Slide P- 1. Chapter P Prerequisites P.1 Real Numbers

Slide P- 1

Chapter P

Prerequisites

P.1

Real Numbers

Slide P- 4

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 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

Slide P- 5

What you’ll learn about

Representing Real Numbers Order and Interval Notation Basic Properties of Algebra Integer Exponents Scientific Notation

… and whyThese topics are fundamental in the study of mathematics and science.

Slide P- 6

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,…}

Slide P- 7

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.

Slide P- 8

The Real Number Line

Slide P- 9

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.

Slide P- 10

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.

Slide P- 11

Example Interpreting Inequalities

Describe the graph of x > 2.

Slide P- 12

Example Interpreting Inequalities

Describe the graph of x > 2.

The inequality describes all real numbers greater than 2.

Slide P- 13

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.

Slide P- 14

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.

Slide P- 15

Graphing Inequalities

x > 2

x < -3 (-,-3]

(2,)

-1< x < 5 (-1,5]

Slide P- 16

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

Slide P- 17

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

Slide P- 18

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

Slide P- 19

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

Slide P- 20

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 Example

3 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 xu

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

Slide P- 21

Example Simplifying Expressions Involving Powers

2 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

Slide P- 22

Example Converting to Scientific Notation

Convert 0.0000345 to scientific notation.

-50.0000345 3.45 10

Slide P- 23

Example Converting from Scientific Notation

Convert 1.23 × 105 from scientific notation.

123,000

P.2

Cartesian Coordinate System

Slide P- 25

Quick Review Solutions

2 2

2 2

2 2

-5 31. Find the distance between and .

4 2Use 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

Slide P- 26


Cartesian Plane Absolute Value of a Real Number Distance Formulas Midpoint Formulas Equations of Circles Applications

… and whyThese topics provide the foundation for the material that will be covered in this textbook.

Slide P- 27

The Cartesian Coordinate Plane

Slide P- 28

Quadrants

Slide P- 29

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

Slide P- 30

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

Slide P- 31

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

Slide P- 32

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 )

Slide P- 33

The Distance Formula using the Pythagorean Theorem

Slide P- 34

Midpoint Formula (Number Line)

The is

.2

a bmidpoint of the line segment with endpoints and a b

Slide P- 35

Midpoint Formula (Coordinate Plane)

The is

, .2 2

a c b d

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

Slide P- 36

Find the distance and midpoint for the line segment joinedby A(-2,3) and B(4,1).

A(-2,3)

B(4,1)

22 )13()42(),( BAd

22 )2()6(),( BAd

40),( BAd 104102

2

)13(,

2

)42(

2

4,

2

2= (1,2)

Distance and Midpoint Example

Slide P- 37

Show that A(4,1), B(0,3), and C(6,5) are vertices of an isosceles triangle.

A(4,1)

B(0,3)

C(6,5)

52)31()04(),( 22 BAd

102)53()60(),( 22 CBd

52)15()64(),( 22 CAd

Since d(AC) = d(AB) , ΔABC is isosceles

Example Problem

Slide P- 38

P is a point on the y-axis that is 5units from the point Q (3,7). Find P.

5)7()03(),( 22 yQPdP

Q(3,7)

(0,y)

549149 2 yy

558142 yy

2558142 yy

033142 yy 0)3)(11( yy y = 3, y = 11

The point P is (0,3) or (0,11)

Example

Slide P- 39

Prove that the diagonals of a rectangle are congruent.

Coordinate Proofs

Given ABCD is a rectangle.Prove AC = BD

A(0,0)

B(0,a)

D(b,0)

C(b,a)

2222 )0()0( ababdAC

2222 )0()0( ababdBD

Since AC= BD, the diagonals of a square are congruent

Slide P- 40

Slide P- 41

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

Slide P- 42

Standard Form Equation of a Circle

Slide P- 43

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

Slide P- 45

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.

3 3

6 4 10

7

24.

4 3

x y

x y

x x y y x

x y

x xx x

x

25.

7 6

122

2 2

y

x

y

y

Slide P- 46


Equations Solving Equations Linear Equations in One Variable Linear Inequalities in One Variable

… and whyThese topics provide the foundation for algebraic techniques needed throughout this textbook.

Slide P- 47

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

Slide P- 48

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.

Slide P- 49

Operations for Equivalent EquationsAn 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

Slide P- 50

Example Solving a Linear Equation Involving Fractions

10 4Solve for . 2

4 4

y yy

10 42

4 410 4

4 2 4 Multiply by the LCD4 4

10y 4 8 Distributive Property

9 12 Simplify

4

3

y y

y y

y

y

y

Slide P- 51

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

Slide P- 52

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 .

P.4

Lines in the Plane

Slide P- 54

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 t

2

1

2

2

he fraction. 1

5

3

3

70 (

45

)

x

x

x x

y

x

x

x

xy

xy

y

x x y y

Slide P- 55


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 whyLinear equations are used extensively in applications involving business and behavioral science.

Slide P- 56

Slope of a Line

Slide P- 57

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

Slide P- 58

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

Slide P- 59

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

Slide P- 60

Point-Slope Form of an Equation of a Line

Slide P- 61

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.

Slide P- 62

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

Slide P- 63

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.

Slide P- 64

Viewing Window

Slide P- 65

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

Slide P- 66

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 5Use point-slope form:

43 2

5

x y

y x

y x

y x

or y = mx + b

5

1

5

4

5

15

83

)2(5

43

5

4

xyb

b

b

bxy

Slide P- 67

Determine the equation of the line (written in standard form) that passes through the point (-2, 3) and is perpendicular to the line 2y – 3x = 5.

Example

P.5

Solving Equations Graphically, Numerically, and Algebraically

Slide P- 69

Quick Review Solutions

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

22

st terms. 2 1 11

5 2

2 1

xx

x x

x

x x

Slide P- 70


Solving Equations Graphically Solving Quadratic Equations Approximating Solutions of Equations Graphically Approximating Solutions of Equations Numerically with

Tables Solving Equations by Finding Intersections

… and whyThese basic techniques are involved in using a graphing utility to solve equations in this textbook.

Slide P- 71

Example Solving by Finding x-Intercepts

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

Slide P- 72

Example Solving by Finding x-Intercepts

2Solve 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

Slide P- 73

Zero Factor Property

Let a and b be real numbers.

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

Slide P- 74

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.

Slide P- 75

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

Slide P- 76

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

Slide P- 77

Example Solving Using the Quadratic Formula

2Solve the equation 2 3 5 0.x x

2

2

2, 3, 5

4

2

3 3 4 2 5

2 2

3 49

43 7

4

5 or 1.

2

a b c

b b acx

a

x x

0)1)(52( xx

01or052 xx

1or 2

5 xx

Slide P- 78

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

Slide P- 79

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.

Slide P- 80

Example Solving by Finding Intersections

Solve the equation | 2 1| 6.x

Slide P- 81

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

Slide P- 83

Quick Review

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

Slide P- 84


Complex Numbers Operations with Complex Numbers Complex Conjugates and Division Complex Solutions of Quadratic Equations

… and why

The zeros of polynomials are complex numbers.

Slide P- 85

Find two numbers whose sum is 10 and whose product is 40.

x = 1st number 10 – x = 2nd number

x(10 – x) = 40

Complex Numbers

Slide P- 86

x(10 – x) = 40 10x – x2 = 40 x2 – 10x = -40 x2 – 10x + 25 = -40 +25 (x – 5)2 = -15

155 x

155155 , x

Complex Numbers

Slide P- 87

)) 15(5 15(5

)) 15(5 15(5

)15(155 15525 40

10

Complex Numbers

Slide P- 88

The imaginary number i is the

square root of –1.

1i 12 i

i416116

1010 i

Complex Numbers

Slide P- 89

Imaginary numbers are not real numbers, so all the rules do not apply.

Example: The product rule does not apply:

100205

100205205 2iii

10101

Complex Numbers

Slide P- 90

If a and b are real numbers, then

a + bi is a complex number. a is the real part. bi is the imaginary part.

The set of complex numbers consist of all the real numbers and all the imaginary numbers

Complex Numbers

Slide P- 91

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.

Complex Numbers

Slide P- 92

Examples of complex numbers: 3 + 2i 8 - 2i 4 (since it can be written as 4 + 0i).

The real numbers are a subset of the complex numbers.

-3i (since it can be written as 0 – 3i).

Complex Numbers

Slide P- 93

i 12i 11

2

3i iiii 12

4i 11 222 i

Complex Numbers

Slide P- 94

5i iiii 14

6i 11124 ii7i iiii 134

8i 11144 ii

Complex Numbers

Slide P- 95

* i -1 -i 1

-1 -i 1 i

-i 1 i -1

1 i -1 -i

i -1 -i 1

i

-1

-i

1

Complex Numbers

Slide P- 96

35i iiii 8384 1

Evaluate:

24i 11 664 i

Complex Numbers

Slide P- 97

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.

Slide P- 98

Example Multiplying Complex Numbers

Find 3 2 4 .i i

Slide P- 99

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

Slide P- 100

Complex Conjugate

The of the complex number is

.

z a bi

z a bi a bi

complex conjugate

Slide P- 101

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

re is a complex pair of solutions.

Slide P- 102

Example Solving a Quadratic Equation

2Solve 2 0.x x

Slide P- 103

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

Slide P- 104

When dividing a complex number by a real number, divide each part of the complex number by the real number.

4

86 i4

8

4

6 i i2

2

3

Complex Numbers

Slide P- 105

The numbers (a + bi ) and (a – bi ) are complex conjugates.

The product (a + bi )·(a – bi ) is the real number a 2 + b 2.

Show: (3 + 2i) (3 – 2i) = 3 2 + 2 2.

Complex Numbers

Slide P- 106

Show: (3 + 2i) (3 – 2i) = 3 2 + 2 2.

(3 + 2i) (3 – 2i) = 3.3 + 3(-2i) + 2i .3 + 2i (-2i)

= 3 2 – 6i + 6i – 2 2i 2

= 3 2 – 2 2(-1)

= 3 2 + 2 2

= 9 + 4

= 13

Complex Numbers

Slide P- 107

When dividing a complex number by a complex number, multiply the denominator and numerator by the conjugate of the denominator.

i

i

1

32

i

i

i

i

1

1

1

32

2

2

1

3322

i

iii

11

132

i

Complex Numbers

Slide P- 108

11

132

i

2

5 i

i2

1

2

5

Complex Numbers

Slide P- 109

Complex Numbers

P.7

Solving Inequalities Algebraically and Graphically

Slide P- 111

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

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

xx

xx

2

2

2

3 22

1

xx

xx x

x

x

Slide P- 112


Solving Absolute Value Inequalities Solving Quadratic Inequalities Approximating Solutions to Inequalities Projectile Motion

… and whyThese techniques are involved in using a graphing utility to solve inequalities in this textbook.

Slide P- 113

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

Slide P- 114

Solve 2x – 3 < 4x + 5 -2x < 8 x > -4

-5 -4 -3

Solve |x – 2| < 1 -1 < x – 2 < 1 1 < x < 3

0 1 2 3 4


Slide P- 115

Solve -1 < 3 – 2x < 5 -4 < -2x < 2 2 > x > -1 -1 < x < 2

Solve |x – 1| > 3 -3 > x – 1 or x – 1 > 3 -2 > x or x > 4 x < -2 or x > 4

-2 -1 0 1 2 3 -2 -1 0 1 2 3 4


Slide P- 116

•|2x – 6| < 4•-4 < 2x – 6 < 4•2 < 2x < 10•1< x < 5

-1 0 1 2 3 4 5( )

•|3x – 1| > 2•3x – 1 < -2 or 3x – 1 > 2•3x < -1 or 3x > 3•x < -1/3 or x > 1

-1 0 1 2 3 4 5

] [


Slide P- 117

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

Slide P- 118

Example Solving a Quadratic Inequality

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

-2 -1

+++0--------0+++

Slide P- 119

Solve x2 – x – 20 < 0

1. Find critical numbers (x + 4)(x - 5) < 0x = -4, x = 5

2. Test Intervals (-∞,-4) (-4,5) and (5, ∞)3. Choose a sample in each interval x = -5 (-5)2 – (-5) – 20 = Positive x = 0 (0)2 - (0) - 20 = Negative x = 6 (6)2 – 3(6) = Positive

Solution is (-4,5)

-4 5

+++0-------0+++


Slide P- 120

Solve x2 – 3x > 0

1. Find critical numbers x(x - 3) > 0x = 0, x = 3

2. Test Intervals (-∞,0) (0,3) and (3, ∞)3. Choose a sample in each interval x = -1 (-1)2 – 3(-1) = Positive x = 1 (1)2 - 3(1) = Negative x = 4 (4)2 – 3(4) = Positive

Solution is (-∞,0) or (3, ∞)

0 3

+++0------0+++


Slide P- 121

Solve x3 – 6x2 + 8x < 0

1. Find critical numbers x(x2 – 6x + 8) < 0x(x – 2)(x – 4) x = 0, x = 2, x = 4

2. Test Intervals (-∞,0) (0,2) (2,4) and (4, ∞)3. Choose a sample in each interval x = -5 (-5)3 – 6(-5)2 + 8(-5) = Negative x = 1 (-1)3 – 6(-1)2 + 8(-1) = Positive x = 3 (3)3 – 6(3)2 + 8(3) = Negative x = 5 (5)3 – 6(5)2 + 8(5) = Positive

Solution is (-∞,0] U [2,4]

0 2 4

-----0++0----0+++


Slide P- 122

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.

Slide P- 123

Chapter Test

-6

2

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.

7 10

5x

2

3. Find the slope 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.

5. Solve the equation al

3 16

3

5

58

ge

2

b

y

yx xy

2

2 5 1raically.

3 2 3

6. Solve the equation algebraically. 6 7 3

9

51 3

or 3 2

xx x

x x x x

Slide P- 124

Chapter Test

2

1 or 1

22

( , 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 wri

3

,

x

x

x

x

x

x

te the result

in standard form. (5 7 ) (3 2 ) 2 5i i i

Documents

Slide P- 1. Chapter P Prerequisites P.1 Real Numbers