(0) =1

0+2=

1

2

e) Points of intersection with the horizontal or oblique asymptotes: None since the remainder is 𝑅(𝑥) = 1 ≠ 0 ever.

f)

𝑥

𝑦 = 𝑓(𝑥) =1

𝑥 + 2

−4 1

−4 + 2= −

1

2

−3 1

−3 + 2= −1

0 1

0 + 2=

1

2

1 1

1 + 2=

1

3


II. 𝑓(𝑥) =3𝑥2+3𝑥−18

𝑥+1

Before we can answer anything let us do long division so it will allows to find the oblique and

horizontal asymptotes easily

3𝑥

𝑥 + 1 3𝑥2 +3𝑥 −18 − (3𝑥2 +3𝑥)

−18

a) Horizontal or Oblique Asymptote Oblique Asymptote is 𝑦 = 3𝑥 Domain of the function: All real numbers except for 𝑥 = −1.

Function rewrite as 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟


𝑓(𝑥) =3𝑥2 + 3𝑥 − 18

𝑥 + 1= 3𝑥 −

18

𝑥 + 1

b) Bottom is zero at 𝑥 = −1 Vertical Asymptotes: 𝑥 = −1

c) d) c) 3𝑥2 + 3𝑥 − 18 = 3(𝑥 − 2)(𝑥 + 3) = 0 giving us 𝑥 = −3 or 𝑥 = 2

𝑥-intercepts: (−3, 0), and (2, 0)

d) 𝑦-intercept is (0, −18), 𝑦 = 𝑓(0) =3(0)2+3(0)−18

0+1=

−18

1= −18

e) Points of intersection with the horizontal or oblique asymptotes: None since 𝑅(𝑥) = −18.

𝑓(𝑥) =3𝑥2 + 3𝑥 − 18

𝑥 + 1= 3𝑥 −

18

𝑥 + 1

Long division gives us


Plot two points on either side of the vertical

asymptote.

𝑥 𝑦 =

3𝑥2 + 3𝑥 − 18

𝑥 + 1

−2 3(−2)2 + 3(−3) − 18

−2 + 1=

−12

−1= 12

−3 3(−3)2 + 3(−3) − 18

−3 + 1= 0

0 3(0)2 + 3(0) − 18

0 + 1=

−18

1= −18

1 3(1)2 + 3(1) − 18

1 + 1=

−12

2= −6

2 3(2)2 + 3(2) − 18

2 + 1=

0

3= 0

f) Plot two points to graph the

horizontal asymptote 𝑦 = 3𝑥.

𝑥 𝑦 = 3𝑥 0 0 2 6


iii. 𝑓(𝑥) =2𝑥2+1

(𝑥−2)(𝑥+3)=

2𝑥2+1

𝑥2+𝑥−6

a) End Behavior: Horizontal asymptote at 𝑦 = 𝑄(𝑥) = 2.

We can also get this from knowing that the

𝑓(𝑥) ≈2𝑥2

𝑥2 = 2 for when 𝑥 approaches ±∞.

Function rewriten as 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟


𝑓(𝑥) = 2 +−2𝑥 + 13

(𝑥 − 2)(𝑥 + 3)

b) Bottom = 0 at 𝑥 = 2 𝑎𝑛𝑑 𝑥 = −3 Vertical Asymptotes: 𝑥 = 2, 𝑥 = −3

c) 𝑥-intercepts: None since 2𝑥2 + 1 ≠ 0 for any real 𝑥.

d) 𝑦-intercept 𝑦 = 𝑓(0) =1

(−2)(3)=

1

−6

e) Points of intersection with the horizontal asymptotes: −2𝑥 + 13 = 0 2𝑥 = 13

𝑥 =13

2

𝑓 (13

2) = 2

f) Points on the graph

𝑥 𝑦 −5 2(−5)2 + 1

(−5)2 − 5 − 6=

51

14

−4 2(−4)2 + 1

(−4)2 − 4 − 6=

33

6

−1 2(−1)2 + 1

(−1)2 − 1 − 6= −

3

6

= −1

2

0 2(0)2 + 1

(0)2 + 0 − 6= −

1

6

1 2(1)2 + 1

(1)2 + 1 − 6= −

3

4

3 2(3)2 + 1

(3)2 + 3 − 6=

19

6

4 2(4)2 + 1

(4)2 + 4 − 6=

33

14


g) Graph with relevant information




1. Quotient Function

2. Rational Function

3. Domain of a rational function

4. Range of rational function

5. Vertical Asymptote

6. Oblique or Horizontal Asymptote

7. Points of intersection of a function with its horizontal, or oblique asymptote



Exercises 2.4

1. In your own words describe what role the numerator function and the denominator function

play in a quotient function. Give examples to demonstrate your ideas.

2. How do you locate vertical asymptotes? Why does your method work?

3. Can the graph of a rational function intersect a vertical asymptote? Why or why not?

4. How do you determine the horizontal or oblique asymptotes?

5. What would the graph of the function 𝑓(𝑥) =1

𝑥+ 𝑥2 look like?

6. If the numerator and denominator have common factors in a rational function, how does that

affect the graph of the rational function?

7. How do you find 𝑥-intercepts and 𝑦-intercepts of a rational function?


8. If you knew all the asymptotes of the rational function, would there be a unique function that

corresponds to these asymptotes? What other information would you need?

9. For the functions 𝑦 = 𝑓(𝑥) below, draw rough sketches their reciprocal function graphs 𝑦 =1

𝑓(𝑥)

and clearly label any horizontal and vertical asymptotes.

a.

b.

This function with a repeating pattern is called a “Periodic function” and is defined on (−∞, ∞).

c.

d.


10. Draw a rough graph the function =𝑓(𝑥)

𝑔(𝑥) , given the graphs of 𝑦 = 𝑓(𝑥), and 𝑦 = 𝑔(𝑥). Locate all

vertical asymptotes and 𝑥-intercepts.

𝑦 = 𝑓(𝑥)

𝑦 = 𝑔(𝑥)

a.

b.

This is not a polynomial but a periodic function defined on (−∞, ∞)

This is not a polynomial but a periodic function defined on (−∞, ∞)

c.


11. For all the rational functions below find the requested information below and then graph the

function. In your graph show all the relevant information and label all parts of the graph. Draw

the asymptotes if any as dotted lines.

I. Perform long division and write the rational function 𝑁𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟

𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟=

𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟

𝐷𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟+ 𝑄𝑢𝑜𝑡𝑖𝑒𝑛𝑡 to

get the end-behavior.

II. Locate the vertical asymptotes.

III. Locate the 𝑥-intercepts and the 𝑦-intercept if possible

IV. Make a T-table to include a couple of points from each piece of the graph in between vertical

asymptotes.

V. Find all points (if any) where the graph intersects the horizontal or oblique asymptotes.

a) 𝑔(𝑥) =1

𝑥−3

b) 𝑓(𝑥) =1

(𝑥−3)2

c) 𝑓(𝑥) =1

(𝑥−3)3

d) 𝑔(𝑥) =2

𝑥−3

e) 𝑓(𝑥) =2

(𝑥−3)2

f) 𝑓(𝑥) =−2

(𝑥−3)3

g) 𝑓(𝑥) =𝑥−1

𝑥−3

h) 𝑓(𝑥) =𝑥−1

(𝑥−3)2

i) 𝑟(𝑥) =1

(𝑥−3)(𝑥+3)

j) ℎ(𝑥) =𝑥+1

(𝑥−3)(𝑥+3)

k) 𝑅(𝑥) =𝑥+1

(𝑥+2)2(𝑥−1)

l) 𝑓(𝑥) =1

𝑥2−4

m) 𝑔(𝑥) =(𝑥+2)2(𝑥−4)

(𝑥−1)

n) 𝑦 =3𝑥2+4𝑥+1

𝑥+2= 3𝑥 − 2 +

5

𝑥+2

o) 𝑓(𝑥) =2𝑥2+1

𝑥−1= 2𝑥 + 2 +

3

𝑥−1

p) 𝑇(𝑥) =3𝑥2+4𝑥+1

(𝑥+2)2 = 3 +−8𝑥−11

(𝑥+2)2

q) 𝑃(𝑥) =3𝑥3+𝑥2+3𝑥+1

𝑥2−4= 3𝑥 + 1 +

15𝑥+5

𝑥2−4

r) 𝑆(𝑥) =3𝑥3−3𝑥−1

(𝑥+1)2 = 3𝑥 − 6 +6𝑥+5

(𝑥+1)2

s) 𝑟(𝑥) =𝑥+1

𝑥2−4

t) 𝑓(𝑥) =3𝑥−2

𝑥2−4

u) 𝑆(𝑥) =𝑥2+2𝑥+1

𝑥2−1

v) 𝑇(𝑥) =𝑥2−1

𝑥2+2𝑥+1

w) 𝐻(𝑥) =𝑥2−1

𝑥+1

x) 𝑄(𝑥) =𝑥3−𝑥2+4𝑥

𝑥−1

y) 𝑅(𝑥) = 𝑥2 −1

𝑥

z) 𝑄(𝑥) =𝑒𝑥

𝑥


12. Choose the graphs below that match the graphs of the rational functions below.

I. 𝑔(𝑥) =3𝑥+3

𝑥2+2𝑥−3

II. ℎ(𝑥) =3

𝑥2+2𝑥−3

Graph A


13. For all the graphs below find a rational function representing it. Use the lowest degree numerator

and denominator to achieve the given vertical asymptotes and 𝑥-axis intercepts. Is there a unique

formula for all of them, or can some of the graphs below could have more than one formula?

Explain your reasoning clearly.

a.

b.


c.

14. For the functions below, please do the following.

I. Find the domain of each function.

II. Determine all the vertical asymptotes.

III. Determine the end behavior of the function.

IV. Use all the information found above to sketch the functions.

a. 𝑦 =√𝑥−1

𝑥+1

b. 𝑦 = √𝑥 − 1 +1

𝑥+1

c. 𝑦 = 2𝑥2 − 4 +1

𝑥−1

d. 𝑦 = 2𝑥2 + 1 −1

𝑥2−1

15. Create a rational function with vertical asymptotes at 𝑥 = −2, 𝑥 = 1 and slant asymptotes at

𝑦 = −3𝑥 + 5.

16. Create a quotient of function 𝑦 =𝑓(𝑥)

𝑔(𝑥) with a vertical asymptote at 𝑥 = 2, and so that as

𝑥 → ±∞, 𝑦 → 3𝑥2 + 5𝑥 − 1.

Solving Equations and Inequalities Page 222

Chapter 3 Solving Equations and Inequalities We saw in chapter two when graphing functions that finding 𝑥-values where a function output or the 𝑦-

coordinate on the graph was equal to zero provided the location of 𝑥-intercepts of the graph of the

function graph. When we know the input to a function, it is often straightforward to just plug in the 𝑥-

value to obtain the output. It turns out that the reverse question comes up in many modeling

situations. That is, we might have some function formula that we have constructed to model a situation

and we want to know what input is required so that the output will be equal to some stated value.

Algebraically this corresponds to solving an equation 𝑓(𝑥) = 𝐶 where we seek 𝑥 so the output is 𝐶.

When finding 𝑥-intercepts, this 𝐶 = 0 and we use the solutions to plot polynomial and rational

functions. In this chapter we investigate and develop tools for solving equations of the form 𝑓(𝑥) = 𝐶

for all of the types of functions 𝑓 we have worked with.

Most of you have been exposed to pre-requisite materials before getting into College Algebra. So we

will assume you are familiar with solving basic linear and some non-linear equations and inequalities. If

you feel you need a refresher, we suggest studying Module 3 in the Developmental and Intermediate

Algebra e-text (Go to page 361).

3.1 Quadratic Equations Recall that

An equation is a statement asserting the equality of two mathematical objects.

An inequality is statement asserting that one mathematical object is larger than or equal to, or

less than or equal to another mathematical object.

So we can create equations as indicated above by setting 𝑓(𝑥) = 𝐶 or we could have both objects be

outputs of different functions e.g., set 𝑓(𝑥) = 𝑔(𝑥).

Examples of equations and inequalities

1. 2𝑥 = 3

2. 2𝑥+3 = 24𝑥−1

3. 𝑥2 − 3𝑥 = 5(𝑥 + 1) − 3

4. 2𝑥 < 3.5

5. 2𝑥+3 ≥ 24𝑥−1

6. 𝑥2 − 3𝑥 ≤ 5(𝑥 + 1) − 3

http://pages.uwc.edu/shubhangi.stalder/Developmental-and-intermediate-Algebra-Textbook2nded.pdf


The symbols above can be used as follows:

1 The equation 𝑎 = 𝑏 signifies that the mathematical object 𝑎 is equal to the mathematical

object 𝑏.

2 The inequality 𝑎 < 𝑏 signifies that the mathematical object 𝑎 is smaller than or less than the

mathematical object 𝑏.

3 The inequality 𝑎 > 𝑏 signifies that the mathematical object 𝑎 is bigger than or greater than the

mathematical object 𝑏.

4 The inequality 𝑎 ≤ 𝑏 signifies that the mathematical object 𝑎 is smaller than or equal to, or less

than or equal to, the mathematical object 𝑏.

5 The inequality 𝑎 ≥ 𝑏 signifies that the mathematical object 𝑎 is bigger than or equal to, or

greater than or equal to, the mathematical object 𝑏.

The symbols above are our shorthand way of writing comparisons between two mathematical objects.

This is another example of the symbolic nature of writing mathematics. It is much easier to write 𝑥 ≤ 2

than to have to say “a number that is less than or equal to 2”, or a number that is not larger than 2”.

Any number(s) when substituted for the variable(s) in the original equation or inequality that

results in a true statement is called a solution to that equation or inequality.

The process in which we use mathematical properties of equality or inequality respectively to

isolate the variable by itself is called solving the equation or inequality.

All real number solutions to an equation or inequality in one variable can be represented on a real

number line. When we equate two non-zero functions like 𝑓(𝑥) = 𝑥2 − 3𝑥 = 3𝑥 − 1 = 𝑔(𝑥) we are

trying to see at what 𝑥 values can the two functions have the same output. Thus solving 𝑓(𝑥) = 𝑔(𝑥)

corresponds to finding where the two graphs intersect each other. Equations in one variable can also be

rearranged so that the non-zero terms are on the left side and zero is on the right side. Thus if we set

ℎ(𝑥) = “the left hand side”, solving this equation reduces to finding the zeroes of ℎ, i.e., solving the

equation ℎ(𝑥) = 0. You can see that equations of the type 𝑓(𝑥) = 𝑔(𝑥) can be also be then written as

𝑓(𝑥) − 𝑔(𝑥) = 0, or 𝑔(𝑥) − 𝑓(𝑥) = 0.

Real solutions to an equation or an inequality in two variables can be represented as points or regions in

a 2-dimensional Cartesian coordinate system. Solutions to equations of the type 𝑓(𝑥, 𝑦) = 𝑔(𝑥, 𝑦) can

be thought of as points of intersection of the surfaces 𝑧 = 𝑓(𝑥, 𝑦), and 𝑧 = 𝑔(𝑥, 𝑦) represented in the

3-dimensional Cartesian coordinate system.

An equation that is true for all values of the variable is called an identity.

The equation 𝑥 × 𝑥 = 𝑥2 is an identity. Both sides of the equal sign yield the same number for any real

number value of 𝑥.

An equation in which you end up with a false statement for all values of the variable is said to

have “No Solution”.

The equation 𝑥 = 𝑥 + 2 has no solution since a quantity is never equal to one more than itself.


Solving an equation or an inequality is like untying a knot or undoing what was done to the variable to

get it into its current state. We operate on an equation with certain mathematical tools to convert the

equation to a simpler equation, that has the same solutions as the original. (These are called equivalent

equations.) Thus solving an equation involves using tools to work the original equation through a

sequence of equivalent equations to eventually get to a statement like 𝑥 = “answer”.

Certain tools used in the process of isolating the variable can sometimes lead us to a value of

the variable that makes the original equation false. Such a solution is called an extraneous

solution and we will need to watch out for these “false” solutions whenever we use those tools.

Two examples of such tools are:

1. Raising both sides of an equation to a power, e.g., square both sides of an equation.

2. Multiplying or dividing both sides of an equation by an expression that has the potential to be equal

to zero.

When making use of the tools mentioned above, it is best to check our final answers by substituting

their values back into the original equation to see if they really are the solutions.

You can see why these tools may cause an equation to have an extraneous solution by working with a

linear equation say, 𝑥 = 3. The only solution to this equation is 𝑥 = 3

a. Look what happens when we square both sides, we get 𝑥2 = 9.

This equation has two solutions 𝑥 = 3, 𝒙 = −𝟑.

b. Look what happens if we multiply both sides by (𝑥 + 2).

𝑥(𝑥 + 2) = 3(𝑥 + 2)

𝑥2 + 2𝑥 = 3𝑥 + 6

𝑥2 − 𝑥 − 6 = 0

(𝑥 − 3)(𝑥 + 2) = 0

𝑥 = 3 and 𝒙 = −𝟐.

Even though we did not really need to square or multiply both sides of the equation 𝑥 = 3 to get the

solutions to it, the process demonstrates how additional solutions got inserted by doing these

operations.

Sometimes when solving polynomial equations like 𝑥2 = −4 we get solutions that are not real. These

solutions are referred to as complex numbers. In case you are not familiar with complex numbers below

a review of complex numbers.

Review of Complex Numbers

For a long time people believed that the set of real numbers accounted for all the numbers that there

are. However in 1545, Girolamo Cardano and others made progress on solving cubic equations such as

𝑥3 − 15𝑥 = 4. Their technique gave a solution to this problem that involved √−121 . The value of this

square root cannot be any real number since any real number when squared will be positive. However

their solution method simplified to a final result of 𝑥 = 4 which is easily seen to make the above

equation true. This lead to a deeper study of the square roots of negative numbers. Today these kinds of

numbers are absolutely essential in higher mathematics, engineering and physics. A simpler problem

that shows the deficiency of the real numbers is to try to find a real number 𝑥 such that 𝑥2 = −1. Play


with this for a while to convince yourself that no real number when multiplied by itself will produce the

result of negative one!

So a new kind of number system evolved where 𝑖 = √−1 is designated to represent the unit imaginary

(non-real) number that is a solution to the equation 𝑥2 = −1. Even though it is hard to imagine what 𝑖

is, we know that its square is −1 or 𝑖2 = 𝑖 × 𝑖 = −1. With this definition of 𝑖 we expand the set of real

numbers to the set of complex numbers.

Set of complex numbers is a collection of all numbers of the form 𝑤 = (𝑎 + 𝑏𝑖), where 𝑎, 𝑏

are real numbers and are called the real part and imaginary part respectively of 𝑤. Another

way to represent this sentence in mathematical notation is

𝐶 = {𝑎 + 𝑏𝑖 |𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑎𝑛𝑑 𝑖2 = −1}.

Note: The set of all real numbers is a subset of the set of complex numbers because every real number can be written as 𝑎 + 0𝑖, e.g., 2.4 = 2.4 + 0𝑖. In set notation it will look like 𝑅 ⊂ 𝐶.

Any time mathematicians find new objects a natural questions to ask is can we can do all the arithmetic

operations we do with real numbers with complex numbers, and do the properties of arithmetic work

the same way. The most natural way to add or subtract would be to combine like terms. To multiply we

use distributive property of multiplication over addition/subtraction. The division is a little more

complex but uses properties of radicals. See examples below to see how the different operations work

with the complex numbers.

Practice Problems

1. Simply the following and write all your answers in the 𝑎 + 𝑏𝑖 form where 𝑎, and 𝑏 are real

numbers.

a) (3 + 5𝑖) + (2 − 3𝑖)

Adding like terms we get (3 + 5𝑖) + (2 − 3𝑖) = (3 + 2) + (5 − 3)𝑖 = 5 + 2𝑖

b) (3 + 5𝑖) − (2 − 3𝑖)

(3 + 5𝑖) − (2 − 3𝑖) = (3 − 2) + (5 − (−3))𝑖 = 1 + 8𝑖

c) (3 + 5𝑖)(2 − 3𝑖)

Using distributive property and adding like terms we get

(3 + 5𝑖)(2 − 3𝑖) = 3(2) + 3(−3𝑖) + 5𝑖(2) + 5𝑖(−3𝑖)

= 6 − 9𝑖 + 10𝑖 − 15𝑖2 (recall 𝑖2 = −1) giving us

= 6 + 𝑖 + 15

= 21 + 𝑖

d) (2 + 3𝑖)(2 − 3𝑖)

= 4 − 6𝑖 + 6𝑖 − 9𝑖2

= 4 + 9

= 13

Such pairs of numbers 2 + 3𝑖 and 2 − 3𝑖 are called complex conjugates of each other.


Complex Conjugates: The pair of complex numbers 𝑎 + 𝑏𝑖 and 𝑎 − 𝑏𝑖 are called complex

conjugates of each other and their product is always a real number.

(𝑎 + 𝑏𝑖)(𝑎 − 𝑏𝑖) = 𝑎2 + 𝑏2

This result allows us to simplify fractions (representing division) of complex numbers.

e) 3+5𝑖

2−3𝑖 Simplify to the form 𝑎 + 𝑏𝑖.

Since we need to write the division in 𝑎 + 𝑏𝑖 form we multiply numerator and denominator

by the conjugate of 2 − 3𝑖 which is 2 + 3𝑖.

3 + 5𝑖

2 − 3𝑖=

(3 + 5𝑖)(2 + 3𝑖)

(2 − 3𝑖)(2 + 3𝑖)=

6 + 9𝑖 + 10𝑖 + 15𝑖2

4 + 6𝑖 − 6𝑖 − 9𝑖2

=6 + 19𝑖 − 15

4 + 9=

−9 + 19𝑖

13= −

9

13+

19

13𝑖

f) Simplify the powers of 𝑖 below. Do you notice any patterns

𝑖1 = 𝑖 𝑖2 = −1 𝑖3 = −𝑖 𝑖4 = 1

𝑖5 = 𝑖 𝑖6 = −1 𝑖7 = −𝑖 𝑖8 = 1

𝑖9 = 𝑖 𝑖10 = −1 𝑖11 = −𝑖 𝑖12 = 1

Note that the powers 𝑖𝑛 start to repeat as 𝑛 increases by 4. So we have in general

𝑖𝑛 = 1 if 𝑛 is a multiple of 4.

𝑖𝑛 = 𝑖 if 𝑛 divided by 4 leaves a remainder of 1.

𝑖𝑛 = −1 if 𝑛 divided by 4 leaves a remainder of 2.

𝑖𝑛 = −i if 𝑛 divided by 4 leaves a remainder of 3.

g) 𝑖401 = 𝑖400 ⋅ 𝑖

Since 401 divided by 4 leaves a remainder of 1 we get 𝑖401 = 𝑖, because 𝑖401 = 𝑖400 ⋅ 𝑖 =

(𝑖4)100𝑖 = 𝑖 because 𝑖4 = 1.

h) 𝑖39 = 𝑖36 ⋅ 𝑖2 ⋅ 𝑖 = −𝑖, since 𝑖36 = (𝑖4)9 = 1 and 𝑖2 = −1.

i) √−81

√−81 = √−1√81 = 9𝑖

j) √−8

√−8 = √8 𝑖 = 2√2 𝑖

Note that if you had √81 and we rewrote it as 9 = √81 = √(−9)(−9) = √−9√−9 = (3𝑖)(3𝑖) =

9𝑖2 = −9. You can see that in splitting a root you have be careful.


Before we go into solving non-linear equations let us review the zero product property.

Zero Product Property of Complex Numbers: Let 𝐴, and 𝐵 be two complex numbers. Then 𝐴𝐵 = 0, if

and only if either 𝐴 = 0 or 𝐵 = 0.

Quadratic Equations

Lectures

Quadratic Equations (15 min)

https://www.youtube.com/watch?v=29_SBzxChMw

Playing

We have seen how to solve quadratic equations that can be factored. We will next look at a series of

examples (some of which don’t factor) that we have been able to solve using square roots. We will look

carefully at what makes these solvable and build from there.

Examples:

Solving Quadratic Equations using the square root operation.

Find all solutions of the following quadratic equations.

Equations With Real Zeros Equations with Complex Zeros

1. 𝑥2 = 4

𝑥 = ±√4

Square roots undo squares (remember even

root property)

𝑥 = 2, or 𝑥 = −2

𝑥2 = −4

Square roots undo squares (remember even root

property)

𝑥 = ±√−4 (recall √−1 = 𝑖)

𝑥 = 2𝑖, or 𝑥 = −2𝑖

2. 𝑥2 = 5

𝑥 = ±√5


root property)

𝑥 = √5, or 𝑥 = −√5

𝑥2 = −5

𝑥 = ±√−5


property)

𝑥 = √5𝑖, or 𝑥 = −√5𝑖

3. (𝑥 + 3)2 = 5

𝑥 + 3 = ±√5


root property)

𝑥 = −3 + √5, or 𝑥 = −3 − √5

(𝑥 + 3)2 = −5

𝑥 + 3 = ±√−5


property)

𝑥 = −3 + √5𝑖, or 𝑥 = −3 − √5𝑖

https://www.youtube.com/watch?v=29_SBzxChMw


Completing the Square

In all the previous examples we undid squares by taking square roots. A question we may want to ask

then is what makes a quadratic polynomial a perfect square. Also, can we modify any quadratic

polynomial so it becomes a perfect square?

Examples: (Assume all variables take on positive real values to make sense of the polynomials

as areas of the rectangles shown)

1. 𝑥2 + 6𝑥

You can visualize the polynomial above as shown in figure 1 below. In order to change the rectangle into a perfect square using the pieces we have, we will have to move some of the 1 by 𝑥 rectangular pieces to see what is needed for this polynomial to be part of a perfect square. Since 𝑥2 already is a square, it would make sense to move the 1 by 𝑥 rectangles. Since a square has the same length and width, it would make sense to take half of the 6 rectangles, turn them and put them under the blue square which would look as in figure 2. You can see that a corner square is missing. To complete the square then we need to add nine 1 by 1 squares as seen in figure 3. This process of adding 9 squares to the existing quadratic polynomial is called completing the square.

𝑥2 + 6𝑥 + ____ = (𝑥+? )2, we can see the process is to add (6

2)

2= 32 = 9

𝑥2 + 6𝑥 + (6

2)

2

= 𝑥2 + 6𝑥 + 9 = (𝑥 + 3)2

From the above diagram, we see that the 9 red squares needed to complete the square are

obtained by taking half of the number of 𝑥 pieces and squaring that, i.e., (6

2)

2= 9.

4. 4(𝑥 + 3)2 = 5

(𝑥 + 3)2 =5

4

𝑥 + 3 = ±√5

4

𝑥 = −3 +√5

2, or 𝑥 = −3 −

√5

2

Square roots undo squares (remember even root property)

4(𝑥 + 3)2 = −5

(𝑥 + 3)2 = −5

4

𝑥 + 3 = ±√−5

4

𝑥 = −3 +√5

2𝑖, or 𝑥 = −3 −

√5

2𝑖

Square roots undo squares (remember even root property)

𝑥

𝑥 6

Figure 1

𝑥

𝑥

6

2= 3

6

2= 3

Figure 2

𝑥

𝑥

6

2= 3

6

2= 3 v

vvv

vv

3

3

Figure 3


2. 𝑥2 − 10𝑥

𝑥2 − 10𝑥 + (−10

2)

2

= 𝑥2 − 10𝑥+25 = (𝑥 − 5)2

3. 𝑥2 + 5𝑥

𝑥2 + 5𝑥 + (5

2)

2

= 𝑥2 + 5𝑥+25

4= (𝑥 +

5

2)

2

We can now use this new completing the square process to help us solve any quadratic equation. We

can make all quadratic equations look like the examples 1-3 by making them perfect squares as shown

below.

Examples

Find solutions to the quadratic equations below.

Equations With Real Zeros Equations with Complex Zeros

1. 𝑥2 + 6𝑥 − 14 = 0

𝑥2 + 6𝑥 = 14

𝑥2 + 6𝑥 + 9 = 14 + 9 (𝑥 + 3)2 = 23

𝑥 + 3 = ±√23

𝑥 = −3 + √23, or 𝑥 = 3 − √23

We start by keeping just the 𝑥2 and 𝑥 terms on one side.

Now, to complete the square on the left hand side

we have to add (6

2)

2= 9 to both sides giving us

Solving an equation in this manner is called solving

the equation using the completing the square

method.

𝑥2 + 6𝑥 + 14 = 0

𝑥2 + 6𝑥 = −14

𝑥2 + 6𝑥 + 9 = −14 + 9 (𝑥 + 3)2 = −5

𝑥 + 3 = ±√−5

𝑥 = −3 + √5𝑖, or 𝑥 = −3 − √5𝑖

We get just the variable terms on one side.

and then to complete the square on the left

hand side we have to add (6

2)

2= 9 to both

sides giving us

2. 2𝑥2 − 10𝑥 − 10 = 0

2𝑥2 − 10𝑥 = 10

𝑥2 − 5𝑥 = 5

𝑥2 − 5𝑥 + (−5

2)

2

= 5 + (−5

2)

2

(𝑥 −5

2)

2

= 5 +25

4

𝑥 −5

2= ±√

20

4+

25

4

Completing the squares is easiest with the coefficient of the 𝑥2 as 1. Rewriting the equation so that the coefficient of the square term is 1 and the constant term is on the right hand side we get

(divide both sides by 2)

2𝑥2 − 10𝑥 + 15 = 0

2𝑥2 − 10𝑥 = −15

𝑥2 − 5𝑥 = −15

2

𝑥2 − 5𝑥 + (−5

2)

2

= −15

2+ (−

5

2)

2

(𝑥 −5

2)

2

= −15

2+

25

4

𝑥 −5

2= ±√

−30

4+

25

4

Rewriting the equation so that the coefficient of the square term is 1 and the constant term is on the right hand side we get

(divide both sides by 2)


Quadratic Formula

Using the method of completing the square, we can solve any quadratic equation. Since the problem of

solving quadratic equations comes up frequently and to save us some time later, we will generalize the

completing the squares process to a generic quadratic equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 to arrive at the

“Quadratic Formula.” We recommend knowing the process of completing the square as well as

memorization of the quadratic formula. Sometimes completing the square is simpler.

A generic quadratic equation can be written as 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0, where 𝑎 ≠ 0, 𝑎𝑛𝑑 𝑏, and 𝑐 are any

real numbers. Applying the completing the square process we get

𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

𝑎𝑥2 + 𝑏𝑥 = −𝑐 (Get the variable terms alone on one side.)

𝑥2 +𝑏

𝑎𝑥 = −

𝑐

𝑎 (Make the 𝑥2 term have coefficient of 1.)

𝑥2 +𝑏

𝑎𝑥 + (

𝑏

2𝑎)

2

= −𝑐

𝑎+ (

𝑏

2𝑎)

2

(Make the left side a perfect square by adding the square of half of the 𝑥-coefficient.)

(𝑥 +𝑏

2𝑎)

2

= −𝑐

𝑎+

𝑏2

4𝑎2

(𝑥 +𝑏

2𝑎)

2

= −4𝑎(𝑐)

4𝑎(𝑎)+

𝑏2

4𝑎2 (Get a common denominator on the right.)

(𝑥 +𝑏

2𝑎)

2

=−4𝑎𝑐 + 𝑏2

4𝑎2

𝑥 +𝑏

2𝑎= ±√

𝑏2 − 4𝑎𝑐

4𝑎2 (Take square root of both sides.)

𝑥 = −𝑏

2𝑎±

√𝑏2 − 4𝑎𝑐

√4𝑎2 (Simplify the radical and combine the two terms.)

𝑥 =5

2± √

45

4

𝑥 =5

2±

√45

√4

𝑥 =5

2±

3√5

2

𝑥 =5

2+

3√5

2 𝑜𝑟 𝑥 =

5

2−

3√5

2

𝑥 =5

2± √

−5

4

𝑥 =5

2±

√5

√4𝑖

𝑥 =5

2±

√5

2𝑖

𝑥 =5

2+

√5

2𝑖 𝑜𝑟 𝑥 =

5

2−

√5

2𝑖


Quadratic Formula: The solution set to the quadratic equation 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎, 𝒂 ≠ 𝟎 and is given by

𝑥 = −𝑏

2𝑎±

√𝑏2−4𝑎𝑐

2𝑎=

−𝑏±√𝑏2−4𝑎𝑐

2𝑎.

Note: This means that the solutions to the quadratic equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 are given by

𝑥 =−𝒃+√𝒃𝟐−𝟒𝒂𝒄

𝟐𝒂, or 𝑥 =

−𝒃−√𝒃𝟐−𝟒𝒂𝒄

𝟐𝒂.

Terminology

The radicand under the square root 𝒃𝟐 − 𝟒𝒂𝒄 is called the discriminant.

Note: Depending on the discriminant’s value we can identify what kinds of solutions we will get. In fact:

1. If the discriminant 𝑏2 − 4𝑎𝑐 > 0, then the quadratic equation has two distinct real solutions.

2. If the discriminant 𝑏2 − 4𝑎𝑐 = 0, then the quadratic equation has one real solution.

3. If the discriminant 𝑏2 − 4𝑎𝑐 < 0, then the quadratic equation has two distinct complex solutions.

Examples

Find all solutions to the quadratic equations below.

1. 3𝑥2 − 𝑥 + 3 = 0

Note that 𝑎 = 3, 𝑏 = −1, 𝑐 = 3

Using the quadratic formula we get

𝑥 =−(−1) ± √(−1)2 − 4(3)(3)

2(3)

𝑥 =1 ± √1 − 36

6

𝑥 =1 ± √−35

6=

1 ± √35𝑖

6

𝑥 =1

6+

√35

6𝑖 𝑜𝑟 𝑥 =

1

6−

√35

6𝑖

Note that here 𝑏2 − 4𝑎𝑐 < 0.

2. 4𝑥2 − 20𝑥 + 25 = 0

𝑥 =−(−20) ± √(−20)2 − 4(4)(25)

2(4)

𝑥 =20 ± √400 − 400

8

𝑥 =20 ± 0

8=

20

8=

5

2

Or by factoring:


4𝑥2 − 20𝑥 + 25 = 0

4𝑥2 − 10𝑥 − 10𝑥 + 25 = 0

2𝑥(2𝑥 − 5) − 2𝑥(2𝑥 − 5) = 0

(2𝑥 − 5)2 = 0

2𝑥 − 5 = 0

2𝑥 = 5 or 𝑥 =5

2

Note that here 𝑏2 − 4𝑎𝑐 = (−20)2 − 4 ⋅ 4 ⋅ 25 = 0 and hence only one solution.

3. 4𝑥2 − 𝑥 − 5 = 0 Using the quadratic formula

𝑥 =−(−1) ± √(−1)2 − 4(4)(−5)

2(4)=

1 ± √1 + 80

8

𝑥 =1 ± √81

8=

1 ± 9

8

𝑥 =1 + 9

8=

10

8=

5

4 𝑜𝑟 𝑥 =

1 − 9

8=

−8

8= −1

Or by factoring:

4𝑥2 − 𝑥 − 5 = 0

4𝑥2 − 5𝑥 + 4𝑥 − 5 = 0

𝑥(4𝑥 − 5) + 1(4𝑥 − 5) = 0

(𝑥 + 1)(4𝑥 − 5) = 0

𝑥 + 1 = 0 or 4𝑥 − 5 = 0

𝑥 = −1 or 4𝑥 = 5

𝑥 = −1 or 𝑥 =5

4

Note that here 𝑏2 − 4𝑎𝑐 = (−1)2 − 4 ⋅ 4 ⋅ (−5) = 81 is positive, so two real solutions.

4. 3𝑥4 + 14𝑥2 + 8 = 0

Here we can let 𝑢 = 𝑥2 and then 𝑥4 = 𝑢2 so we have 3𝑢2 + 14𝑢 + 8 = 0.

Factoring we get (3𝑢 + 2)(𝑢 + 4) = 0 and with 𝑢 = 𝑥2,

𝑥2 = −4 or 3𝑥2 = −2 or

𝑥 = ±√−4 = ±2𝑖 or 𝑥 = ±√2

3𝑖 = ±

√2

√3𝑖 = ±

√2√3

√3√3𝑖

𝑥 = 2𝑖, −2𝑖, √6

3𝑖, −

√6

3𝑖

5. (𝑥 + 3)2/3 − 3(𝑥 + 3)1/3 − 4 = 0

We set 𝑢 = (𝑥 + 3)1/3 and then have 𝑢2 − 3𝑢 − 4 = 0.

Factoring, we have (𝑢 − 4)(𝑢 + 1) = 0

Thus 𝑢 = 4, 𝑜𝑟 𝑢 = −1. Now we go back to 𝑥.

(𝑥 + 3)1/3 = 4 or (𝑥 + 3)1/3 = −1. Now cube both sides of each.


(𝑥 + 3) = 64 or (𝑥 + 3) = −1. Now subtract 3.

𝑥 = 61 or 𝑥 = −4 . Both solutions work in the original Equation.

6. Find the points of intersection the linear graph 𝑦 = 𝑓(𝑥) =1

2𝑥 + 2 and the parabola

graph of 𝑦 = 𝑔(𝑥) = −1

3𝑥2 + 3𝑥 + 4.

We need to find all 𝑥-values so that the output from both functions is the same 𝑦-value. In other words, we find all 𝑥, for which 𝑓(𝑥) = 𝑔(𝑥), i.e., find any 𝑥 such that:

1

2𝑥 + 2 = −

1

3𝑥2 + 3𝑥 + 4

Multiply both sides by 6 to clear fractions 3𝑥 + 12 = −2𝑥2 + 18𝑥 + 24

Then bring all terms to one side 2𝑥2 − 15𝑥 − 12 = 0

Substituting into the quadratic formula, the 𝑥-coordinates

are: 𝑥 =15±√321

4

𝑥 =15+√321

4 or 𝑥 =

15−√321

4 are the exact coordinates

or 𝑥1 ≈ 8.23 𝑜𝑟 𝑥2 ≈ −0.73 are the approximations. We can obtain the 𝑦-coordinates by plugging into 𝑓 or 𝑔,

Thus 𝑦1 = 𝑓(𝑥1) =1

2(

15+√321

4) + 2 =

31

8+

√321

8≈ 6.11,

and

𝑦2 = 𝑓(𝑥2) =1

2(

15−√321

4) + 2 =

31

8−

√321

8≈ 1.635. The

graph shows the two graphs and their intersection points.

7. Find the points of intersection intersection of the curves with equations 𝑦 = 3𝑥 − 2

and 𝑥2

9+

𝑦2

25= 1.

We expect two solutions if the line actually crosses the ellipse. Since the 𝑦 from the first equation must be the same as the 𝑦 in the second equation, at a solution value for 𝑥, we can substitute in the expression 3𝑥 − 2 into the second equation for 𝑦 to get: 𝑥2

9+

(3𝑥−2)2

25= 1. Expand, multiply by 9 ⋅ 25 = 225 and

collecting terms, we get: 25𝑥2 + 9(9𝑥2 − 12𝑥 + 4) = 225 and 106𝑥2 − 108𝑥 − 189 =

0. Thus 𝑥 =(108±√91800)

212≈ −0.920 𝑜𝑟 1.949. Substituting

into 𝑦 = 3𝑥 − 2, 𝑦 ≈ −4.76 𝑜𝑟 𝑦 = 3.816. The graph illustrates the curves and the intersection points.

Remember, the graph of a quadratic function 𝑦 = 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 is a parabola. Finding real

solutions to quadratic equations𝑓(𝑥) = 0 locates the 𝑥-intercepts of the parabola if they exist.


Section 3.1 Worksheet Date:______________ Name:__________________________


1. Equation

2. Inequality

3. Solutions, or zeros of an equation

4. Solutions of an inequality

5. Quadratic Formula

6. Discriminant

7. Complex numbers

8. Maximum or Minimum Value of a quadratic function



Exercises 3.1

1. What is the difference between 𝑥-intercepts, 𝑦-intercepts of a function, and solutions of an

equation?

2. How would you find the maximum or minimum of a parabola?

3. What is the difference between solutions of an equation and that of an inequality?

4. Can you use your knowledge of solving quadratic equations, exponential and logarithmic

functions to solve the following problems?

a. 22𝑥 − 2𝑥 − 2 = 0

b. 𝑥4 − 3𝑥2 − 4 = 0


5. Simplify the following and write your answer in standard 𝑎 + 𝑏𝑖 form.

a. √−72 b. √−66

√−6 c. √−121√144

d. (3 + 5𝑖)(−5 + 6𝑖) e. 4−2𝑖

−2−5𝑖 f. 𝑖35

6. Solve the following equations for the given variable. If there is more than one solution, separate them with commas.

a. (5𝑦 + 4)(2𝑦 − 3) = 0 b. 𝑢2 − 10𝑢 + 21 = 0 c. 5𝑤2 = 17𝑤 − 6

d. 𝑥2 − 10𝑥 + 10 = 0 (by completing the square) Form:

o (𝑥 + _____ )2 = ________ o (𝑥 − _____ )2 = ________

Solution 𝑥 = __________

e. 2𝑥2 + 5𝑥 − 1 = 0 f. 2𝑥2 − 3𝑥 + 6 = 0


7. Find the discriminant and determine the number of real solutions of the quadratic equation. Then find the actual solutions (real or complex).

a. 4𝑥2 − 12𝑥 + 9 = 0 Discriminant: Number of solutions: Actual solutions:

b. −2𝑥2 − 6𝑥 + 8 = 0 Discriminant: Number of solutions: Actual solutions:

c. 2𝑥2 − 5𝑥 + 8 = 0

Discriminant: Number of solutions: Actual solutions:

8. Determine all the solutions to the equations below. If there is more than one solution, separate them with commas.

a. 𝑥4 − 5𝑥2 − 6 = 0 b. 𝑥4/3 − 5𝑥2/3 − 6 = 0 c. 𝑦6 − 5𝑦3 − 6 = 0

d. 1

𝑥2 −37

𝑥+ 36 = 0 e. (𝑥 − 1)2 − 37(𝑥 − 1) + 36 = 0 f. 𝑢4 + 2𝑢2 + 1 = 0


g. 2𝑥(𝑥 − 1)3(2𝑥 + 3)4(5 − 𝑥)2(𝑥2 + 4) = 0

Solve the following problems. If there is no solution, please state so.

9. The length of a rectangle is 5 yd less than twice the width, and the area of the rectangle is 33 𝑦𝑑2. Find the dimensions of the rectangle.

10. A model rocket is launched with an initial velocity of 235 ft/s. The rocket’s height ℎ (in feet ) 𝑡 seconds after it is launched is given by the following. ℎ = 235𝑡 − 16𝑡2

a. Find all the values of 𝑡 for which the rocket’s height is 151 feet. Round your answer(s) to the nearest hundredth.

b. Also locate the time when the rocket is at its maximum height.

c. The rocket deploys a parachute just as it reaches maximum height and after that, it falls at a constant

rate of 20 𝑓𝑡

𝑠. Determine how long after the launch

the rocket will hit the ground.


11. The cost 𝐶 in (dollars) of manufacturing 𝑥 wheels at Ravi’s Bicycle Supply is given by the function

𝐶(𝑥) = 0.5𝑥2 − 170𝑥 + 25,850. What is the minimum cost of manufacturing wheels? Do not

round your answer.

12. Consider the parabola shaped graph of the quadratic function 𝑓(𝑥) = −2𝑥2 + 16𝑥 − 34.

a. Complete the square to locate the vertex. Then find the maximum or minimum of the function

𝑓(𝑥) and indicate which it is a maximum, or a minimum.

b. Locate the focus and 𝑥- and 𝑦-axis intercepts and plot the graph of 𝑦 = 𝑓(𝑥) with these features

labled.


13. Find the intersection points of the graphs of functions below. Then sketch graph of both functions, and label the intersection points clearly. 𝑦 = 𝑓(𝑥) = 𝑥2 − 6𝑥 + 4 and 𝑦 = 𝑔(𝑥) = 1 − 2𝑥

14. Find the intersection points of the graphs of both relations below. Then the sketch graph of both the relations, and label the intersection points clearly. 𝑦 = 𝑓(𝑥) = 𝑥2 − 5 and the circle given by 𝑥2 + 𝑦2 = 5.


3.2 Polynomial Equations of Degree Three or Higher When trying to solve higher degree polynomial equations, we resort to the zero-product property of real

numbers just as we did when solving quadratic equations by factoring. The word “trying” is intentional

here because solving polynomial equations can be a very difficult problem and sometimes exact

solutions cannot be found. By taking all terms to one side (with the other side = 0) and factoring (when

possible) into two lower degree polynomials, we reduce the problem of solving a given polynomial

equation to solving two polynomial equations of lower degree. In this chapter, we build on our

observations from chapter 2 where we saw that 𝑥-intercepts of a graph at 𝑥 = 𝑎 correspond to (𝑥 − 𝑎)

being a factor of the polynomial. The key idea is to somehow find a number 𝑥 = 𝑎 where 𝑓(𝑎) = 0.

We call 𝑎 a zero of the function 𝑓 and for polynomials, we’ll see that (𝑥 − 𝑎) is a factor. We’ll use this

idea extensively in what follows. We focus mainly on polynomials with integer coefficients.

We start by considering some observed patterns and rigorously explaining these patterns to establish

Theorems or Laws that will always hold true. One observation concerns the nature of rational solutions

to polynomials with integer coefficients. Consider the polynomial equation (7𝑥 − 5)(3𝑥 + 4) = 0. It

is clear that 𝑥 =5

7 and 𝑥 = −

4

3 are the only solutions by the zero-product property of real numbers. If

we look at the unfactored form of the equation 21𝑥2 + 13𝑥 − 20 = 0, we notice that the numerators

of the zeros are factors of the constant term −20 and the denominators of the zeros are factors of the

leading coefficient 21. Formally stated this pattern is called the Rational Zeros Theorem.

Rational Zeros Theorem:

Let 𝑓(𝑥) = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + 𝑎3𝑥3 + ⋯ + 𝑎𝑛𝑥𝑛 be a non-constant polynomial function of

degree 𝑛 with no common factor between all the integer coefficients 𝑎0, 𝑎1, 𝑎2, 𝑎3, … 𝑎𝑛.

Any rational zero of 𝑓, 𝑎 =𝑝

𝑞 in reduced form must have 𝑝 as factor of the constant term 𝑎0

and 𝑞 as a factor of the leading coefficient 𝑎𝑛.

Thus for 𝑓(𝑥) = 3𝑥3 + 4𝑥 − 5 = 0, this theorem says any rational number zero must be of the form

𝑎 = ±1,5

1,3 . It is easy to check that none of 𝑎 = ±1,

1

3, 5,

5

3 make 𝑓(𝑎) = 0, so all zeros of 𝑓 are

irrational or complex numbers.

Proof: To prove this statement we what is given, i.e., that 𝑓(𝑥) = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + 𝑎3𝑥3 + ⋯ +

𝑎𝑛𝑥𝑛, and that 𝑥 =𝑝

𝑞 is a rational zero of this polynomial.

That means 𝑓 (𝑝

𝑞) = 0 or that 𝑎0 + 𝑎1 (

𝑝

𝑞) + 𝑎2 (

𝑝

𝑞)

2+ 𝑎3 (

𝑝

𝑞)

3+ ⋯ 𝑎𝑛−1𝑥𝑛−1 + 𝑎𝑛 (

𝑝

𝑞)

𝑛= 0 .

Multiply both sides 𝑞𝑛. We get

𝑎0𝑞𝑛 + 𝑎1𝑝𝑞𝑛−1 + 𝑎2𝑝2𝑞𝑛−2 + 𝑎3𝑝3𝑞𝑛−3 + ⋯ 𝑎𝑛−1𝑞 ⋅ 𝑝𝑛−1 + 𝑎𝑛𝑝𝑛 = 0.

Now we can take the first or the last term to the right side to get the equations below.


𝑎0𝑞𝑛 + 𝑎1𝑝𝑞𝑛−1 + 𝑎2𝑝2𝑞𝑛−2 + 𝑎3𝑝3𝑞𝑛−3 + ⋯ 𝑎𝑛−1𝑞 ⋅ 𝑝𝑛−1 = −𝑎𝑛𝑝𝑛 .

And,

𝑎1𝑝𝑞𝑛−1 + 𝑎2𝑝2𝑞𝑛−2 + 𝑎3𝑝3𝑞𝑛−3 + ⋯ 𝑎𝑛−1𝑞 ⋅ 𝑝𝑛−1 + 𝑎𝑛𝑝𝑛 = −𝑎0𝑞𝑛.

In the first equation each term on the left is divisible by q and hence the right side must be too. Since 𝑝

𝑞

is in simplest form, 𝑝 and 𝑞 don’t share any common factors, and all the factors in 𝑞 must be in 𝑎𝑛 or 𝑞

must be a divisor of 𝑎𝑛 .

Likewise, in the second equation above each term on the left is divisible by p and hence the right side

must be too. Since 𝑝

𝑞 is in simplest form, 𝑝 and 𝑞 don’t share any common factors, and all the factors in

𝑝 must be in 𝑎0 or 𝑝 must be a divisor of 𝑎0. This proves our theorem.

Division and Factor Theorems

Recall that when doing long division of numbers or polynomials we can use the division algorithm to

write the dividend in terms of the divisor, the quotient, and the remainder.

Division Algorithm: Let 𝑓(𝑥) be a polynomial of degree 𝑛. On long division of 𝑓(𝑥) ÷ (𝑥 − 𝑎)

for any fixed real number 𝑎, we obtain a remainder of 𝑅 that is a constant since it must be

lower degree than the divisor (𝑥 − 𝑎) and a quotient 𝑄(𝑥) of degree one less than 𝑛. Thus we

can write

𝑓(𝑥) = 𝑄(𝑥)(𝑥 − 𝑎) + 𝑅

We can see by this algorithm then that 𝑓(𝑎) = 𝑄(𝑎)(𝑎 − 𝑎) + 𝑅 = 𝑅. This observation is stated as:

Remainder Theorem: Let 𝑓(𝑥) be a polynomial with real coefficients. Then the remainder when

dividing 𝑓(𝑥) by (𝑥 − 𝑎) is equal to the value of the function at 𝑥 = 𝑎.

𝑅 = 𝑓(𝑎)

Note that when the remainder is zero, then from the division algorithm above, 𝑓(𝑥) = 𝑄(𝑥)(𝑥 − 𝑎) + 0

or 𝑓(𝑥) = 𝑄(𝑥)(𝑥 − 𝑎) and (𝑥 − 𝑎) is a factor of the polynomial. This important result is stated as:

Factor Theorem: Let 𝑓(𝑥) be a polynomial. Then (𝑥 − 𝑎) is a factor of 𝑓(𝑥) if and only if

𝑓(𝑎) = 0.

Complex Zeros of Real Polynomials

We have already seen that not all polynomials have real zeros, e.g., show that 𝑓(𝑥) = 𝑥2 + 4 is never

zero for any real value of 𝑥. Even when polynomials do have real zeros, they may not be rational zeros.

For example, 𝑥2 − 2 = 0 has zeros 𝑎 = √2 and 𝑎 = −√2 both real, but irrational. The quadratic

formula provides complex zeros of any quadratic function when the discriminant is negative. For


example 𝑓(𝑥) = 𝑥2 − 6𝑥 + 13 has zeros 𝑥 =6±√−16

2= 3 ± 2𝑖. When a polynomial 𝑓 has real

coefficients, it turns out that the outputs 𝑓(𝑎 + 𝑏𝑖) and 𝑓(𝑎 − 𝑏𝑖) are complex conjugates of each

other. Thus if either one of these outputs is equal to 0 + 0𝑖, the other must be 0 − 0𝑖 which are both

zero.

Complex Conjugate Zeros Theorem: If 𝑓 is a polynomial with real coefficients then all non-real

zeros come in pairs 𝑎 = 𝛼 + 𝛽𝑖 and �̅� = 𝛼 − 𝛽𝑖.

Fundamental Theorem of Algebra (FTA): Let 𝑓(𝑥) = 𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + 𝑎3𝑥3 + ⋯ + 𝑎𝑛𝑥𝑛, be

a polynomial with coefficients that are real or complex and is of degree 𝑛 ≥ 1. Then 𝑓(𝑥) has

at least one zero in the complex number system. It may be that the zero is actually real.

The proof of this theorem was established by Carl Friedrich Gauss about 200 years ago and is not

obvious. If we look at a selection of polynomials and their zeros, we can see that even for innocent

looking polynomials, the set of real numbers does not always contain a zero of a polynomial. Consider

the polynomials below with real coefficients.

i. 𝑥2 − 𝑥 − 6 = 0 → This factors as (𝑥 − 3)(𝑥 + 2) and the zeros are both integers.

ii. 4𝑥2 − 9 = 0 → This also factors as (2𝑥 − 3)(2𝑥 + 3) and the zeros are rational

numbers 𝑥 = ±3

2.

iii. 𝑥2 − 5 = 0 → Using the square root property, we have 𝑥 = ±√5 . Here we have to

go beyond rational numbers to irrational real numbers to find the zeros.

iv. 𝑥2 − 𝑥 + 6 = 0 → The quadratic formula provides the solutions 𝑥 =1±𝑖√23

2. Here we

needed to go beyond the real numbers to the complex numbers to find a zero of this

innocent looking polynomial.

So it would seem plausible that with a little tinkering, we might come up with a polynomial where we

have to go beyond the complex numbers to find a zero. Well, the Fundamental Theorem of Algebra

says that can’t be done and that every polynomial will have at least one zero somewhere in the complex

plane.

An easy consequence of the FTA is that any polynomial of degree 𝑛 can be factored into linear

factors and written as 𝑓(𝑥) = 𝑎𝑛(𝑥 − 𝑧1)(𝑥 − 𝑧2) … (𝑥 − 𝑧𝑛) where the 𝑧𝑖 are the zeros of the

polynomial.

Some of the 𝑧𝑖 may be repeated. Some may be real and some may be complex.

The proof of this result follows from repeated use of the Factor Theorem and the FTA. Let’s say we

start with an 𝑛𝑡ℎ degree polynomial. Then the FTA says there is some zero, say 𝑥 = 𝑧1. Then the Factor

Theorem says that we can write 𝑓(𝑥) = 𝑄1(𝑥)(𝑥 − 𝑧1) with the degree of 𝑄1 being 𝑛 − 1. Now if 𝑛 >

1, then the FTA insures that 𝑄1 has a zero say 𝑥 = 𝑧2 . Now the Factor Theorem on 𝑄1 says 𝑄1(𝑥) =

𝑄2(𝑥)(𝑥 − 𝑧2) with the degree of 𝑄2 being 𝑛 − 2. Thus we have 𝑓(𝑥) = 𝑄2(𝑥)(𝑥 − 𝑧2)(𝑥 − 𝑧1). We


can keep doing this 𝑛 time until we get to 𝑄𝑛 which will be of degree zero and will be equal to the

leading coefficient of 𝑓(𝑥).

In the case where the coefficients or 𝑓 are real numbers, then the Conjugate Zeros Theorem says that all

the non-real complex zeros appear in conjugate pairs of the form 𝑧𝑖 = 𝛼 + 𝛽𝑖 and 𝑧𝑖+1 = 𝛼 − 𝛽𝑖. Thus

if 2 + 3𝑖 is a zero, then 2 − 3𝑖 is a zero too. One immediate consequence of this is that real

polynomials must always have an even number of non-real complex zeros. If one takes all the complex

pair factors of the form (𝑥 − (𝛼 + 𝛽𝑖))(𝑥 − (𝛼 − 𝛽𝑖)) and expands them, you get real quadratic factors

of the form (𝑥2 − 2𝛼𝑥 + 𝛼2 + 𝛽2). Thus any real polynomial can be written in terms of the product of

real linear factors (𝑥 − 𝑧𝑖) for each real zero and real quadratic factors of the form (𝑥2 − 2𝛼𝑖𝑥 + 𝛼𝑖2 +

𝛽𝑖2) for each pair of non-real complex zeros.

Practice Problems

1. Find the factored form of each polynomial below

a. 𝑥2 − 4𝑥 + 1

We find the zeros as 𝑥 =(4±√16−4)

2=

4±2√3

2= 1 ± √3.

Thus the factored form is (𝑥 − (1 + √3)) (𝑥 − (1 − √3)) 𝑜𝑟 (𝑥 − 1 − √3)(𝑥 − 1 + √3)

b. 𝑥3 − 8

We see that 𝑥 = 2 is a real zero. This factors as a difference of cubes as (𝑥 − 2)(𝑥2 +

2𝑥 + 4).

Now find the zeros of 𝑥2 + 2𝑥 + 4 as 𝑥 =−2±√4−16

2=

−2±2√3 𝑖

2= −1 ± √3𝑖.

Finally, we have:𝑥3 − 8 = (𝑥 − 2)(𝑥 + 1 − 𝑖√3)(𝑥 + 1 + 𝑖√3)

c. 𝑥4 − 3𝑥2 − 4

We can factor this as (𝑥2 − 4)(𝑥2 + 1) and the zeros of the first are ±2 and of the second

are ±𝑖. Thus the factored form is 𝑥4 − 3𝑥2 − 4 = (𝑥 − 2)(𝑥 + 2)(𝑥 − 𝑖)(𝑥 + 𝑖)

2. Find real polynomials given the following information about their zeros.

a. 𝑓(𝑥) is degree 3 and has zeros 2, 3, and 3

5.

With these zeros, 𝑓 must be factored as 𝑎(𝑥 − 1)(𝑥 − 3) (𝑥 −3

5) where 𝑎 is the leading

coefficient that is not known. We might multiply the last factor by 5 and make the 𝑎 out

front 1

5 as big to have 𝑓(𝑥) = 𝑐(𝑥 − 1)(𝑥 − 3)(5𝑥 − 3). We’d need to know the value of 𝑓

at some other point to figure out what the leading coefficient should be. For example if we

were given that 𝑓(2) = 5, we substitute 2 and 5 for the input and output to get:

5 = 𝑐(1)(−1)(7) so that 𝑐 = −5

7 and 𝑓(𝑥) = −

5

7(𝑥 − 1)(𝑥 − 3)(5𝑥 − 3).

b. 𝑓(𝑥) is degree 5 with zeros -4, 2 − 3𝑖 and −1 + 4𝑖.

The conjugates 2 + 3𝑖 and −1 − 4𝑖 are also zeros. The factored form of 𝑓(𝑥) is

𝑓(𝑥) = 𝑎(𝑥 + 4)(𝑥 − (2 − 3𝑖))(𝑥 − (2 + 3𝑖))(𝑥 − (−1 + 4𝑖))(𝑥 − (−1 − 4𝑖))


If we multiply the conjugate pair factors we get two quadratic factors and a linear factor and

𝑓(𝑥) = 𝑎(𝑥 + 4)(𝑥2 − 4𝑥 + 13)(𝑥2 + 2𝑥 + 17).

c. Suppose 𝑓(𝑥) is a real polynomial of degree 11 and its zeros include: 3, 5𝑖, −2 − 4𝑖.

i. List another zero of 𝑓.

For sure, the conjugates −5𝑖 and −2 + 4𝑖 must also be zeros of 𝑓.

ii. What is the maximum number of real zeros 𝑓 could have?

Since the total number of zeros can be no more than 11 and at least 4 of them are

non-real, we could have at most 7 real zeros.

iii. What is the maximum number of non-real zeros that 𝑓 could have?

With 11 zeros total and one known real zero at 𝑎 = 3, the other 10 zeros could

potentially all be non-real, (5 pairs of conjugates).

iv. If the leading coefficient of 𝑓 is −8, give several possible expressions for 𝑓(𝑥).

With the 5 known zeros, there are 6 that could be chosen, so several examples of

what 𝑓 might look like in factored form are:

𝑓(𝑥) = −8(𝑥 − 3)(𝑥 − 5𝑖)(𝑥 + 5𝑖)(𝑥 + 2 + 4𝑖)(𝑥 + 2 − 4𝑖)𝑥6

𝑓(𝑥) = −8(𝑥 − 3)7(𝑥 − 5𝑖)(𝑥 + 5𝑖)(𝑥 + 2 + 4𝑖)(𝑥 + 2 − 4𝑖)

𝑓(𝑥) = −8(𝑥 − 3)(𝑥 − 5𝑖)(𝑥 + 5𝑖)(𝑥 + 2 + 4𝑖)(𝑥 + 2 − 4𝑖)(𝑥 + 6)3(𝑥 − 9)3

Using Division to Find Zeros

When presented with finding zeros of a complicated polynomial function, often one or more zeros are

known or can be found to a high degree of accuracy. In this scenario the Factor Theorem says our

polynomial can be written as 𝑓(𝑥) = (𝑥 − 𝑘𝑛𝑜𝑤𝑛 𝑧𝑒𝑟𝑜)𝑄(𝑥) and further zeros of 𝑓 must be zeros of

𝑄(𝑥) which is one lower in degree than 𝑓. We can find this function 𝑄 by long division of 𝑓(𝑥) by

(𝑥 − 𝑘𝑛𝑜𝑤𝑛 𝑧𝑒𝑟𝑜). If we know a zero of 𝑄 we can divide out the corresponding factor and reduce the

problem to one lower in degree again. If we can get down to a quadratic, we can always resort to the

quadratic formula to find the last two zeros. We’ll take a brief look at long division and develop a

streamlined version called “Synthetic Division” which is helpful for our present purpose in dividing by

expressions of the form (𝑥 − 𝑎).

We used long division of polynomials to find oblique asymptotes of rational functions. See example

below for (6𝑥2 + 7𝑥 + 4) ÷ (𝑥 − 2). As you can see in long division, the first term of the quotient will

always have the leading coefficient of the dividend but 𝑥 will be to one degree less. Subtracting out the

second term amounts subtract (6) × (−2) or adding 6 × 2. When dividing by 𝑥 − 𝑎 these second

terms will always be subtracting (#) × (−𝑎). The steps under Synthetic Division column convert all

these (subtracting negatives) to adding and we end up with the synthetic division algorithm described

just after the diagram.


Long Division

Synthetic Division

Synthetic Division Steps for dividing a polynomial by a divisor of the form (𝑥 − 𝑎).

Step 1: The coefficients of the dividend polynomial go in the boxes in the top row in descending powers

with zeros inserted when powers are missing. In the example above we put 6 7 4 to indicate the

dividend.

Step 2: When dividing by 𝑥 − 𝑎 put 𝒂 ( Here 𝑎 = 2. ) in the second row goes in the second row to Left

of the vertical bar.

Step 3: Then bring the leading coefficient down to row three and this is the first coefficient of the

quotient.

Step 4: Multiply 𝑎 by the first number in the bottom row and put the result in the second row under the

coefficient of the next term in the dividend. Add the two terms in that column and put that resulting

number in the last row in that column. Then multiply that number by 𝑎 = 2 and put that number in the

second row just below the third coefficient of the dividend and add. Continue the process and the final

entry in the bottom row is the remainder. The last number in the last row is the remainder. The

remaining numbers are the coefficients of the quotient polynomial which is one degree lower than the

dividend.

This method is very fast and the addition avoids the common mistakes of subtracting negatives. Now

we will utilize this to solve higher degree polynomial equations.

6𝑥 + 19

(𝑥 − 2) 6𝑥2 + 7𝑥 + 4

−(6𝑥2 − 12𝑥)

19𝑥 + 4

−(19𝑥 − 38)

42

Quotient 6𝑥 + 19

Remainder is 42

6 7 4 2 12 38

6

19 42

Quotient 6𝑥 + 19

Remainder is 42


Practice Example:

1. Find all the zeros of 3𝑥4 + 2𝑥3 − 4𝑥2 − 29𝑥 + 10 = 0

Solution:

The potential rational zeros are ±𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 10

𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 3.

Thus the possible rational zeros could be ±1, ±1

3, ±2, ±

2

3, ±5, ±

5

3, ±10, ±

10

3,

We can now use synthetic division to see if any of these are zeros. We could also just substitute

each of these into the function to see if we get zero for any of the outputs. The advantage of doing the

division is that it gives you the quotient which can be used to find additional zeros.

3 2 −4 −29 10

−2 −6 8 −8 74

3 −4 4 −37 84

3 2 −4 −29 10

2 6 16 24 −10

3 8 12 −5 0

𝑥 = 2 is a zero so we can work with the degree-3 polynomial 3𝑥3 + 8𝑥2 + 12𝑥 − 5 which is our quotient

3 8 12 −5 1

3

1 3 5

3 9 15 0

𝑥 =1

3 is a zero also and now the quotient is

3𝑥2 + 9𝑥 + 15 = 0. Note that we can factor 3 out of the quotient before plugging into the quadratic formula. Thus we seek zeros of 𝑥2 + 3𝑥 + 5. We get the remaining zeros to be

𝑥 =−3±√9−4(1)(5)

2 after simplifying you will get 𝑥 = −

3

2+

√11

2𝑖, −

3

2−

√11

2𝑖

So all our solutions of 3𝑥4 + 2𝑥3 − 4𝑥2 − 29𝑥 + 10 = 0 are

𝑥 =1

3, 2,

3

2+

√11

2𝑖, −

3

2−

√11

2𝑖. We could also use this to write the polynomial in factored form as:

3𝑥4 + 2𝑥3 − 4𝑥2 − 29𝑥 + 10 = 3(𝑥 −1

3)(𝑥 − 2)(𝑥 −

3 + √11𝑖

2)(𝑥 −

3 − √11𝑖

2)

2. Find the zeros of 𝑓(𝑥) = 4𝑥5 − 13𝑥3 − 8𝑥2 + 3𝑥 + 2 and write it in factored form. Solution:

The potential rational zeros are: 𝑎 = ±1 𝑜𝑟 2

1 𝑜𝑟 2 𝑜𝑟 4= ± {1,

1

2,

1

4, 2} .

Synthetic division shows that 𝑥 = 2, −1

2 and

1

2 are zeros and we get:

4 0 -13 -8 3 2 We are left with the quadratic 4(𝑥2 + 2𝑥 + 1) = 0

which factors as 4(𝑥 + 1)2 = 0. Thus −1 is a zero of multiplicity 2.

The zeros are thus 𝑥 = −1, −1

2,

1

2, 𝑎𝑛𝑑 2

The factored form of 𝑓 is:

𝑓(𝑥) = 4(𝑥 + 1)2 (𝑥 +1

2) (𝑥 −

1

2) (𝑥 − 2).

𝑎 = 2 8 16 6 -4 -2

4 8 3 -2 -1 0

𝑎 =1

2

2 5 4 1

4 10 8 2 0

𝑎 = −1

2

-2 -4 -2

4 8 4 0


3. Given the graph of the polynomial 𝑓(𝑥) = 2𝑥3 + 𝑥2 + 4𝑥 − 15. Find all the zeros of 𝑓 and the complete factored form of 𝑓.

Solution:

From the graph we see that 𝑥 =3

2 is a zero

and by synthetic division we get: 2 1 4 -15 3/2 3 6 15

2 4 10 0 Thus the quadratic quotient is: 2(𝑥2 + 2𝑥 + 5) which by the quadratic formula has complex zeros 𝑥 = −1 ± 2𝑖

The zeros are: 𝑥 =3

2, −1 + 2𝑖, −1 − 2𝑖

𝑓(𝑥) = 2 (𝑥 −3

2) (𝑥 + 1 − 2𝑖)(𝑥 + 1 + 2𝑖)




Rational Zeros Theorem

Factor Theorem

Remainder Theorem

Complex Conjugates Zeros Theorem

Real Zeros

Complex Zeros

Synthetic Division

Fundamental Theorem of Algebra



Exercises 3.2

1. Let 𝑃(𝑥) be a polynomial.

a. What can you say about the non-real zeros of the equation 𝑃(𝑥) = 0, if the coefficients

of 𝑃(𝑥) are real?

b. What is the maximum number of non-real zeros a polynomial of odd degree can have?

c. Can a polynomial of odd degree have all non-real zeros?

d. Why must an odd degree polynomial with real coefficients have at least one real zero?

e. Can you have a polynomial with non-real coefficients of odd degree with all non-real

zeros?


2. Suppose 𝑅(𝑥) is a polynomial of degree 13 whose coefficients are real numbers. Also, suppose that 𝑅(𝑥) has the following zeros: 7, −8 , 5𝑖, −2 − 4𝑖.

a) Find another zero of 𝑅(𝑥).

b) What is the maximum number of real zeros that 𝑅(𝑥) can have?

c) What is the maximum number of non-real zeros that 𝑅(𝑥) can have?

d) If the leading coefficient of the polynomial is −3, list three possible formulas for 𝑅(𝑥) with the degree and zeros given above.

3. Find a polynomial 𝑓(𝑥) of degree 4 that has the following zeros −2, 1, −6, 𝑎𝑛𝑑 0. Leave your

answer in factored form.

4. Find the real polynomial of degree 4 with zeros 𝑎 = 3, −1

2, 𝑎𝑛𝑑 2 − 3𝑖 and leading coefficient


5. Perform the following division to find the quotient and the remainder. Write your answer in the

Quotient +𝑅𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟

𝑥−4 form. When possible use synthetic division.

a. (6𝑥2 + 37𝑥 + 39) ÷ (𝑥 + 5)

b. (3𝑥2 − 𝑥3 + 6𝑥 − 8) ÷ (𝑥 − 4)

c. (24𝑥3 + 4𝑥2 + 14𝑥 + 3) ÷ (6𝑥 − 2)


6. Use the rational zeros theorem to list all possible zeros of the following. Be sure that no value in

your list appears more than once.

a. 𝑔(𝑥) = −5𝑥4 − 𝑥3 − 3𝑥2 − 3𝑥 + 3

b. 𝑔(𝑥) = −6𝑥4 − 𝑥3 − 3𝑥2 − 3𝑥 + 9


7. The function below has at least one rational zero. Use this fact to find all the zeros of the function.

𝑔(𝑥) = 4𝑥3 + 12𝑥2 − 𝑥 − 3

If there is more than one zero, separate them with commas. Write exact values, not decimal

approximations.

8. The function below has a rational zero. Use this fact to find all the zeros of the function.

𝑓(𝑥) = 2𝑥3 − 3𝑥2 + 𝑥 − 6



Write exact values, not decimal approximations.

𝑔(𝑥) = 7𝑥4 + 20𝑥3 + 10𝑥2 − 5𝑥 − 2

10. For the polynomial below −2 is a zero. Express ℎ(𝑥) as a product of linear factors.

ℎ(𝑥) = 𝑥3 + 8𝑥2 + 30𝑥 + 36.



ℎ(𝑥) = 5𝑥4 − 29𝑥3 − 40𝑥2 − 13𝑥 − 7

12. Find all the other zeros of 𝑃(𝑥) given that 3−3𝑖 is a zero. (Hint: find a quadratic factor with real

zeros, and then use long division to obtain a second quadratic factor.)

𝑃(𝑥) = 𝑥4 − 7𝑥3 + 22𝑥2 − 6𝑥 − 36


13. Find all the zeros of 𝑓(𝑥) = 12𝑥4 + 5𝑥3 + 46𝑥2 + 20𝑥 − 8 whose graph is given below.

14. Find the exact values of the points of intersection of the two functions below.

𝑓(𝑥) = 3𝑥5 − 2𝑥4 − 4𝑥3 − 5𝑥2 + 6𝑥 − 10 and 𝑔(𝑥) = 𝑥5 + 𝑥4 − 2𝑥3 − 8𝑥2 + 10𝑥 − 16

15. Create two polynomial functions that intersect each other at 𝑥 = 2, 𝑥 = −3, 𝑥 = 2√3 , and

𝑥 = −2√3.


3.3 Exponential and Logarithmic Equations

In logarithmic and exponential equations, the unknown appears in the equation as an input to a

logarithmic or exponential function. In this section we deal with equations where the variable appears

only in a logarithm or only in an exponent and not both in the same equation. The standard method of

solving these equations is to isolate the logarithmic or exponential function expression on one side of

the equation and then operate on both sides of the equation with the inverse function. Thus for the

logarithmic equation log2(𝑥 − 4) = 7 we operate on both sides with the 2𝑥 function to get (𝑥 − 4) =

27. This is referred to as exponentiating both sides of the equation. Likewise, we’d operate on both

sides of an exponential equation with a logarithmic function to undo the exponential function. This is

much like what you do with radical or power equations, e.g., for 𝑥3 = 7, you undo the cube function by

operating on both sides with the cube root function to get 𝑥 = √73

.

Logarithmic Equations and Inequalities

The basic idea for solving logarithmic equations is to isolate the logarithm expression on one side and

then exponentiate both sides. This is equivalent to converting the logarithm statement into its

exponential form as shown below:

loga (𝑥-expression) = # → 𝑎loga (𝑥−expression) = 𝑎# → (𝑥-expression)= 𝑎#

Once the logarithm is undone, further steps may be needed to isolate the variable 𝑥.

Practice Example

1. log(2𝑥 − 1) = 5

Solution:

The common logarithm here is base-10 and we exponentiate by taking 10𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 to get.

2𝑥 − 1 = 105 2𝑥 = 100,001 𝑥 = 50,000.5 Check log(2 ⋅ 50,000.5 − 1) = log(100,001 − 1) = log 100,000 = 5 It is important to check the answer since the domain of a logarithmic function is restricted.

2. log3(𝑥 − 1) = −2

Solution:

Here we operate on both sides with the 3( ) function to undo the log3( ).

𝑥 − 1 = 3−2

𝑥 = 1 +1

9=

10

9

Check that the answer works.

Convert from logarithmic to exponential form.


When more than one logarithmic term appears in an equation, we can often use properties of

logarithms to combine them into a single logarithm and then proceed as above. We list these properties

below:

𝑙𝑜𝑔𝑎(𝑢𝑣) = 𝑙𝑜𝑔𝑎(𝑢) + 𝑙𝑜𝑔𝑎(𝑢)

𝑙𝑜𝑔𝑎 (𝑢

𝑣) = 𝑙𝑜𝑔𝑎(𝑢) − 𝑙𝑜𝑔𝑎(𝑢)

𝑙𝑜𝑔𝑎(𝑢𝑛) = 𝑛𝑙𝑜𝑔𝑎𝑢

3. Solve log2(𝑥 + 7) + log2 𝑥 = 3

Solution:

Combine the log terms to get: log2((𝑥 + 7)𝑥) = 3

Exponentiate taking 2𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒: 𝑥2 + 7𝑥 = 23 = 8.

Take to one side and factor: 𝑥2 + 7𝑥 − 8 = (𝑥 + 8)(𝑥 − 1) = 0

Answers: 𝑥 = −8 𝑜𝑟 𝑥 = 1.

Checking the answers:

𝑥 = −8 → log2(−1) + log2(−8) these are undefined and hence this solution is extraneous!

𝑥 = 1 → log2(8) + log2 1 = 3 + 0 = 3 so, The solution is just 𝑥 = 1.

4. Solve log4(𝑥² − 15𝑥 − 16) = log4(𝑥 + 1) + 3

Solution:

Taking the log terms to the left side and combining, we get: log4𝑥²−15𝑥−16

𝑥+1= 3

Exponentiating we get 𝑥²−15𝑥−16

𝑥+1= 64 .

The left side reduces to (𝑥 − 16) and solving we get 𝑥 − 16 = 64 → 𝑥 = 80

Check this answer in the original equation to get: log4(5184) = log4(81) + 3.

Subtract log4 81 and combine to get log45184

81= log4 64 = 3 . So 𝑥 = 80 is the solution.

Solving logarithmic inequalities is done much like for equalities except we must be careful when

undoing the logarithm, i.e, taking 𝑎 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 on both sides of an inequality. The question is whether

the new inequality will be true exactly when the original inequality was true. The answer comes

down to whether the undoing function 𝑎𝑥 is increasing for all 𝑥. Note that the exponential function

𝑎𝑥 is increasing when 𝑎 > 1, but decreasing when 0 < 𝑎 < 1 . Consider operating with the 𝑒𝑥

function to undo a natural logarithm in an inequality. If we start with 𝑎 < 𝑏 will it be true that

𝑒𝑎 < 𝑒𝑏 ? The answer is clearly yes since the 𝑦 = 𝑒𝑥 graph increases as we move from 𝑥 = 𝑎 to the

right to 𝑥 = 𝑏. Note that if we were to operate with (1

2)

𝑥 on both sides, this is a decreasing function

and would thus reverse the order of the inequality. For example 2 < 5, but (1

2)

2=

1

4> (

1

2)

5=

1

32.

Luckily mostly the logarithms we deal with are either the common logarithm with base 10 or the


natural logarithm with base 𝑒 and thus we can exponentiate both sides with these bases and not

worry about switching the direction of the inequality. One must also check that the expression

contained within the logarithm is in the domain of the logarithm.

Aside: We saw this switching of inequalities before when we multiplied or divided both sides of an

inequality by (-1). That operation of ÷ (−1) can be thought of as operating with the decreasing

function 𝑓(𝑥) = −𝑥 function on both sides.

5. log3(𝑥 − 1) ≤ −2

Solution:

First off, we must have 𝑥 − 1 > 0 (or 𝑥 > 1) in order for the logarithm to be defined.

Next operate on both sides with 3𝑥 (which preserves the direction of the inequality since 𝑦 = 3𝑥 is

an an increasing function).

𝑥 − 1 ≤ 3−2

𝑥 ≤ 1 +1

9 or 𝑥 ≤

10

9 and also 𝑥 > 1

Thus the solutions is the interval 1 < 𝑥 ≤ 11

9.

Exponential Equations and Inequalities

In exponential equations, the unknown appears within the exponent on some base. These equations

show up in financial problems and growth and decay problems where exponential models are often

used.

Consider the annual compounding problem of determining how long it would take an investment to

double in value given that it grows by 5.6% each year. This is annual compounding of interest and is

describe by 𝐴 = 𝑃(1.056)𝑡 where 𝑃 is the initial value and 𝐴 is the value 𝑡 years later. The doubling

problem is to find what 𝑡 should be so that 𝐴 = 2𝑃, or 2𝑃 = 𝑃(1.056)𝑡. We could divide out the 𝑃 so

that our doubling problem comes down to solving 2 = (1.056)𝑡 which is a typical exponential

equation. To undo the exponential, one might operate on both sides with the base 1.056 logarithm.

The drawback to doing this is that it is not readily apparent how to evaluate log1.056 2 = 𝑡. A more

practical method is to just take the natural or common logarithm on both sides since these are available

on any scientific calculator. The right side is then simplified using the third property of logarithms

above.

Thus we’d solve 2 = (1.056)𝑡 as: ln 2 = ln 1.056𝑡 = 𝑡 ln 1.056 and 𝑡 =ln 2

ln 1.056≈ 12.72 .

Or, using the common logarithm, log 2 = log 1.056𝑡 = 𝑡 and 𝑡 =log 2

log 1.056≈ 12.72 .

The basic tool for solving exponential equations is to isolate the exponential expression on one side and

take either the natural or common logarithm of both sides. For inequalities, we isolate the exponential

so both sides are a positive quantity and then operate on both sides with either the common or natural

logarithm functions. Since both of these are increasing functions, we don’t need to alter the direction

of the inequality in this step of taking logs on both sides.


Sample Problems:

1. Solve the equations and inequalities below.

a) 2𝑥 = 3 Solution: We take the common (or could use natural log) of both sides to get: log 2𝑥 = log 3 𝑥 log(2) = log (3) and therefore

𝑥 =log 3

log 2

This is the exact solution. An approximate solution by a scientific calculator is

𝑥 =𝑙𝑜𝑔3

𝑙𝑜𝑔2≈ 1.585

Graphically that means the two functions 𝑦 = 2𝑥 and

𝑦 = 3 intersect at 𝑥 =𝑙𝑜𝑔3

𝑙𝑜𝑔2≈ 1.585. Or that the 𝑥-

intercept of the function 𝑓(𝑥) = 2𝑥 − 3 is at

𝑥 =𝑙𝑜𝑔3

𝑙𝑜𝑔2≈ 1.585

a) 2𝑥 < 3 Solution: We take ln (or could use log ) on both sides.

𝑥 ln 2 < ln 3 As with the equation, we divide by the positive number ln 2 to get:

𝑥 <ln 3

ln 2≈ 1.585

Solution: (−∞,ln 3

ln 2)

b) The half-life of penicillin (a form of antibiotic) is about 35 minutes for an adult with normal renal function (or kidneys). A typical dose for an adult is between 250-3000mg 2 to 4 times a day to treat different kinds of bacterial infections. An adult was accidentally given an overdose of 6000𝑚𝑔. How long will it take for the amount in this person to come down to 200mg? Assume this person has normal kidney function and a half-life of 35 minutes.

Solution:

The model is: 6000 (1

2)

𝑡

35= 200.

To undo here we first divide both sides by 6000 and then take natural logarithms on both sides.

(1

2)

𝑡

35=

200

6000 or 𝑙𝑛 (

1

2)

𝑡

35= 𝑙𝑛 (

200

6000)

𝑡

35ln (

1

2) = 𝑙𝑛 (

200

6000)

𝑡 =35𝑙𝑛 (

2006000)

ln (12)

≈ 171.74 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 ≈ 2 ℎ𝑜𝑢𝑟𝑠 51.6 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

So assuming you drink a lot of fluids to flush things out of the body it will take just a little under 3 hours.

(1.585,3)


c) 2𝑥−5 = 45−𝑥 Solution:

Method 1: We might notice that 4 = 22, so we can rewrite the right side as:

2𝑥−5 = (22)5−𝑥

2𝑥−5 = 210−2𝑥 That means the exponents must be the same. 𝑥 − 5 = 10 − 2𝑥 3𝑥 = 15

Solution: 𝑥 = 5

Method 2: We can take natural logarithm of both sides to get: (𝑥 − 5) ln 2 = (5 − 𝑥) ln 4 Now expand and collect the 𝑥-terms so:

𝑥 ln 2 − 5 ln 2 = 5 ln 4 − 𝑥 ln 4 →

𝑥(ln 2 + ln 4) = 5(ln 4 + ln 2) → Solution: 𝑥 = 5

d) 2𝑥−5 = 34−𝑥 Solution: Here we have to use logarithmic functions.

ln (2𝑥−5) = 𝑙𝑛(34−𝑥) (𝑥 − 5) ln 2 = (4 − 𝑥) ln 3 Since 𝑙𝑛2 and 𝑙𝑛3 are just numbers we treat them like it and solve the equation above as a linear equation can be solved. 𝑥 ln 2 − 5 ln 2 = 4 ln 3 − 𝑥 ln 3 𝑥 ln 2 + 𝑥 ln 3 = 4 ln 3 + 5 ln 2 𝑥(ln 2 + ln 3) = 4 ln 3 + 5 ln 2

𝑥 =4 ln 3 + 5 ln 2

ln 2 + ln 3

This is the exact solution to the equation and approximate solution can be found by using your calculators which would give you

𝑥 =4𝑙𝑛3+5𝑙𝑛2

𝑙𝑛2+𝑙𝑛3≈ 4.387

e) Amy invested $4000 at 5% interest compounded quarterly. How many years will she will have

to wait for her money to grow to $5000. Solution:

The quarterly compounding formula is: 4000 (1 +0.05

4)

4𝑡= 5000.

We use the natural logarithmic function to isolate the unknown in the exponent.

(1 +0.05

4)

4𝑡

=5000

4000

4𝑡 ⋅ ln (1 +0.05

4) = ln (

5

4)

𝑡 =ln(

5

4)

4 ln(1+0.05

4)

≈ 4.49 yrs

She would have to wait about 4 years and approximately 6 months.


f) log3(𝑥 − 1) = log3(5𝑥 − 7)

Since the logarithmic function is one-to-one we get: 𝑥 − 1 = 5𝑥 − 7 Or −4𝑥 = −6

𝑥 =−6

−4=

3

2

Check your answer please….

g) log3(𝑥 − 1) − log3(5𝑥 − 7) = 2 We can use properties of logarithms to solve this problem.

log3 (𝑥 − 1

5𝑥 − 7) = 2

Now using the fact that logarithmic function is the inverse function of the exponential function we get

𝑥 − 1

5𝑥 − 7= 32

𝑥 − 1 = 9(5𝑥 − 7) 𝑥 − 1 = 45𝑥 − 63

62 = 44𝑥

𝑥 =62

44=

31

22

We need to check this solution, so go ahead do that.

h) log 2 + log(𝑥 − 1) = log(3𝑥 − 1) We can use properties of logarithms to solve this problem

log(2(𝑥 − 1)) = log(3𝑥 − 1)

Using the fact that log functions are one-to-one we get 2𝑥 − 2 = 3𝑥 − 1

𝑥 = −1 Note that the domain of log (𝑥 − 1) is 𝑥 > 1 so we have an extraneous solution There is no solution to this problem.


Application Problems 1. A sample from a spruce tree that was discovered buried under 40 feet of clay while digging a

well in eastern Wisconsin was found to contain 20% of its original carbon-14. This tree was

buried thousands of years ago by glaciers advancing out of Lake Michigan and pushing the clay

lake-bottom ahead of them. It is known that the half-life of carbon-14 is 5730 years. Estimate

how long ago the tree lived.

Solution:

The amount of a radioactive compound remaining is given in terms of the initial amount and

half-life by: 𝐴(𝑡) = 𝐴0 (1

2)

𝑡

5730.

It is known that we now have only 0.2𝐴0 remaining, so: 0.2 𝐴0 = 𝐴0 (1

2)

𝑡

5730.

We divide out the 𝐴0 and take natural logarithms on both sides to get:

ln 0.2 =𝑡

5730ln

1

2

Isolate 𝑡, by multiplying by 5730 and dividing by ln1

2 to get: 𝑡 =

5730 ln 0.2

ln1

2

≈ 13,300 𝑦𝑒𝑎𝑟𝑠.

2. How long will it take for $5000 earning 6% interest compounded quarterly to have the same

value as $400 which earns 11% interest compounded continuously?

Solution:

We set the value for each account equal to get:

5000 (1 +0.06

4)

4𝑡

= 400𝑒0.11𝑡

We can divide by 400 and then take logarithms of both sides. On the left, we have a product

and we will use the product property of logarithms to separate the two factors since the

exponent 4𝑡 is not applied to the factor 12.5.

÷ 400 → 12.5 (1.015)4𝑡 = 𝑒0.11𝑡

Take logarithms → ln 12.5 + ln 1.0154𝑡 = 0.11𝑡

Bring down the 4𝑡 and collect the 𝑡-terms on the right →

→ ln 12.5 + 4𝑡 ln 1.015 = 0.11𝑡

→ ln 12.5 = 𝑡(0.11 − 4 ln 1.015)

→ 𝑡 =ln 12.5

0.11 − 4 ln 1.015≈ 50 𝑦𝑒𝑎𝑟𝑠


3. Determine the 𝐻+ concentration range if the 𝑝𝐻 is to be kept between 4.5 and 6.8 .

Solution:

The definition of 𝑝𝐻 is: 𝑝𝐻 = − log[𝐻+] with [𝐻+] being the 𝐻+concentration, we require that:

4.5 < − log[𝐻+] < 6.8

We first divide by (-1) and then operate with the 10𝑥 function to undo the log .

−4.5 > log[𝐻+] > −6.8

10−4.5 > [𝐻+] > 10−6.8

Approximating: 0.000000158< [𝐻+] < 0.0000316

4. A cooked turkey’s temperature is modeled by the function 𝑇(𝑡) = 74 + 115𝑒−0.02𝑡 where the

input 𝑡 is the number of minutes after it was taken out of the oven and the temperature is in

degrees Fahrenheit. Determine the time range when the bird temperature will be between 140

and 160 degrees.

Solution:

We require: 140 ≤ 74 + 115𝑒−0.02𝑡 ≤ 160

Subtract 74, then ÷ by 115, then take ln to get:

66 ≤ 115𝑒−0.02𝑡 ≤ 86 → 66

115≤ 𝑒−0.02𝑡 ≤

86

115 → ln

66

115≤ −0.02𝑡 ≤ ln

86

115

Dividing by −0.02 and approximating we get: 14.5 𝑚𝑖𝑛 ≤ 𝑡 ≤ 27.88 𝑚𝑖𝑛.




List different methods of solving exponential equations

List different methods of solving exponential equations



Exercises 3.3

1. How are the product, quotient and power properties of logarithmic functions used to solve

logarithmic or exponential equations?

2. List at least three applications of being able to solve exponential and logarithmic equations not

listed in the section.

Find all the solutions to the following equations.

3. 𝑙𝑜𝑔3(𝑥 + 1) = 2

4. 𝑙𝑜𝑔3(𝑥 + 1) − 𝑙𝑜𝑔3(𝑥) = 2

5. 𝑙𝑜𝑔3(2𝑥 − 1) + 𝑙𝑜𝑔3(𝑥 + 1) = 2


6. 𝑙𝑜𝑔4(−1 − 2𝑥) = −1

7. 4 + log(2𝑥 − 1) = 5

8. 𝑙𝑜𝑔5(𝑥 − 3) = 1 − 𝑙𝑜𝑔5(𝑥 − 7)

9. ln(𝑥 + 4) − ln 18 = ln 5


10. 125 = 25−𝑥−2

11. 2𝑥2−61𝑥 = 643−9𝑥

12. 15−8𝑦 = 6

(Round your answer to the nearest hundredth. Do not round any intermediate computations.)


13. 𝑒−8𝑢 = 6

(Round your answer to the nearest hundredth. Do not round any intermediate computations.)

14. 17−𝑥−3 = 16−8𝑥

(Write the exact answer using base-10 logarithms)

15. 3𝑥−1 = 52𝑥−1

(Write the exact answer using natural logarithms)


16. 200𝑒0.05𝑡 = 50

17. 3𝑒2𝑥 − 5𝑒𝑥 + 2 = 0 (Hint a substitution 𝑢 = 𝑒𝑥 will be useful.)

18. 9𝑥 − 3𝑥 − 2 = 0 (Hint a substitution 𝑢 = 3𝑥 will be useful

19. 1500 (1 +0.05

4)

4𝑡= 3000


20. Write a word problem that would give rise to the exponential equation below and solve the

equation

2000(1.02 )4𝑡 = 5000


21. Write a word problem that would give rise to the exponential equation below and solve the

equation.

20 = 250 (1

2)

𝑡

5730

22. A car is purchased for $28,500. After each year the resale value decreased by 35%. What will

be the resale value be after 4 years? Also determine when the resale value will be less than

$5000. Round your answers to the nearest whole number.

23. A loan of $39,000 is made at 5% interest, compounded annually. Assuming no repayment is

made, after how many years will the amount due reach $63,000 or more? (Use a calculator if

necessary.) Write the smallest possible whole number answer.


24. The number of bacteria in a certain population increases according to a continuous exponential

growth model, with a growth rate parameter of 4.1% per hour. How many hours will it take for

the sample to double?

Note: This is a continuous growth model. (𝐴(𝑡) = 𝐴0𝑒𝑟𝑡)

Do not round any intermediate computations, and round your answer to the nearest hundredth of an

hour.

25. An initial amount of $1800 is invested in an account at an interest rate of 2% per year

compounded continuously. Find the amount in the account after 6 years. Round your answer to

nearest cent. Also how many years are required for the value to reach $10,000.

26. The light intensity in Lake Superior decreases due to absorption by 12% for each meter of depth.

If the intensity just below surface is at 150𝑤

𝑚2. Write a formula for the intensity at dept 𝑥

meters. Also determine how deep one must go for the light intensity to be 0.05𝑊

𝑚2.

27. A model for world population in billions 𝑡 years after the year 2000 is given by (𝑡) =24

1+3𝑒−0.04𝑡 .

Determine in what year the population is projected to be 12 billion by this model.

Also, what does the model predict as 𝑡 gets very large?


28. The value of an exponential growth stock fund increased from $10,000 to $18,000 over a 12

year period.

a. Find the continuous growth rate and model 𝐴(𝑡) = 𝑃𝑒𝑟𝑡 that describes this growth.

b. Find the quarterly compounding growth rate and model 𝐴(𝑡) = 𝑃 (1 +𝑟

4)

𝑟𝑡that describes this

growth.

29. A sample of radioactive iodine decreased from 50 micrograms to 10 micrograms over a 19 day

period.

a. Find a decay model 𝐴(𝑡) = 𝐴0𝑒𝑟𝑡 that models this decay.

b. Determine the half-life of this radioactive iodine.

30. Extra Credit: You want to buy a house that might be in range of $180,000 to $200,000 in 10

years. You know you will need to save at least a 20% down payment cost. How much should you

invest per month starting now in an account that pays 5% interest so that in 10 years you will

have enough for your down payment at that time? Please explain carefully how you computed

this amount. You might start with $100/month and see how that does and adjust this amount

accordingly.


3.4 Systems of Equations We know how to solve some types of equations in one variable. But very often there is a need to solve a collection of more than one equation in

two or more variables. So by now you are familiar on how a mathematician uses prior knowledge to play with the objects at hand to deliver the

solutions. So let us look at certain types of systems of equations which is a collection of equations in two or more variables and see how we

could find the solutions if they exist.

A system of linear equations in two or more variables is a set of two or more linear equations. A system of nonlinear equations in two or more

variables is a set of two or more nonlinear equations. A solution to a system of equations is a point in the plane that satisfies all the equations in

the system.

Recall that working with a system of linear equations in two variables means working with lines. If two lines intersect each other they are said to

form a consistent independent system of equations with a unique solution, which is the point of intersection. If two lines overlap each other, or

are the same line, then all points on the line are solutions to the system of equations. Such a system of equations is called a consistent

dependent system of equations with infinitely many solutions. If two lines are parallel to each other they are said to form an inconsistent system

of equations with no solution. Graphically, the three scenarios are depicted below.

Consistent Independent Consistent Dependent Inconsistent

Unique Solution Infinitely many solutions No Solutions

Example

Example

Example

The two lines are not parallel and therefore intersect in one point. The point of intersection will be the unique solution to the system of equations.

The lines are parallel and have the same 𝑦-intercept or the lines overlap. In this situation all points on the line are solutions to the system of equations and therefore has infinitely many solutions.

The lines are parallel and have different 𝑦-intercepts. In this situation no point in the plane can be the solution and therefore the system has no solutions.


When working with a system of nonlinear equations, knowing what shape the graphs of each of the

equations look like may also help in understanding the nature of the solutions or how many to expect.

Methods to solve a system of equations and inequalities

Graphical Method: Plot the graphs of all the equations or inequalities in the system on the same

coordinate system and then from the graph locate the solution or the solution set if one exists.

Depending on the graph though you may at times only be able to get an approximate value of the 𝑥 and

𝑦 coordinates of the solution or the point that satisfies all the equations, or inequalities in the system.

Graphing method is not always the best but can give you an insight into the solutions.

Substitution Method: Solve one of the equations for either the 𝑥 or 𝑦 and substitute this value into the

other equation making it then an equation in one variable. This will allow you to solve the new equation

in one variable. Then replace that value in one of the original equations to find the other coordinate.

Elimination Method: Align the equations one above the other so that all the variables and degrees of

each of the equations are lined up. Then you multiply one of both of the equations by the required

constants so that when you add the two equations it results in a one variable equation and you can then

proceed as in the substitution method.

When dealing with more than two variables the first two methods will work, the elimination method

becomes a little more complicated but can still come in handy. In this case we will see what our other

options are later.

Practice examples

Solve the systems of equations if possible.

1. {3𝑥 − 𝑦 = 22𝑥 + 𝑦 = 3

Substitution Method Solve the first equation for 𝑦, giving us 𝑦 = 3𝑥 − 2 Substitute this value of 𝑦 in the second equation 2𝑥 + (3𝑥 − 2) = 3 or 5𝑥 = 5, or 𝑥 = 1

𝑦 = 3(1) − 2 = 3 − 2 = 1 Solution: (1,1) Elimination Method Add the two equations and we get

5𝑥 = 5 𝑥 = 1

3(1) − 𝑦 = 2 𝑦 = 1

Solution: (1,1)

Graphical Method


2. {2𝑥 + 𝑦 = 24𝑥 + 2𝑦 = 3

Elimination Method Multiply the first equation by −2

{−4𝑥 − 2𝑦 = −44𝑥 + 2𝑦 = 3

add the two equations and we get

0 = −1 a false statement therefore this system of equations has no solution or is an inconsistent system.

A question to ask is why this method works. Remember that when working with equations, you can create equivalent equations (equations with same solutions as the original equation) by multiplying/dividing both sides by a constant or adding/subtracting equal quantities from both sides. So essentially we are manipulating the two equations to get what we want.

3. {𝑥2 − 𝑦2 = 4𝑥 + 3𝑦 = 5

We can see from our knowledge of graphs that the first equation is a hyperbola and the second a line so we expect two solutions. As you can see it is not easy to read the solutions from their graphs but we got an insight into how many solutions to expect. So here substitution method will come in handy. Solve the second equation for 𝑥 and we get 𝑥 = −3𝑦 + 5. Now substitute this value in the first equation we get

(−3𝑦 + 5)2 − 𝑦2 = 4 9𝑦2 − 30𝑦 + 25 − 𝑦2 = 4 8𝑦2 − 30𝑦 + 21 = 0

Using quadratic formula we get

𝑦 =30 ± √(−30)2 − 4(8)(21)

16=15 ± √57

8

𝑦 =15

8+√57

8 will give us 𝑥 = −3(

15

8+√57

8) + 5 Or 𝑥 = −

5

8−3√57

8

𝑦 =15

8−√57

8 will give us 𝑥 = −3(

15

8−√57

8) + 5 Or 𝑥 = −

5

8+3√57

8

Solutions are

(−5

8−3√57

8,15

8+√57

8) and (−

5

8+3√57

8,15

8−√57

8)


4. {𝑥2 + 2𝑦 = 4𝑥 − 2𝑦 = −2

Elimination Method Adding the two equations we get 𝑥2 + 𝑥 = 2 or 𝑥2 + 𝑥 − 2 = 0

(𝑥 − 1)(𝑥 + 2) = 0 𝑥 = 1 or 𝑥 = −2

𝑥 = 1 gives us 1 − 2𝑦 = −2 or 𝑦 =3

2

𝑥 = −2 gives us −2 − 2𝑦 = −2 or 𝑦 = 0

Solutions are (−2,0) and (1,3

2).

Graphing does give us solutions here as

(−2,0) and (1,3

2).

Solutions to system inequalities are all points in a region that satisfy both inequalities. What this means

is that if we have a strict inequality the points on that curve are automatically not part of the solution

and we draw the curve as dotted lines to indicate this. When solving inequalities in one variable the

solutions are regions on a number line, now when working with inequalities in two variables we will

work with regions in the plane. Finding solutions to inequalities in two variables is really asking for a

particular relationship be satisfied between the (𝑥, 𝑦) coordinates. For example, if we say we want a

solution to the inequality 3𝑥 − 𝑦 > 5 that is the same as saying find all coordinates (𝑥, 𝑦), so that

3𝑥 − 5 > 𝑦 or that the 𝑦-coordinate is always smaller than five less than three times the 𝑥-coodinate.

That means all points in the solution sets are coordinates whose 𝑦-coordinate satisfies that relationship.

See below for an example of how (1,1) is not a solution to the inequality, but (1, −5) is a solution.

All points in the region below the line shown by the shaded blue region are the solutions as all the 𝑦-coordinates in this region satisfy the required condition.

To extend this to a system of inequalities would mean finding solutions to each of the inequalities in the system and seeing where all the solutions sets intersect. That is pretty much how you would solve system of inequalities and find the points in the regions that are solutions to all the inequalities in the system.


5. {3𝑥 − 𝑦 > 22𝑥 + 𝑦 < 3

Step 1: Plot each of the lines in the system. Determine whether each of the lines is dotted or solid based (dotted if working with a inequality, and solid if working with either ≤, or ≥). Step 2: Take a test point on the lines to see if it is part of the solution or not, then shade the appropriate region. In this case each of the lines are dotted lines since we have strict inequalities. Let us pick (0,0) as our test point. In the first inequality we see 0 > 2 is a false statement so all the points in the region under the line 3𝑥 − 𝑦 = 2 are solutions to this inequality. For the other inequality (0,0) as our test point we see 0 < 3 which is true so all points in the place above the line 2𝑥 + 𝑦.

6. {𝑥2 + 2𝑦 < 4𝑥 − 2𝑦 ≥ −2

Plot each inequality and the overlap is the solution as you see in the picture to the right.




1. System of Equations

2. System of Inequalities

3. Solutions to system of equations

4. Solutions to system of inequalities

5. Graphing Method

6. Elimination Method

7. Substitution Method



Exercises 3.4

1. What is a system of equations?

2. When would you use the graphing method?

3. When would you use the elimination method?

4. When would you use the substitution method?

5. What is the difference between solving system of equations versus system of inequalities?

6. What is most number of solutions you expect from of a system of equations where both

equations are second degree polynomials?

7. What is least number of solutions you expect from of a system of equations where both

equations are second degree polynomials?


8. Find the solutions to system of equations and inequalities below.

a. {5𝑥 − 𝑦 = 4𝑥 − 2𝑦 = 3

b. {1

2𝑥 − 𝑦 = 4

𝑥 − 2𝑦 = 3

c. {𝑥 − 2𝑦 = 4

−4𝑥 + 16 = −8𝑦

d. {1.5𝑥 − 2𝑦 = 1𝑥 − 5𝑦 = −3

e. {2𝑦 = 1 − 𝑥4𝑥 − 5𝑦 = −3

f. {𝑦 = 1 − 3𝑥6𝑥 + 2𝑦 = 5

g. {𝑥 = 4 − 3𝑦2𝑥 + 8 = 5𝑦

h. {

2

3𝑦 − 3𝑥 =

1

21

3𝑥 −

2

5𝑦 =

2

5

i. {0.55𝑥 − 1.01𝑦 = 4.131.2 + 3.24𝑦 = −7.32

j. {𝑥2 + 𝑦2 = 4

𝑥 − 𝑦2 = −2

k. {𝑥2 + 𝑦2 = 4

𝑥2 − 𝑦2 = 1

l. {𝑥2 + 𝑦2 = 4

𝑥 − 𝑦2 = −6

m. {9𝑥2 + 4𝑦2 = 36𝑥 = 3

n. {9𝑥2 + 4𝑦2 = 36𝑦 = 3𝑥 − 1

o. {9𝑥2 + 4𝑦2 = 363𝑥 − 𝑦 = −12

p. {𝑥 − 𝑦 < 4𝑥 − 2𝑦 ≥ 3

q. {𝑥2 + 𝑦2 ≤ 4

𝑥 − 𝑦2 < 1

r. {𝑥2 + 𝑦2 > 4

𝑥2 − 𝑦2 < 1

s. {𝑥2 + 𝑦2 ≥ 4

𝑥 − 𝑦2 ≥ −6

9. Use systems of equations or inequalities to solve the following.

a. Paul invests $100,000 in three stocks that pay dividends of 6%, 8%, and 10%. The

amount invested in the 10% is twice the amount at 6% and the return on the whole

investment is $8600 per year. How much is invested in each stock type?

b. Tickets at a play were sold for either $5 for children, $7 for college students and $9 for

general adult admission. The total number of tickets sold was 400 and the total ticket

revenue was $2500. How many tickets of each type were sold if the number of general

adult tickets was 1/7th of the total of the child and student tickets sold.


3.5 Introduction to Matrices and Gauss Elimination Method We saw in the previous section how to solve a system of equations. The first thing to notice in the

examples we solved was for a system of linear equations we created equivalent systems of equations by

performing what are called elementary row operations which is basically allowing you to multiply or

divide one of the equations in the system by a constant, or add two or more equations together creating

a new equivalent system of equations. Two equivalent systems of equations have the same set of

solutions.

Matrix: A matrix is a collection of numbers organized into rows and columns.

The size of a matrix is determined by the number of rows and columns it has.

For counting numbers 𝑚, and 𝑛, a 𝑚 × 𝑛 matrix is said to have 𝑚 rows and 𝑛 columns.

For example,

1. [3 −1−2 4

] is a 2 by 2matrix (with 2 rows and 2 columns).

2. [1 −2 30 1 47 −1 2

] is a 3 by 3 matrix (with 3 rows and 3 columns)

3. [−1 0 43 −2 5

] is a 2 by 3 matrix (with 2 rows and 3 columns).

4. [2 3 4−1 3 5−2 1 5

|6−21] this matrix is called an augmented matrix and is a 3 by 4 matrix. Such matrices

can be used to represent system of equations. This particular matrix could represent the system of

equations given by {

2𝑥 + 3𝑦 + 4𝑧 = 6−𝑥 + 3𝑦 + 5𝑧 = −2−2𝑥 + 𝑦 + 5𝑧 = 1

as you can see the coefficients of 𝑥, 𝑦, and 𝑧 are in the

first, second, third columns respectively and the constant term are in the last column. The equality

symbol is replaced with the bar between the third and the fourth column.

Playing

As with any new object mathematicians define, we may wonder how to do arithmetic with these new

objects. Can you think of a way we could add and subtract two matrices?

Just like any objects we can add and subtract like objects which in this case would mean two matrices

must have the same size to be able to add and subtract. And then a natural way to perform these

operations would be to do it term by term.

[3 −1−2 4

] + [2 53 −6

] = [3 + 2 −1 + 5−2 + 3 4 + (−6)

] = [5 41 −2

]

Subtraction would be similar.

What do you think would the value of 2 [3 −1−2 4

]. Trust your intuition before peeking on the next

page.


2 [3 −1−2 4

] = [2 × 3 2 × −12 × −2 2 × 4

] = [6 −2−4 8

]

This is referred to as multiplying the matrix by a scalar.

We can also multiply two matrices but in this case it is not a very intuitive how the multiplication

works. See if you can come up with a clever way to multiply matrices and decide on which matrices we

could multiply together and how.

Matrices as such may have been used earlier but became mostly popular around 1850’s. Matrix theory

is a subject in mathematics that deals with intricacies of matrices. They are used in many fields like

quantum mechanism, chemistry to describe atomic structures, computer graphics, and many other

interesting fields.

We can use matrices to solve very complex systems of equations. For now we will only focus on how to

use the elementary row operations to solve system of equations of three or more variables.

The Gauss Elimination Method is basically taking the augmented matrix of a system of linear equations

and using elementary row operations bring the matrix into a triangular form that has ones in the

diagonals and zeros under them. For example, for a system of linear equations in three variables to get

it in the form [ 1 −1 20 1 −30 0 1

|4−12] this will allow us to get the solutions to the system of equations

easily since here for example we will get {𝑥 − 𝑦 + 2𝑧 = 4𝑦 − 3𝑧 = −1𝑧 = 2

and then we can proceed to get the solutions

to system of equations.

Elementary Row Operations

Let us see why this works. We will use a two by two system of linear equations and then show you the

beauty of this approach as we can then solve systems of linear equations that has three or more

variables. We will represent rows by writing 𝑅𝑛 or 𝑅𝑛 to mean nth row. Elementary row operations will

be denoted as 𝑅2 = −3𝑅1 + 2𝑅2 means the new row 2 is found by adding −3 times row1 to 2 times

old row2.

Pay careful attention to the solving of the system of linear equations in two variables and the

equivalent step by step use of matrices to solve the same system. This careful attention will pay

dividends later to solve more complex system of linear equations.


Practice Examples

Solve the system of equations below.

Solving by elimination method Solving using matrices

1. {𝟑𝒙 − 𝟐𝒚 = 𝟔𝟐𝒙 − 𝟓𝒚 = 𝟕

So first multiply the first equation by 𝟐 and second equation by −𝟑 creating the equivalent system of

{𝟔𝒙 − 𝟒𝒚 = 𝟏𝟐

−𝟔𝒙 + 𝟏𝟓𝒚 = −𝟐𝟏

Add the two equations 𝟏𝟏𝒚 = −𝟗 or

𝒚 =−𝟗

𝟏𝟏

Use this value of 𝒚 into one of the equations to get the value of 𝒙. So we get

𝟑𝒙 − 𝟐(−𝟗

𝟏𝟏) = 𝟔 or

𝟑𝒙 = 𝟔 −𝟏𝟖

𝟏𝟏=𝟔𝟔−𝟏𝟖

𝟏𝟏=𝟒𝟖

𝟏𝟏

Or 𝒙 =𝟏

𝟑(𝟒𝟖

𝟏𝟏) =

𝟏𝟔

𝟏𝟏.

The solution of the system is

𝒙 =𝟏𝟔

𝟏𝟏, 𝒚 = −

𝟗

𝟏𝟏

1. {𝟑𝒙 − 𝟐𝒚 = 𝟔𝟐𝒙 − 𝟓𝒚 = 𝟕

The augmented matrix to represent this system of equations will be

[ 3 −22 −5

| 6 7]

𝑅1=2𝑅1𝑅2=−3𝑅2→ [

6 −4−6 15

| 12−21

]

𝑅2=𝑅2+𝑅1→ [

6 −40 11

| 12−9]

𝑅2=

1

11𝑅2

→ [6 −40 1

| 12

−9

11

]

𝑅1=4𝑅2+𝑅1→ [

6 00 1

| 12 −

36

11

−9

11

]

𝑅1=1

6𝑅1

→ [1 00 1

| 2 −

6

11

−9

11

] = [1 00 1

|

16

11

−9

11

]

𝑥 =16

11 and 𝑦 = −

9

11

It may look like a lot of work compared to use matrices but when working with more than two variables

you will see the efficiency of it. So let’s attempt to solve a three by three system of linear equations.


2. {

2𝑥 − 3𝑦 + 𝑧 = 73𝑥 + 2𝑦 − 2𝑧 = −3−𝑥 + 𝑦 + 3𝑧 = 4

Our goal is to get ones in the diagonals and zeros under it so we can start with interchanging

rows one and three represented as below. All that means is we are writing the third equation

first and the first equation third. Interchanging equations creates an equivalent system of

equations. Then continue with the elementary row operations.

[2 −3 13 2 −2−1 1 3

|7−34]𝑅1↔𝑅3→ [

−1 1 33 2 −22 −3 1

|4−37](−1)𝑅1→𝑅1→ [

1 −1 −33 2 −22 −3 1

|−4−37]

𝑅2−3𝑅1→𝑅2𝑅3−2𝑅1→𝑅3→ [

1 −1 −3

0 5 7

0 −1 7

|−4

9

15

] 𝑅2↔𝑅3→ [

1 −1 −3

0 −1 7

0 5 7

|−4

15

9

]

−1(𝑅2)→𝑅2→ [

1 −1 −3

0 1 −7

0 5 7

|−4

−15

9

]𝑅3−5𝑅2→𝑅3𝑅1+𝑅2→𝑅1→ [

1 0 −10

0 1 −7

0 0 42

|−19

−15

84

]

1

42𝑅3→𝑅3

→ [1 0 −10

0 1 −7

0 0 1

|−19

−15

2

]𝑅1+10𝑅3→𝑅1𝑅2+7𝑅3→𝑅2→ [

1 0 0

0 1 0

0 0 1

|1

−1

2

]

Solution: 𝑥 = 1, 𝑦 = −1, 𝑧 = 2

Interchange R1 and R3

+(−3 3 9 12)

0 5 7 9

𝑅2 − 3𝑅1→ 𝑅2

3 2 −2 −3

+(−2 2 6 8)

0 −1 7 15

𝑅3 − 2𝑅1 → 𝑅3

2 −3 1 7

Interchange R2 and R3

+(0 −5 35 75)

0 0 42 84

𝑅3 − 5𝑅2 → 𝑅3

0 5 7 9

+( 0 0 10 20)

0 0 0 1

𝑅1 + 10𝑅3 → 𝑅1

1 0 −10 −19


3. {

𝑥 + 𝑦 − 𝑧 = 22𝑥 + 2𝑦 − 2𝑧 = 43𝑥 − 2𝑦 + 𝑧 = 1

Our augmented matrix will be

[1 1 −12 2 −23 −2 1

|241]

𝑅2−2𝑅1→𝑅2𝑅3−3𝑅1→𝑅3→ [

1 1 −1

0 0 0

0 −5 4

|2

0

−5

]𝑅2↔𝑅3→ [

1 1 −1

0 −5 4

0 0 0

|2

−5

0

]

1

−5𝑅2→𝑅2

→ [1 1 −1

0 1 −4/5

0 0 0

| 2

1

0

]𝑅1−𝑅2→𝑅1→ [

−1 0 −1/5

0 1 −4/5

0 0 0

| 1

1

0

]

Since the bottow row is always true we can set 𝑧 = 𝑡 where 𝑡 = 𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟

That makes 𝑦 =4

5𝑡 + 1 and 𝑥 =

1

5𝑡 + 1 .

There are infinitely many solutions to this sytem.

4. {

𝑥 + 𝑦 − 𝑧 = 22𝑥 + 2𝑦 − 2𝑧 = 6−𝑥 − 2𝑦 + 𝑧 = −3

Our augmented matrix will be

[1 1 −12 2 −2−1 −2 1

|26−3]

𝑅2−2𝑅1→𝑅2𝑅3+𝑅1→𝑅3→ [

1 1 −1

0 0 0

0 −1 0

|2

2

−1

]

In row 2 we notice the equation 0 = 2 which is never true so this system of equations has no solutions.




1. Matrix

2. Elementary Row Operations

3. Gauss Elimination Method



Exercises 3.5

1. Please answer the following without looking at other resources.

a. Is it possible to add two different matrices? If yes, explain when and how.

b. Is it possible to subtract two different matrices? If yes, explain when and how.

c. Is it possible to multiply two different matrices? If yes, explain when and how.

d. Is it possible to divide two different matrices? If yes, explain when and how.

2. Why should we bother learning about the Gauss Elimination method?

3. What are the advantages of writing the system of equations using matrices?


4. Do you think we use matrices for a two by two system of linear equations?

5. What about using matrices for two by two nonlinear system of equations? If you think we can,

explain how?

6. Solve the following systems using Gauss Elimination Method.

a. {

𝑥 − 2𝑦 + 𝑧 = 43𝑥 − 𝑦 + 6𝑧 = −14𝑥 + 3𝑦 − 7𝑧 = 5

b. {4𝑥 + 5𝑦 + 2𝑧 = −5𝑥 + 𝑦 − 3𝑧 = −4

−3𝑥 + 2𝑦 − 𝑧 = −13

c. {

𝑥 + 𝑦 + 2𝑧 = 42𝑥 + 2𝑦 + 4𝑧 = −4−3𝑥 + 2𝑦 − 𝑧 = −13

d. {

𝑥 − 𝑦 + 2𝑧 = −5𝑥 + 𝑦 − 3𝑧 = −4

−2𝑥 + 2𝑦 − 4𝑧 = 10

e.

{

1

2𝑥 −

1

3𝑦 +

1

5𝑧 =

11

30

𝑥 −2

3𝑦 + 𝑧 =

4

3

4𝑥 − 𝑦 +1

2𝑧 =

7

2

f. {

0.21𝑥 − 0.2𝑦 − 1.4𝑧 = 2.01𝑥 − 0.33𝑦 + 1.4𝑧 = 0.26

−0.4𝑥 + 3.1𝑦 + 2.1𝑧 = −8.7

g. {

3𝑥 − 2𝑧 = 7−𝑥 + 4𝑦 + 5𝑧 = −7−2𝑥 − 4𝑦 − 3𝑧 = 0

h. {

𝑥 + 𝑦 + 𝑧 + 𝑤 = 12𝑥 − 𝑦 + 3𝑧 − 𝑦 = −9𝑥 − 𝑦 − 𝑧 + 𝑤 = 1

𝑥 − 2𝑦 + 𝑧 − 2𝑤 = −8

i. {

𝑥 − 2𝑦 + 3𝑧 − 𝑤 = 12−𝑥 + 3𝑦 − 2𝑧 − 3𝑤 = 1−4𝑥 + 6𝑦 + 𝑧 − 3𝑤 = 1𝑥 − 2𝑧 + 𝑤 = −6

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