MTH 209 Week 4 - Biker John€¦ · MTH 209 Week 4 . Final Exam logistics Here is what I've found...

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MTH 209 Week 4

Final Exam logistics

Here is what I've found out about the final exam in

MyMathLab (running from the end of class this

week (week 4 at 10pm) to 11:59pm five days after

the last day of class.

Final Exam logistics There will be 50 questions.

You have only one attempt to complete the exam.

Once you start the exam, it must be completed in that sitting. (Don't start until you have

time to complete it that day or evening.)

You may skip and get back to a question BUT return to it before you hit submit.

You must be in the same session to return to a question.

There is no time limit to the exam (except for 11:59pm five nights after the last class).

You will not have the following help that exists in homework:

Online sections of the textbook

Animated help

Step-by-step instructions

Video explanations

Links to similar exercises

You will be logged out of the exam automatically after 3 hours of inactivity.

Your session will end.

IMPORTANT! You will also be logged out of the exam if you use your back button on

your browser. You session will end.

Section 6.6

Solving

Equations by

Factoring I

(Quadratics)

Objectives

• The Zero-Product Property

• Solving Quadratic Equations

• Applications

Zero-Product Property

To solve equations we often use the zero-product

property, which states that if the product of two

numbers is 0, then at least one of the numbers must

Example

Solve each equation.

Solution

3 ( 4) 0x x (4 1)(3 4) 0x x

3 ( 4) 0x x

04( )3x x

3 0 or 4 0 x x

0 o r 4 x x

(4 1)(3 4) 0x x

3( 4 01)(4 )xx

0 or 31 044 xx

4 1 or 3 4x x

1 4 or

4 3x x

Try Q’s pg 402 13,15,19,23

Solving Quadratic Equations

Any quadratic polynomial in the variable x can be

written as ax2 + bx + c with a ≠ 0.

Any quadratic equation in the variable x can be

written as ax2 + bx + c = 0, with a ≠ 0. This form of

quadratic equation is called the standard form of a

quadratic equation.

Example

Solve each quadratic equation. Check your answers.

a. 36 + x2 = –12x b. x2 − 49 = 0

Solution

a. 36 + x2 = –12x

x2 + 12x + 36 = 0

(x + 6)(x + 6) = 0

x = −6

The only solution is −6.

To check this value,

substitute −6 for x in

the given equation.

36 + x2 = –12x

36 + (−6)2 = –12(−6)

72 = 72

Example (cont)

b. x2 − 49 = 0

(x + 7)(x − 7) = 0

x + 7 = 0 x − 7 =

0 x = 7

To check these values,

substitute 7 and −7 for

x in the given equation. 72 − 49 = 0

(−7)2 − 49 =

0 0 = 0

The solutions are −7 and

Try Q’s pg 403 27,35,43,49

Example

Solve 2x2 − 7x = −5

Solution

The solutions are

(2x – 5)(x − 1) = 0

2x – 5 = 0 or x – 1 = 0

or x = 1

2x2 − 7x = −5

2x2 − 7x + 5 = 0

51 and .

Try Q’s pg 403 51

Example

If a model rocket is launched at 48 feet per second,

then its height, h, after t seconds is h = 48t – 16t2.

After how long does the rocket strike the ground?

Solution

The rocket strikes the ground when the height is 0.

48t – 16t2 = 0

16t(3 – t) = 0

16t = 0 3 – t = 0

The rocket strikes the

ground after 3

seconds.

Try Q’s pg 403 67a

Example

A frame surrounding a picture is 2 inches wide. The

picture inside the frame is 7 inches longer than it is

wide. If the overall area of the picture and frame is

198 square inches, find the dimensions of the picture

inside the frame.

Solution

Let x be the width of the picture and x + 7 be its

length.

x + 11

Example (cont)

(x + 4)(x + 11) =198

x2 + 15x + 44 = 198

x2 + 15x − 154 = 0

(x − 7)(x + 22) = 0

x − 7 = 0 or x + 22 = 0

x = 7 or x = −22

The only valid solution for x is 7 inches. Because the

length is 7 inches more than the width, the

dimensions are 7 inches and 14 inches.

Try Q’s pg 404 73

Section 8.4

Functions and

Their Properties

Objectives

• Expressing Domain and Range in Interval Notation

• Absolute Value Function

• Polynomial Functions

• Rational Functions (Optional)

• Operations on Functions

Expressing Domain and Range in

Interval Notation

The set of all valid inputs for a function is called the

domain, and the set of all outputs from a function is

called the range.

Rather than writing “the set of all real numbers” for

the domain of f, we can use interval notation to

express the domain as (−∞, ∞).

Example

Write the domain for each function in interval

notation.

a. f(x) = 3x b.

Solution

a. The expression 3x is defined for all real numbers

x. Thus the domain of f is

b. The expression is defined except when x – 4 = 0

x = 4. Thus the domain of f includes all real

numbers except 4 and can be written

,4 4, .

Try Q’s pg 563 13,21,25

Absolute Value Function

We can define the absolute value function

by f(x) = |x|.

To graph y = |x|, we begin by making a table of

values.

–2 2

–1 1

Example

Sketch the graph of f(x) = |x – 3|. Write its domain

and range in interval notation.

Solution

Start by making a table of values.

-3 -2 -1 1 2 3 4 5 6 7 8

The domain of f is , .

The range of f is [0, ).

Try Q’s pg 563 39

Polynomial Functions

The following expressions are examples of

polynomials of one variable.

As a result, we say that the following are symbolic

representations of polynomial functions of one

variable.

2 3, , and 51 515 3x xx x

32( ) 3, , and ( 5 1 ( )) 1 5 5f g x x xx xx hx

Example

Determine whether f(x) represents a polynomial

function. If possible, identify the type of polynomial

function and its degree.

3( ) 6 2 7f x x x

3.5( ) 4f x x

cubic polynomial, of degree 3

not a polynomial function

because the exponent on the

variable is negative

not a polynomial

Try Q’s pg 564 43,45,47,51

Example

A graph of is shown. Evaluate f(1)

graphically and check your result symbolically.

Solution

3( ) 5f x x x

To calculate f(–1) graphically

find –1 on the x-axis and

move down until the graph of

f is reached. Then move

horizontally to the y-axis.

f(1) = –4 3( 1) 5( ( )1 1)f

Try Q’s pg 564 59,73

Example

Evaluate f(x) at the given value of x.

Solution

3 2( ) 4 3 7, 2f x x x x

3 2( 2) 4( 2) 3( 2) 7f

( 2) 4( 8) 3(4) 7f

( 2) 32 12 7f

( 2) 27f

Try Q’s pg 564 55-64

Example

Use and to evaluate each of

the following.

Solution

2( ) 3 1f x x 2( ) 6g x x

a. ( )(1) b. ( )( 2) c. 0f

f g f gg

2a. ( ) 3 11 (1)

2( ) 61 ( )

( )(1) (1) (1)

f g f g

Example (cont)

the following.

Solution

2( ) 3 1f x x 2( ) 6g x x

a. ( )(1) b. ( )( 2) c. 0f

f g f gg

2b. ( ) 3( ) 1

f2( ) 6 (

( )( 2) ( 2) ( 2)

f g f g

Example (cont)

the following.

Solution

2( ) 3 1f x x 2( ) 6g x x

a. ( )(1) b. ( )( 2) c. 0f

f g f gg

3( ) 1 =

Try Q’s pg 564 64-70

Section 11.1

Quadratic

Functions and

Their Graphs

Objectives

• Graphs of Quadratic Functions

• Min-Max Applications

• Basic Transformations of Graphs

• More About Graphing Quadratic Functions

(Optional)

The graph of any quadratic function is a

parabola.

The vertex is the lowest point on the graph of

a parabola that opens upward and the highest

point on the graph of a parabola that opens

downward.

The graph is symmetric with respect to the y-

axis. In this case the y-axis is the axis of

symmetry for the graph.

Example

Use the graph of the quadratic function to identify the

vertex, axis of symmetry, and whether the parabola

opens upward or downward.

Vertex (0, 2)

Axis of symmetry: x = –2

Open: up

Vertex (0, 4)

Axis of symmetry: x = 0

Open: down

Try Q’s pg 725 29,31

Example

Find the vertex for the graph of

Support your answer graphically.

Solution

a = 2 and b = 8

Substitute into the equation to find the y-value.

2( ) 2 8 3.f x x x

2( ) 2( 2) 8( 2) 3

8 16 3

The vertex is (2, 11), which

is supported by the graph.

Try Q’s pg 724 15

Example

Identify the vertex, and the axis of symmetry on the

graph, then graph.

Solution

Begin by making a

table of values.

Plot the points and

sketch a smooth curve.

The vertex is (0, –2)

axis of symmetry x = 0

2( ) 2f x x

x f(x) = x2 –

Example

graph, then graph.

Solution

Begin by making a

table of values.

Plot the points and

sketch a smooth

curve.

The vertex is (2, 0)

2( ) ( 2)g x x

x g(x) = (x –

Example

graph, then graph.

Solution

Begin by making a

table of values.

Plot the points and

sketch a smooth

curve.

The vertex is (2, 0)

2( ) 2 3h x x x

x h(x) = x2 – 2x –

Example

Find the maximum y-value of the graph of

Solution

The graph is a parabola that opens downward

because a < 0. The highest point on the graph

is the vertex.

a = 1 and b = 2

2( ) 2 3.f x x x

2 2( 1)

2( ) ( 1) 2( 1) 3

Try Q’s pg 726 81

Example

A baseball is hit into the air and its height h in feet

after t seconds can be calculated by

a. What is the height of the baseball when it is hit?

b. Determine the maximum height of the baseball.

Solution

a. The baseball is hit when t = 0.

b. The graph opens downward

because a < 0. The maximum height occurs at

the vertex. a = –16 and b = 64.

2( ) 16 64 2.h t t t

2( ) 16 64 2h t t t 2(0) 16(0) 64(0) 2

64 642

2 2( 16) 32

Example (cont)

2( ) 16 64 2 h t t t

2(2) 16(2) 64(2) 2

The maximum height is 66 feet.

2( ) 16 64 2 h t t t

Try Q’s pg 726 87

Basic Transformations of Graphs

The graph of y = ax2, a > 0.

As a increases, the resulting parabola becomes

narrower.

When a > 0, the graph of y = ax2 never lies below the

x-axis.

Section 11.2

Parabolas and

Modeling

Objectives

• Vertical and Horizontal Translations

• Vertex Form

• Modeling with Quadratic Functions (Optional)

Vertical and Horizontal Translations

The graph of y = x2 is a parabola opening upward

with vertex (0, 0).

All three graphs have the same shape.

y = x2

y = x2 + 1 shifted upward 1 unit

y = x2 – 2 shifted downward 2 units

Such shifts are called translations because they do

not change the shape of the graph only its position

with vertex (0, 0).

y = x2

y = (x – 1)2

Horizontal shift to the right 1 unit

with vertex (0, 0).

y = x2

y = (x + 2)2

Horizontal shift to the left 2 units

Example

Sketch the graph of the equation and identify the

vertex.

Solution

The graph is similar to y = x2 except it has been

translated 3 units down.

The vertex is (0, 3).

2 3y x

Try Q’s pg 739 15,19,27

Example

vertex.

Solution

translated left 4 units.

2( 4)y x

Example

vertex.

Solution

translated down 2 units and

right 1 unit.

2( 1) 2y x

Try Q’s pg 739 39,37

Example

Compare the graph of y = f(x) to the graph of y = x2.

Then sketch a graph of y = f(x) and y = x2 in the

xy-plane.

Solution

The graph is translated to the right

2 units and upward 3 units.

The vertex for f(x) is (2, 3) and the

vertex of y = x2 is (0, 0).

The graph opens upward

and is wider.

21( ) ( 2) 3

4f x x

Try Q’s pg 739 41

Example

Write the vertex form of the parabola with a = 3 and

vertex (2, 1). Then express the equation in the form

y = ax2 + bx + c.

Solution

The vertex form of the parabola is

where the vertex is (h, k).

a = 3, h = 2 and k = 1

To write the equation in y = ax2 + bx + c, do the

following:

2( ) ,y a x h k

2)3( 12y x

2( 4 43 1)y x x

23 12 12 1y x x 23 12 13y x x

Try Q’s pg 740 65,68

Example

Write each equation in vertex form. Identify the

vertex.

Solution

a. Because , add and subtract 16

on the right.

2 8 13y x x 22 8 7y x x

2 8 13y x x

2 16 1 68 13y x x

4 3y x

Example (cont)

b. This equation is slightly different because the

leading coefficient is 2 rather than 1. Start by

factoring 2 from the first two terms on the right

22 8 7y x x

2( 4 ) 7

2 42 4 74y x x

2 42 4 7 8y x x

2 2 1y x The vertex is ( 2, 1).

Try Q’s pg 739 53

Section 11.3

Quadratic

Equations

Objectives

• Basics of Quadratic Equations

• The Square Root Property

• Completing the Square

• Solving an Equation for a Variable

• Applications of Quadratic Equations

Basics of Quadratic Equations

Any quadratic function f can be represented by

f(x) = ax2 + bx + c with a 0.

Examples:

2 2 21( ) 2 1, ( ) 2 , and ( ) 2 1

3f x x g x x x h x x x

Basics

The different types of solutions to a quadratic

equation.

Example

Solve each quadratic equation. Support your results

numerically and graphically.

a. b. c.

Solution

a. Symbolic: Numerical: Graphical:

23 2 0x 2 9 6x x 2 2 8 0x x

The equation has no real

solutions because x2 ≥ 0 for

all real numbers x.

Example (cont)

b. 2 9 6x x

( 3)( 3) 0

3 0 or 3 0

3 or 3

The equation has

one real solution.

Example (cont)

c. 2 2 8 0x x

2 2 8 0

( 2)( 4) 0

2 0 or 4 0

2 or 4

The equation has two real

solutions.

Try Q’s pg 752 29,37,39

The Square Root Property

The square root property is used to solve quadratic

equations that have no x-terms.

Example

Solve each equation.

a. b. c.

Solution

2 10x 225 16 0x 2( 3) 16x

b. 225 16 0x

225 16x

c. 2( 3) 16x

( 3) 16x

1 or 7x

Try Q’s pg 752 51,53,57

Real World Connection

If an object is dropped from a height of h feet, its

distance d above the ground after t seconds is given

by 2( ) 16d t h t

Example

Find the term that should be added to to form

a perfect square trinomial.

Solution

Coefficient of x-term is –8, so we let b = –8. To

complete the square we divide by 2 and then square

the result.

2 8x x

2 2168 ( 4)x x x

Try Q’s pg 752 67

Example

Solve the equation

Solution

Write the equation in x2 + bx = d form.

2 8 13 0x x

2 8 13x x

2 168 3 61 1x x

2( 4) 3x

5.73 or 2.27x

Try Q’s pg 752 73

Example

Solve the equation

Solution

Write the equation in x2 + bx = d form.

20 2 8 7x x

22 8 7x x

2 1( 2)

1.29 or 2.71x

Try Q’s pg 752 81

Example

Solve the equation for the specified variable.

Solution

216 for w x x216w x

Try Q’s pg 753 105,107

Section 11.4

Quadratic

Formula

Objectives

• Solving Quadratic Equations

• The Discriminant

• Quadratic Equations Having Complex Solutions

, 0 and 1,xf x a a a

The solutions to ax2 + bx + c = 0 with a

≠ 0 are given by

QUADRATIC FORMULA

b b acx

Example

Solve the equation 4x2 + 3x – 8 = 0. Support your

results graphically.

Solution

Symbolic Solution

Let a = 4, b = 3 and c = − 8.

b b acx

23 3 4 4 8

1.1x 1.8x or

Example (cont)

4x2 + 3x – 8 = 0

Graphical Solution

Try Q’s pg 764 9

Example

Solve the equation 3x2 − 6x + 3 = 0. Support your

result graphically.

Solution

Let a = 3, b = −6 and c = 3.

b b acx

26 6 4 3 3

Try Q’s pg 764 11

Example

Solve the equation 2x2 + 4x + 5 = 0. Support your

result graphically.

Solution

Let a = 2, b = 4 and c = 5.

b b acx

24 4 4 2 5

There are no real solutions

for this equation because

is not a real number. 24

Try Q’s pg 764 13

To determine the number of solutions to

the quadratic equation ax2 + bx + c = 0,

evaluate the discriminant b2 – 4ac.

1. If b2 – 4ac > 0, there are two real solutions.

2. If b2 – 4ac = 0, there is one real solution.

3. If b2 – 4ac < 0, there are no real solutions;

there are two complex solutions.

THE DISCRIMINANT AND QUADRATIC

EQUATIONS

Example

Use the discriminant to determine the number of

solutions to −2x2 + 5x = 3. Then solve the equation

using the quadratic formula.

Solution

−2x2 + 5x − 3 = 0

Let a = −2, b = 5 and c = −3.

Thus, there are two

solutions.

b2 – 4ac

= (5)2 – 4(−2)(−3) = 1

b b acx

Try Q’s pg 764 39a, b

If k > 0, the solution to x2 + k = 0 are given by

THE EQUATION x2 + k = 0

.x i k

Example

Solve x2 + 17 = 0.

Solution

The solutions are

17 or 17.x i i

Try Q’s pg 764 57

Example

Solve 3x2 – 7x + 5 = 0. Write your answer in standard

form: a + bi.

Solution

Let a = 3, b = −7 and c = 5.

b b acx

27 7 4 3 5

6 6x i

Try Q’s pg 765 73

Example

Solve Write your answer in standard

a + bi.

Solution

Begin by adding 2x to each side of the equation and

then multiply by 5 to clear fractions.

Let a = −2, b = 10 and c = −15.

223 2 .

22 10 15 0x x

Example (cont)

Let a = −2, b = 10 and c = −15.

223 2 .

b b acx

210 10 4 2 15

10 2 5

2 2x i

Try Q’s pg 765 79

Example

Solve by completing the square.

Solution

After applying the distributive property, the equation

becomes

Since b = −4 ,add to each side of the

equation.

4 5x x

2 4 5.x x

4 2 4 4 5 4x x

The solutions are

2 + i and 2 − i.

Try Q’s pg 765 91

Section 11.5

Quadratic

Inequalities

Objectives

• Basic Concepts

• Graphical and Numerical Solutions

• Symbolic Solutions

Example

Determine whether the inequality is quadratic.

a. 6x + 2x2 – x3 ≥ 0 b. 8 + 7x2 + 2 < 6x2 + x

Solution

a. The inequality 6x + 2x2 – x3 ≥ 0 is not quadratic

because it has an x3-term.

b. Write the inequality as follows.

8 + 7x2 + 2 < 6x2 + x

10 + 7x2 < 6x2 + x

x2 – x + 10 < 0

The inequality is quadratic because it can be written

in the form ax2 + bx + c > 0 with a = 1, b = −1, and c

Try Q’s pg 775 7,11

Example

Make a table of values for x2 − 2x – 15 and then

sketch the graph. Use the table and graph to solve

x2 – 2x – 15 ≤ 0. Write your

answer in interval notation.

Solution

Interval notation [−3, 5]

x y = x2 – 2x – 15

−3 0

−2 −7

−1 −12

0 −15

1 −16

2 −15

3 −12

4 −7

Example (cont)

x2 − 2x – 15

Try Q’s pg 775 19,27

Example

Solve x2 > 4. Write your answer in interval notation.

Solution

The graph of y = x2 – 4 is shown with intercepts −2

and 2. Thus the solution set is given by

x < −2 or x > 2, which can be

written in interval notation as

, 2 2, .

Try Q’s pg 776 49

Example

Solve each of the inequalities graphically.

a. x2 + 3 > 0 b. x2 + 3 < 0 c. (x − 3)2 ≤ 0

Solution

a. Because the graph is always

above the x-axis, x2 + 3 is

always greater than 0. The

solution set includes all real

numbers, or (−∞,∞).

b. Because the graph never

goes below the x-axis, x2 + 3

is never less than 0. Thus

there are no real solutions.

Example (cont)

Solve each of the inequalities graphically.

a. x2 + 3 > 0 b. x2 + 3 < 0 c. (x − 3)2 ≤ 0

Solution

a. Because the graph never

goes below the x-axis, (x −

3)2 ≤ 0 is never less than 0.

When x = 3, y = 0, so 3 is

the only solution to the

inequality (x − 3)2 ≤ 0.

Try Q’s pg 776 51,55,57

Example

Solve each inequality symbolically. Write your

a. x2 – 4x – 12 ≥ 0 b. 10 – x2 > 9x

Solution

x2 – 4x – 12 = 0

(x + 2)(x – 6) = 0

x = −2

x + 2 = 0 x – 6 = 0

The solutions lie

outside the values. In

interval notation the

solution set is

( , 2] [6, ).

Example (cont)

Solve each inequality symbolically. Write your

a. x2 – 4x – 12 ≥ 0 b. 10 – x2 > 9x

Solution

– x2 − 9x + 10 < 0

– x2 − 9x + 10 = 0

(x + 10)(x – 1) = 0

x + 10 = 0 x – 1 = 0

x = −10 x = 1

The solutions lie

between these two

values. In interval

notation the solution set

is (−10, 1).

x2 + 9x − 10 = 0

Try Q’s pg 776 31,37

Example

A rectangular building needs to be 9 feet longer than

it is wide. The area of the building must be at least

532 square feet. What widths x are possible for this

building?

Solution

x2 + 9x = 532

x(x + 9) ≥ 532

x2 + 9x – 532 = 0

29 9 4 1 532

9 2209

28, 19

The width is positive, so

the building must be 19

feet or more, x ≥ 19

Try Q’s pg 776 67

Due for this week…

Homework 4 (on MyMathLab – via the Materials

Link) The fifth night after class at 11:59pm.

Read Chapter 12.1, 12.3 and 14.1-14.3

Do the MyMathLab Self-Check for week 4.

Learning team hardest problem assignment.

Complete the Week 4 study plan after submitting

week 4 homework.

Participate in the Chat Discussions in the OLS

End of week 4

You again have the answers to those problems not

assigned

Practice is SOOO important in this course.

Work as much as you can with MyMathLab, the

materials in the text, and on my Webpage.

Do everything you can scrape time up for, first the

hardest topics then the easiest.

You are building a skill like typing, skiing, playing a

game, solving puzzles.

Next Week: Composite Functions, Inverse

Functions, Logarithmic functions,

Arithmetic/Geometric sequences/series.

MTH 209 Week 4 - Biker John€¦ · MTH 209 Week 4 . Final Exam logistics Here is what I've found...

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