Preliminary Chapter Powerpoints · 3 Copyright © 2007 Pearson Education, Inc. Publishing as...

1

Slide P- 1Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Chapter P

Prerequisites

2


P.1

Real Numbers


Quick Review

( )

1. List the positive integers between -4 and 4.

2. List all negative integers greater than -4.

3. Use a calculator to evaluate the expression

2 4.5 3. Round the value to two decimal places.

2.3 4.5

4. Eva

−

−

3

luate the algebraic expression for the given values

of the variable. 2 1, 1,1.5

5. List the possible remainders when the positive integer

is divided by 6.

x x x

n

+ − = −

3


Quick Review Solutions

{ }

{ }

( )

1. List the positive integers between -4 and 4.

2. List all negative integers greater than -4.

3. Use a calculator to evaluate the expression

2 4.5 3. Round the value to two deci

2.3 4.

1

5

,2,3

-3,-2,-1

−

−

{ }3

mal places.

4. Evaluate the algebraic expression for the given values

of the variable. 2 1, 1,1.5

5. List the possible remainders when the positive integer

i

2.73

-4,

s divid

5.375

1,2,ed by 6.

x x x

n

+

−

− = −

3,4,5


What you’ll learn about

� Representing Real Numbers

� Order and Interval Notation

� Basic Properties of Algebra

� Integer Exponents

� Scientific Notation

… and why

These topics are fundamental in the study of

mathematics and science.

4


Real Numbers

A real number is any number that can be written as a decimal.

Subsets of the real numbers include:

� The natural (or counting) numbers: {1,2,3…}

� The whole numbers: {0,1,2,…}

� The integers: {…,-3,-2,-1,0,1,2,3,…}


Rational Numbers

Rational numbers can be represented as a ratio a/b where a and b

are integers and b≠0.

The decimal form of a rational number either terminates or is

indefinitely repeating.

5


The Real Number Line


Order of Real Numbers

Let a and b be any two real numbers.

Symbol Definition Reada>b a – b is positive a is greater than b

a<b a – b is negative a is less than b

a≥b a – b is positive or zero a is greater than or equal to b

a≤b a – b is negative or zero a is less than or equal to b

The symbols >, <, ≥, and ≤ are inequality symbols.

6


Trichotomy Property

Let a and b be any two real numbers.

Exactly one of the following is true:

a < b, a = b, or a > b.


Example Interpreting Inequalities

Describe the graph of x > 2.

7


Example Interpreting Inequalities

Describe the graph of x > 2.

The inequality describes all real numbers greater than 2.


Bounded Intervals of Real Numbers

Let a and b be real numbers with a < b.

Interval Notation Inequality Notation

[a,b] a ≤ x ≤ b

(a,b) a < x < b

[a,b) a ≤ x < b

(a,b] a < x ≤ b

The numbers a and b are the endpoints of each interval.

8


Unbounded Intervals of Real Numbers

Let a and b be real numbers.

Interval Notation Inequality Notation

[a,∞) x ≥ a

(a, ∞) x > a

(-∞,b] x ≤ b

(-∞,b) x < b

Each of these intervals has exactly one endpoint,

namely a or b.


Properties of Algebra

Let , , and be real numbers, variables, or algebraic expressions.

Addition:

Multiplication

Addition: ( ) ( )

Multiplication: ( )

u v w

u v v u

uv vu

u v w u v w

uv w u

+ = +

=

+ + = + +

=

1. Communative Property

2. Associative Property

( )

Addition: 0

Multiplication: 1

vw

u u

u u

+ =

⋅ =

3. Identity Property

9


Properties of Algebra


Addition: (- ) 0

1Mulitiplication: 1, 0

Multiplication over addition:

( )

u v w

u u

u uu

u v w uv uw

+ =

⋅ = ≠

+ = +

4. Inverse Property

5. Distributive Property

( )

Multiplication over subtraction:

( )

( )

u v w uw vw

u v w uv uw

u v w uw vw

+ = +

− = −

− = −


Properties of the Additive Inverse


1. ( ) ( 3) 3

2. ( ) ( ) ( 4)3 4( 3) 12

u v w

u u

u v u v uv

− − = − − =

− = − = − − = − = −

Property Example

3. ( )( ) ( 6)( 7) 42

4. ( 1) ( 1)5 5

5. ( ) ( ) ( ) (7 9) ( 7) ( 9) 16

u v uv

u u

u v u v

− − = − − =

− = − − = −

− + = − + − − + = − + − = −

10


Exponential Notation

n factors... ,

Let be a real number, variable, or algebraic expression and

a positive integer. Then where is the

, is the , and is the ,

read as " to

n

n

a a a a

a n

a n

a a

a

⋅ ⋅ ⋅ ⋅=

exponent base th power of n a

the th power."n


Properties of Exponents

Let and be a real numbers, variables, or algebraic expressions

and and be integers. All bases are assumed to be nonzero.

1. m n m n

u v

m n

u u u +

=

Property Example3 4 3 4 7

9

9 4 5

4

0 0

- -3

3

5 5 5 5

2.

3. 1 8 1

1 14.

5. (

m

m n

n

n

n

u xu x x

u x

u

u yu y

+

− −

⋅ = =

= = =

= =

= =

5 5 5 5

2 3 2 3 6

77

7

) (2 ) 2 32

6. ( ) ( )

7.

m m m

m n mn

mm

m

uv u v z z z

u u x x x

u u a a

v v b b

⋅

= = =

= = =

= =

11


Example Simplifying Expressions

Involving Powers2 3

1 2Simplify .

u v

u v

−

−


Example Simplifying Expressions

Involving Powers2 3

1 2Simplify .

u v

u v

−

−

2 3 2 1 3

1 2 2 3 5

u v u u u

u v v v v

−

−= =

12


Example Converting to Scientific

Notation

Convert 0.0000345 to scientific notation.


Example Converting to Scientific

Notation

Convert 0.0000345 to scientific notation.

-5

0.0000345 3.45 10= ×

13


Example Converting from Scientific

Notation

Convert 1.23 × 105 from scientific notation.


Example Converting from Scientific

Notation

Convert 1.23 × 105 from scientific notation.

123,000

14


P.2

Cartesian Coordinate System


Quick Review

( ) ( )

2 2

2 2

2 2

-5 31. Find the distance between and .

4 2

Use a calculator to evaluate the expression. Round answers

to two decimal places.

2. 8 6

-12 83.

2

4. 3 5

5. 2 5 1 3

+

+

+

− + −

15



( ) ( )

2 2

2 2

2 2

-5 31. Find the distance between and .

4 2

Use a calculator to evaluate the expression. Round answers

to two decimal places.

2. 8 6

-12 83.

2

4. 3 5

5. 2

2.75

10

-2

5.83

3.5 1 3 61

+

+

+

− + −



� Cartesian Plane

� Absolute Value of a Real Number

� Distance Formulas

� Midpoint Formulas

� Equations of Circles

� Applications

… and why

These topics provide the foundation for the material that will be

covered in this textbook.

16


The Cartesian Coordinate Plane


Quadrants

17


Absolute Value of a Real Number

The is

, if 0

| | if 0.

0, if 0

a a

a a a

a

>

= − < =

absolute value of a real number a


Properties of Absolute Value

Let and be real numbers.

1. | | 0

2. | - | | |

3. | | | || |

| |4. , 0

| |

a b

a

a a

ab a b

a ab

b b

≥

=

=

= ≠

18


Distance Formula (Number Line)

Let and be real numbers. The is | | .

Note that | | | | .

a b a b

a b b a

−

− = −

distance between and a b


Distance Formula (Coordinate Plane)

( ) ( )2 2

1 2 1 2

The in the

coordinate plane is .d x x y y= − + −

distance between points and 1 1 2 2

d P(x , y ) Q(x , y )

19


The Distance Formula using the

Pythagorean Theorem


Midpoint Formula (Number Line)

The is

.2

a b+

midpoint of the line segment with endpoints and a b

20


Midpoint Formula (Coordinate Plane)

The is

, .2 2

a c b d+ +

midpoint of the line segment with endpoints ( ) and ( )a,b c,d


Standard Form Equation of a Circle

2 2 2

The with center ( , )

and radius is ( ) ( ) .

h k

r x h y k r− + − =

standard form equation of a circle

21


Standard Form Equation of a Circle


Example Finding Standard Form

Equations of Circles

Find the standard form equation of the circle with center

(2, 3) and radius 4.−

22


Example Finding Standard Form

Equations of Circles

Find the standard form equation of the circle with center

(2, 3) and radius 4.−

2 2 2

2 2

( ) ( ) where 2, 3,and 4.

Thus the equation is ( 2) ( 3) 16.

x h y k r h k r

x y

− + − = = = − =

− + + =


P.3

Linear Equations and Inequalities

23


Quick Review

Simplify the expression by combining like terms.

1. 2 4 2 3

2. 3(2 2) 4( 1)

Use the LCD to combine the fractions. Simplify the

resulting fraction.

3 43.

24.

4 3

25. 2

x x y y x

x y

x x

x x

y

+ − − −

− + −

+

++

+



Simplify the expression by combining like terms.

1. 2 4 2 3

2. 3(2 2) 4( 1)

Use the LCD to combine the fractions. Simplify the

resulting fraction.

3 43.

3 3

6 4 10

7

24.

4 3

x y

x y

x x y y x

x y

x x

x x

x

+ − − −

− + −

+

+

−

+

+ −

25.

7 6

12

22

2

y

x

y

y+

+

+

24



� Equations

� Solving Equations

� Linear Equations in One Variable

� Linear Inequalities in One Variable

… and why

These topics provide the foundation for algebraic

techniques needed throughout this textbook.


Properties of Equality

Let , , , and be real numbers, variables, or algebraic expressions.

If , then .

If

u v w z

u u

u v v u

=

= =

1. Reflexive

2. Symmetric

3. Transitive , and , then .

If and , then .

If and , then .

u v v w u w

u v w z u w v z

u v w z uw vz

= = =

= = + = +

= = =

4. Addition

5. Multiplication

25


Linear Equations in x

A linear equation in x is one that can be

written in the form ax + b = 0, where a and b

are real numbers with a ≠ 0.


Operations for Equivalent Equations

An equivalent equation is obtained if one or more of the following

operations are performed.

1. Combine like terms,

Operation Given Equation Equivalent Equation

3 1 2 3

9 3

reduce fractions, and

remove grouping symbols

2. Perform the same

operation on both sides.

(a) Add ( 3) 3 7 4

(

x x x

x x

+ = =

− + = =

b) Subtract (2 ) 5 2 4 3 4

(c) Multiply by a

nonzero constant (1/3) 3 12 4

(d) Divide by a constant

x x x x

x x

= + =

= =

nonzero term (3) 3 12 4x x= =

26


Example Solving a Linear Equation

Involving Fractions

10 4Solve for . 2

4 4

y yy

−= +


Example Solving a Linear Equation

Involving Fractions

10 4Solve for . 2

4 4

y yy

−= +

10 42

4 4

10 44 2 4 Multiply by the LCD

4 4

10y 4 8 Distributive Property

9 12 Simplify

4

3

y y

y y

y

y

y

−= +

− = +

− = +

=

=

27


Linear Inequality in x

A is one that can be written in the form

0, 0, 0, or 0, where and are

real numbers with 0.

ax b ax b ax b ax b a b

a

+ < + ≤ + > + ≥

≠

linear inequality in x


Properties of Inequalities

Let , , , and be real numbers, variables, or algebraic expressions,

and a real number.

If , and , then .

If

u v w z

c

u v v w u w

u v

< < <

<

1. Transitive

2. Addition then .

If and then .

If and 0, then .

If and 0, then .

T

u w v w

u v w z u w v z

u v c uc vc

u v c uc vc

+ < +

< < + < +

< > <

< < >

3. Multiplication

he above properties are true if < is replaced by . There are

similar properties for > and .

≤

≥

28


P.4

Lines in the Plane


Quick Review

Solve for .

1. 50 100 200

2. 3(1 2 ) 4( 2) 10

Solve for .

3. 2 3 5

4. 2 3( )

7 25. Simplify the fraction.

10 ( 3)

x

x

x x

y

x y

x x y y

− + =

− + + =

− =

− + =

−

− − −

29



Solve for .

1. 50 100 200

2. 3(1 2 ) 4( 2) 10

Solve for .

3. 2 3 5

4. 2 3( )

7 25. Simplify t

2

1

2

2

he fraction. 1

5

3

3

70 (

4

5

)

x

x

x x

y

x

x

x

xy

xy

y

x x y y

− + =

− + + =

− =

− + =

−

− − −

= −

=

−=

−=

−



� Slope of a Line

� Point-Slope Form Equation of a Line

� Slope-Intercept Form Equation of a Line

� Graphing Linear Equations in Two Variables

� Parallel and Perpendicular Lines

� Applying Linear Equations in Two Variables

… and why

Linear equations are used extensively in applications involving

business and behavioral science.

30


Slope of a Line


Slope of a Line

1 1

2 1

2 2

2 1

1 2

The slope of the nonvertical line through the points ( , )

and ( , ) is .

If the line is vertical, then and the slope is undefined.

x y

y yyx y m

x x x

x x

−∆= =

∆ −

=

31


Example Finding the Slope of a Line

Find the slope of the line containing the points (3,-2) and (0,1).


Example Finding the Slope of a Line

Find the slope of the line containing the points (3,-2) and (0,1).

2 1

2 1

1 ( 2) 31

0 3 3

Thus, the slope of the line is 1.

y ym

x x

− − −= = = = −

− − −

−

32


Point-Slope Form of an Equation of a Line

1 1 1 1

The of an equation of a line that passes through

the point ( , ) and has slope is ( ).x y m y y m x x− = −

point - slope form


Point-Slope Form of an Equation of a Line

33


Slope-Intercept Form of an Equation of a

Line

The slope-intercept form of an equation of a line with slope m

and y-intercept (0,b) is y = mx + b.


Forms of Equations of Lines

General form: Ax + By + C = 0, A and B not both zero

Slope-intercept form: y = mx + b

Point-slope form: y – y1 = m(x – x1)

Vertical line: x = a

Horizontal line: y = b

34


Graphing with a Graphing Utility

To draw a graph of an equation using a grapher:

1. Rewrite the equation in the form y = (an

expression in x).

2. Enter the equation into the grapher.

3. Select an appropriate viewing window.

4. Press the “graph” key.


Viewing Window

35


Parallel and Perpendicular Lines

1 2

1. Two nonvertical lines are parallel if and only if their

slopes are equal.

2. Two nonvertical lines are perpendicular if and only

if their slopes and are opposite reciprocals.

That is, if and only

m m

1

2

1 if .m

m= −


Example Finding an Equation of a

Parallel Line

Find an equation of a line through (2, 3) that is parallel to

4 5 10.x y

−

+ =

36


Example Finding an Equation of a

Parallel Line

Find an equation of a line through (2, 3) that is parallel to

4 5 10.x y

−

+ =

( )

Find the slope of 4 5 10.

5 4 10

4 42 The slope of this line is .

5 5

Use point-slope form:

43 2

5

x y

y x

y x

y x

+ =

= − +

= − + −

+ = − −


P.5

Solving Equations Graphically,

Numerically, and Algebraically

37


Quick Review

( )

( )( )

2

3 2

4 2

Expand the product.

1. 2

2. 2 1 4 3

Factor completely.

3. 2 2

4. 5 36

5. Combine the fractions and reduce the resulting fraction

2to lowest terms.

2 1 1

x y

x x

x x x

y y

x

x x

+

+ −

+ − −

+ −

−+ −



( )

( )( )

( )( )( )

( )( )( )

2 2

3 2

4 2 2

2

2

4 4

8 2 3

1

Expand the product.

1. 2

2. 2 1 4 3

Factor completely.

3. 2 2

4. 5 36

5. Combine the fractions and reduce the resulting fraction

to low

2

9 2

e

1

2

x xy y

x x

x x x

y y

x y

x x

x x

y y

x

y

+

+ −

+ − −

+

+ +

− −

+ − +

+ − +−

( )( )

2

2st terms.

2 1 11

5 2

2 1

xx

x x

x

x x−

+ −

− +

+ −

38



� Solving Equations Graphically

� Solving Quadratic Equations

� Approximating Solutions of Equations Graphically

� Approximating Solutions of Equations Numerically with Tables

� Solving Equations by Finding Intersections

… and why

These basic techniques are involved in using a graphing utility to

solve equations in this textbook.


Example Solving by Finding x-Intercepts

2

Solve the equation 2 3 2 0 graphically.x x− − =

39


Example Solving by Finding x-Intercepts

2

Solve the equation 2 3 2 0 graphically.x x− − =

2Find the -intercepts of 2 3 2.

Use the Trace to see that ( 0.5,0) and (2,0) are -intercepts.

Thus the solutions are 0.5 and 2.

x y x x

x

x x

= − −

−

= − =


Zero Factor Property

Let a and b be real numbers.

If ab = 0, then a = 0 or b = 0.

40


Quadratic Equation in x

A quadratic equation in x is one that can be written in

the form ax2 + bx + c = 0, where a, b, and c are real

numbers with a ≠ 0.


Completing the Square

2 2

2 2

2

22

To solve by completing the square, add ( / 2) to

both sides of the equation and factor the left side of the new

equation.

2 2

2 4

x bx c b

b bx bx c

b bx c

+ =

+ + = +

+ = +

41


Quadratic Equation

2

2

The solutions of the quadratic equation 0, where

0, are given by the

4.

2

ax bx c

a

b b acx

a

+ + =

≠

− ± −=

quadratic formula


Example Solving Using the Quadratic

Formula

2

Solve the equation 2 3 5 0.x x+ − =

42


Example Solving Using the Quadratic

Formula

2

Solve the equation 2 3 5 0.x x+ − =

( )( )

( )

2

2

2, 3, 5

4

2

3 3 4 2 5

2 2

3 49

4

3 7

4

5 or 1.

2

a b c

b b acx

a

x x

= = = −

− ± −=

− ± − −=

− ±=

− ±=

= − =


Solving Quadratic Equations Algebraically

These are four basic ways to solve quadratic equations

algebraically.

1. Factoring

2. Extracting Square Roots

3. Completing the Square

4. Using the Quadratic Formula

43


Agreement about Approximate Solutions

For applications, round to a value that is

reasonable for the context of the problem. For

all others round to two decimal places unless

directed otherwise.


Example Solving by Finding Intersections

Solve the equation | 2 1 | 6.x − =

44


Example Solving by Finding Intersections

Solve the equation | 2 1 | 6.x − =

Graph | 2 1 | and 6. Use Trace or the intersect feature

of your grapher to find the points of intersection.

The graph indicates that the solutions are 2.5 and 3.5.

y x y

x x

= − =

= − =


P.6

Complex Numbers

45


Quick Review

( )( )

Add or subtract, and simplify.

1. (2 3) ( 3)

2. (4 3) ( 4)

Multiply and simplify.

3. ( 3)( 2)

4. 3 3

5. (2 1)(3 5)

x x

x x

x x

x x

x x

+ + − +

− − +

+ −

+ −

+ +



( )( )

2

2

2

Add or subtract, and simplify.

1. (2 3) ( 3)

2. (4 3) ( 4)

Multiply and simplify.

3. ( 3)( 2)

4. 3 3

5. (2 1)(

6

3 7

6

3

6 135 53 )

x x

x x

x x

x

x

x x

x

x x

x x

x x

+ + − +

− − +

+

+

−

+ −

−

+

+ −

+ + +

−

46



� Complex Numbers

� Operations with Complex Numbers

� Complex Conjugates and Division

� Complex Solutions of Quadratic Equations

… and why

The zeros of polynomials are complex numbers.


Complex Number

A complex number is any number that can be

written in the form a + bi, where a and b are

real numbers. The real number a is the real

part, the real number b is the imaginary part,

and a + bi is the standard form.

47


Addition and Subtraction of Complex

Numbers

If a + bi and c + di are two complex numbers, then

Sum: (a + bi ) + (c + di ) = (a + c) + (b + d)i,

Difference: (a + bi ) – (c + di ) = (a - c) + (b -d)i.


Example Multiplying Complex Numbers

( )( )Find 3 2 4 .i i+ −

48


Example Multiplying Complex Numbers

( )( )Find 3 2 4 .i i+ −

( )( )2

3 2 4

12 3 8 2

12 5 2( 1)

12 5 2

14 5

i i

i i i

i

i

i

+ −

= − + −

= + − −

= + +

= +


Complex Conjugate

The of the complex number is

.

z a bi

z a bi a bi

= +

= + = −

complex conjugate

49


Discriminant of a Quadratic Equation

2

2

2

2

For a quadratic equation 0, where , , and are

real numbers and 0.

if 4 0, there are two distinct real solutions.

if 4 0, there is one repeated real solution.

if 4 0, the

ax bx c a b c

a

b ac

b ac

b ac

+ + =

≠

− >

− =

− <

i

i

i re is a complex pair of solutions.


Example Solving a Quadratic Equation

2Solve 2 0.x x+ + =

50


Example Solving a Quadratic Equation

2Solve 2 0.x x+ + =

( )

( )

2

1, and 2.

1 1 4 1 2

2 1

1 7

2

1 7

2

1 7 1 7So the solutions are and .

2 2

a b c

x

i

i ix x

= = =

− ± −=

− ± −=

− ±=

− + − −= =


P.7

Solving Inequalities Algebraically

and Graphically

51


Quick Review

2

2

2

Solve for .

1. 3 2 1 9

2. | 2 1| 3

3. Factor completely. 4 9

494. Reduce the fraction to lowest terms.

7

25. Add the fractions and simplify.

1

x

x

x

x

x

x x

x x

x x

− < + <

+ =

−

−

+

++

+



( )( )2

2

2

Solve for .

1. 3 2 1 9

2. | 2 1 | 3

3. Factor completely. 4 9

494. Reduce the fraction to lowest terms.

7

5. Add the fractions and

2 4

2 or 1

2 3

simpli

2 3

7

fy.

x

x

x

x

x

x

x x

x x

x

x

x

xx

− < <

= − =

−

− < + <

+ =

−

− +

−

+2

2

2

3 22

1

xx

xx x

x

x

+ ++

+ ++

52



� Solving Absolute Value Inequalities

� Solving Quadratic Inequalities

� Approximating Solutions to Inequalities

� Projectile Motion

… and why

These techniques are involved in using a graphing

utility to solve inequalities in this textbook.


Solving Absolute Value Inequalities

Let be an algebraic expression in and let be a real number

with 0.

1. If | | , then is in the interval ( , ). That is,

| | if and only if .

2. If | | , then is in the interval (

u x a

a

u a u a a

u a a u a

u a u

≠

< −

< − < <

> , ) or ( , ). That is,

| | if and only if or .

The inequalities < and > can be replaced with and ,

respectively.

a a

u a u a u a

−∞ − ∞

> < − >

≤ ≥

53


Example Solving an Absolute Value

Inequality

Solve | 3 | 5.x + <


Example Solving an Absolute Value

Inequality

Solve | 3 | 5.x + <

| 3 | 5

5 3 5

8 2

As an interval the solution in ( 8, 2).

x

x

x

+ <

− < + <

− < <

−

54


Example Solving a Quadratic Inequality

2Solve 3 2 0.x x+ + <


Example Solving a Quadratic Inequality

2Solve 3 2 0.x x+ + <

2

2

Solve 3 2 0

( 2)( 1) 0

2 or 1.

Use these solutions and a sketch of the equation

3 2 to find the solution to the inequality

in interval form ( 2, 1).

x x

x x

x x

y x x

+ + =

+ + =

= − = −

= + +

− −

55


Projectile Motion

Suppose an object is launched vertically from a point

so feet above the ground with an initial velocity of vo

feet per second. The vertical position s (in feet) of the

object t seconds after it is launched is

s = -16t2 + vot + so.


Chapter Test

1. Write the number in scientific notation.

The diameter of a red blood corpuscle is about 0.000007 meter.

2. Find the standard form equation for the circle with center (5, 3)

and radius 4.

3. Find the s

−

lope of the line through the points ( 1, 2) and (4, 5).

4. Find the equation of the line through (2, 3) and perpendicular

to the line 2 5 3.

2 5 15. Solve the equation algebraically.

3 2 3

6. Solve th

x y

x x

− − −

−

+ =

− ++ =

2e equation algebraically. 6 7 3x x+ =

56


Chapter Test

2

7. Solve the equation algebraically. | 4 1 | 3

8. Solve the inequality. | 3 4 | 2

9. Solve the inequality. 4 12 9 0

10. Perform the indicated operation, and write the result

in standard form. (5 7 ) (3

x

x

x x

i

+ =

+ ≥

+ + ≥

− − − 2 )i

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