Slide 6-1 Copyright © 2005 Pearson Education, Inc. SEVENTH EDITION and EXPANDED SEVENTH EDITION.

Slide 6-1 Copyright © 2005 Pearson Education, Inc.

SEVENTH EDITION and EXPANDED SEVENTH EDITION

Copyright © 2005 Pearson Education, Inc.

Chapter 6

Algebra, Graphs and Functions


6.1

Order of Operations


Definitions

Algebra: a generalized form of arithmetic. Variables: used to represent numbers Algebraic expression: a collection of variables,

numbers, parentheses, and operation symbols. Examples:

24 2, 4, 4(3 5), , 8 2

3 5

xx x y y y

x


Order of Operations

1. First, perform all operations within parentheses or other grouping symbols (according to the following order).

2. Next, perform all exponential operations (that is, raising to powers or finding roots).

3. Next, perform all multiplication and divisions from left to right.

4. Finally, perform all additions and subtractions from left to right.


Example: Evaluating an Expression

Evaluate the expression x2 + 4x + 5 for x = 3. Solution:

x2 + 4x + 5

= 32 + 4(3) + 5

= 9 + 12 + 5

= 26


Example: Substituting for Two Variables

Evaluate when x = 3 and y = 4. Solution:

2 24 3 5x xy y

2 2

2 2

4(3) 3(3)(4) 5(4 )

4(9) 36 5(16)

36 36 80

0 80

8

4 3 5

0

x xy y


6.2

Linear Equations in One Variable


Definitions

Like terms are terms that have the same variables with the same exponents on the variables.

Unlike terms have different variables or different

exponents on the variables.

2 22 , 7 5 , 8x x x x

3 22 , 7 5 , 6x x x


Properties of the Real Numbers

Associative property of multiplication

(ab)c = a(bc)

Associative property of addition

(a + b) + c = a + (b + c)

Commutative property of multiplication

ab = ba

Commutative property of addition

a + b = b + a

Distributive propertya(b + c) = ab + ac


Example: Combine Like Terms

8x + 4x

= (8 + 4)x

= 12x

5y 6y

= (5 6)y

= y

x + 15 5x + 9

= (1 5)x + (15+9)

= 4x + 24

3x + 2 + 6y 4 + 7x

= (3 + 7)x + 6y + (2 4)

= 10x + 6y 2


Solving Equations

Addition Property of Equality

If a = b, then a + c = b + c for all real numbers a, b, and c.

Find the solution to the equation

x 9 = 24.

x 9 + 9 = 24 + 9

x = 33

Check: x 9 = 24

33 9 = 24 ? 24 = 24 true


Solving Equations continued

Subtraction Property of Equality

If a = b, then a c = b c for all real numbers a, b, and c.


x + 12 = 31.

x + 12 12 = 31 12

x = 19

Check: x + 12 = 31

19 + 12 = 31 ? 31 = 31 true



Multiplication Property of Equality

If a = b, then a • c = b • c for all real numbers a, b, and c, where c 0.


9.

7

x

1

97

(7 7 9)7

7

x

x

1 7

x63

63x



Division Property of Equality

If a = b, then for all real numbers a, b, and c, c 0.

Find the solution to the equation 4x = 48.

a b

c c

4

4 48

4 48

124

x

x

x


General Procedure for Solving Linear Equations If the equation contains fractions, multiply both sides of

the equation by the lowest common denominator (or least common multiple). This step will eliminate all fractions from the equation.

Use the distributive property to remove parentheses when necessary.

Combine like terms on the same side of the equal sign when possible.


General Procedure for Solving Linear Equations continued Use the addition or subtraction property to

collect all terms with a variable on one side of the equal sign and all constants on the other side of the equal sign. It may be necessary to use the addition or subtraction property more than once. This process will eventually result in an equation of the form ax = b, where a and b are real numbers.


General Procedure for Solving Linear Equations continued Solve for the variable using the division or

multiplication property. This will result in an answer in the form x = c, where c is a real number.


Example: Solving Equations

Solve 3x 4 = 17.

4

3 4 17

3 4 17

3 21

3 21

4

3 37

x

x

x

x

x


Solve 21 = 6 + 3(x + 2)

21 6 3( 2)

21 6 3 6

21 3 12

1221 3 12

9 3

9 3

3

12

3 3

x

x

x

x

x

x

x


Solve 8x + 3 = 6x + 21

8 3 6 21

8 3 6 21

8 6 18

8 6 18

2 18

2 18

9

3 3

6 6

2 2

x x

x x

x x

x x

x

x

x

x

x


Solve 6(x 2) + 2x + 3 = 4(2x 3) + 2

False, the equation has no solution. The equation is inconsistent.

6( 2) 2 3 4(2 3) 2

6 12 2 3 8 12 2

8 9 8 10

8 8 9 8 8 10

9 10

x x x

x x x

x x

x x x x


Solve 4(x + 1) 6(x + 2) = 2(x + 4)

True, 0 = 0 the solution is all real numbers.

4( 1) 6( 2) 2( 4)

4 4 6 12 2 8

2 8 2 8

2 2 8 2 2 8

8 8

8 8 8 8

0 0

x x x

x x x

x x

x x x x


Proportions

A proportion is a statement of equality between two ratios.

Cross Multiplication If then ad = bc, b 0, d 0.,

a c

b d


To Solve Application Problems Using Proportions Represent the unknown quantity by a variable. Set up the proportion by listing the given ratio on the

left-hand side of the equal sign and the unknown and other given quantity on the right-hand side of the equal sign. When setting up the right-hand side of the proportion, the same respective quantities should occupy the same respective positions on the left and right.


To Solve Application Problems Using Proportions continued For example, an acceptable proportion might be

Once the proportion is properly written, drop the units and use cross multiplication to solve the equation.

Answer the question or questions asked.

miles miles

hour hour


Example

A 50 pound bag of fertilizer will cover an area of 15,000 ft2. How many pounds are needed to cover an area of 226,000 ft2?

754 pounds of fertilizer would be needed.

2 2

50 pounds

15,000 ft 226,000 ft

(50)(226,000) 15,000

11,300,000 15,000

11,300,000 15,000

15,000 15,000

753.33

x

x

x

x

x


6.3

Formulas


Definitions

A formula is an equation that typically has a real-life application.


Perimeter

The formula for the perimeter of a rectangle is Perimeter = 2 • length + 2 • width or P = 2l + 2w.

Use the formula to find the perimeter of a yard when l = 150 feet and w = 100 feet. P = 2l + 2w

P = 2(150) + 2(100)

P = 300 + 200

P = 500 feet


Example

The formula for the volume of a cylinder is V = r2h. Use the formula to find the height of a cylinder with a radius of 6 inches and a volume of 565.49 in3.

The height of the cylinder is 5 inches.

2

2565.49 (6 )

565.49 36

565.49 36

36 365.000

V r h

h

h

h

h


Exponential Equations: Carbon Dating

Carbon dating is used by scientists to find the age of fossils, bones, and other items. The formula used in carbon dating is

If 15 mg of C14 is present in an animal bone recently excavated, how many milligrams will be present in 4000 years?

/ 56000 2 .tP P


Exponential Equations: Carbon Dating continued

In 4000 years, approximately 9.2 mg of the original 15 mg of C14 will remain.

/ 56000

4000 / 5600

.71

2

15(2)

15(2)

15(0.61)

9.2 mg

tP P

P

P

P

P


Solving for a Variable in a Formula or Equation Solve the equation 3x + 8y 9 = 0 for y.

9 9

3

3 8 9 0

3 8 9 0

3 8 9

3

8 8

3 8 9

8 9 3

8 9 3

9 3

8

x x

x y

x y

x y

x y

y x

y x

xy


Solve for b2.

1 2( )2

hA b b

1 2

1 2

1 2

1 2

1 2

1 2

( )2

2 2 ( )2

2 ( )

( )2

2

2

hA b b

hA b b

A h b b

h b bA

h hA

b bh

Ab b

h


6.4

Applications of Linear Equations in One Variable


Translating Words to Expressions

3x 9The sum of three times a number decreased by 9.

2x + 8Eight more than twice a number

2xTwice a number

x 5Five less than a number

x + 10Ten more than a number

Mathematical ExpressionPhrase


To Solve a Word Problem

Read the problem carefully at least twice to be sure that you understand it.

If possible, draw a sketch to help visualize the problem. Determine which quantity you are being asked to find.

Choose a letter to represent this unknown quantity. Write down exactly what this letter represents.

Write the word problem as an equation. Solve the equation for the unknown quantity. Answer the question or questions asked. Check the solution.


Example

The bill (parts and labor) for the repairs of a car was $496.50. The cost of the parts was $339. The cost of the labor was $45 per hour. How many hours were billed?

Let h = the number of hours billed Cost of parts + labor = total amount

339 + 45h = 496.50


Example continued

The car was worked on for 3.5 hours.

339 45 496.50

339 339 45 496.50 339

45 157.50

45 157.50

45 453.5

h

h

h

h

h


Example

Sandra Cone wants to fence in a rectangular region in her backyard for her lambs. She only has 184 feet of fencing to use for the perimeter of the region. What should the dimensions of the region be if she wants the length to be 8 feet greater than the width?


continued, 184 feet of fencing, length 8 feet longer than width Let x = width of region Let x + 8 = length P = 2l + 2w x + 8

x

184 2( ) 2( 8)

184 2 2 16

184 4 16

168 4

42

x x

x x

x

x

x

The width of the region is 42 feet and the length is 50 feet.


6.5

Variation


Direct Variation

Variation is an equation that relates one variable to one or more other variables.

In direct variation, the values of the two related variables increase or decrease together.

If a variable y varies directly with a variable x, then y = kx where k is the constant of proportionality (or the variation constant).


Example

The amount of interest earned on an investment, I, varies directly as the interest rate, r. If the interest earned is $50 when the interest rate is 5%, find the amount of interest earned when the interest rate is 7%.

I = rx

$50 = 0.05x

1000 = x


Example continued

x = 1000, r = 7%

I = rx

I = 0.07(1000)

I = $70 The amount of interest earned is $70.


Inverse Variation

When two quantities vary inversely, as one quantity increases, the other quantity decreases, and vice versa.

If a variable y varies inversely with a variable, x, then y = k/x where k is the constant of proportionality.


Example

Suppose y varies inversely as x. If y = 12 when x = 18, find y when x = 21.

Now substitute 216 for k in y = k/x and find y when x = 21.

1218

216

ky

xk

k

216

2110.3

ky

x

y

y


Joint Variation

One quantity may vary directly as the product of two or more other quantities.

The general form of a joint variation, where y, varies directly as x and z, is y = kxz where k is the constant of proportionality.


Example

The area, A, of a triangle varies jointly as its base, b, and height, h. If the area of a triangle is 48 in2 when its base is 12 in. and its height is 8 in., find the area of a triangle whose base is 15 in. and whose height is 20 in.

A = kbh


Example continued

48 (12)(8)

48 (96)

48

961

2

A kbh

k

k

k

k

2

1(15)(20)

2

150 in.

A kbh

A

A


Combined Variation

A varies jointly as B and C and inversely as the square of D. If A = 1 when B = 9, C = 4, and D = 6, find A when B = 8, C = 12, and D = 5.

Write the equation.

2

kBCA

D


Combined Variation continued

Find the constant of proportionality.

Now find A.

2

2

(9)(4)1

636

136

1

kBCA

Dk

k

k

2

2

(1)(8)(12)

696

253.84

kBCA

D

A

A

A


6.6

Linear Inequalities


Symbols of Inequality

a < b means that a is less than b. a b means that a is less than or equal to b. a > b means that a is greater than b. a b means that a is greater than or equal to b.


Example: Graphing

Graph the solution set of x 4, where x is a real number, on the number line.

The numbers less than or equal to 4 are all the points on the number line to the left of 4 and 4 itself. The closed circle at 4 shows that 4 is included in the solution set.


Example: Graphing

Graph the solution set of x > 3, where x is a real number, on the number line.

The numbers greater than 3 are all the points on the number line to the right of 3. The open circle at 3 is used to indicate that 3 is not included in the solution set.


Solve 3x 8 < 10 and graph the solution set.

The solution set is all real numbers less than 6.

3 8 10

3 8 8 10 8

3 18

3 18

3 36

x

x

x

x

x


Compound Inequality

Graph the solution set of the inequality 4 < x 3

a) where x is an integer. The solution set is the integers between 4 and

3, including 3.


Compound Inequality continued

b) where x is a real number The solution set consists of all real numbers

between 4 and 3, including the 3 but not the 4.


Example

A student must have an average (the mean) on five tests that is greater than or equal to 85% but less than 92% to receive a final grade of B. Jamal’s grade on the first four tests were 98%, 89%, 88%, and 93%. What range of grades on the fifth test will give him a B in the course?


Example continued

98 89 88 9385 92

5368

85 925

5(85) 368 92(5)

425 368 460

425 368 368 368 460 368

57 92

x

x

x

x

x

x


6.7

Graphing Linear Equations


Rectangular Coordinate System

x-axis

y-axis

origin

Quadrant IQuadrant II

Quadrant III Quadrant IV

The horizontal line is called the x-axis.

The vertical line is called the y-axis.

The point of intersection is the origin.


Plotting Points

Each point in the xy-plane corresponds to a unique ordered pair (a, b).

Plot the point (2, 4). Move 2 units right Move 4 units up

2 units

4 units


Graphing Linear Equations

Graph the equation

y = 5x + 2

31

02/5

20

yx


To Graph Equations by Plotting Points

Solve the equation for y. Select at least three values for x and find their

corresponding values of y. Plot the points. The points should be in a straight line. Draw a

line through the set of points and place arrow tips at both ends of the line.


Graphing Using Intercepts

The x-intercept is found by letting y = 0 and solving for x. Example: y = 3x + 6

0 = 3x + 6 6 = 3x 2 = x

The y-intercept is found by letting x = 0 and solving for y. Example: y = 3x + 6

y = 3(0) + 6 y = 6


Graph 3x + 2y = 6

Find the x-intercept. 3x + 2y = 6 3x + 2(0) = 6 3x = 6 x = 2

Find the y-intercept. 3x + 2y = 6 3(0) + 2y = 6 2y = 6 y = 3


Slope

The ratio of the vertical change to the horizontal change for any two points on the line.

2 1

2 1

vertical changeSlope =

horizontal change

y ym

x x


Types of Slope

The slope of a vertical line is undefined.

The slope of a horizontal line is zero.

zero

negative

undefined

positive


Example: Finding Slope

Find the slope of the line through the points (5, 3) and (2, 3)

2 1

2 1

3 ( 3)

2 53 3

70

07

y ym

x x

m

m

m


Graphing Equations by Using the Slope and y-Intercept

Slope-Intercept Form of the Equation of the Line

y = mx + b where m is the slope of the line and (0, b) is the y-intercept of the line.


Steps

Solve the equation for y to place the equation in slope-intercept form.

Determine the slope and y-intercept from the equation.

Plot the y-intercept. Obtain a second point using the slope. Draw a straight line through the points.


Example

Graph 2x 3y = 9 Write in slope-intercept

form.

The y-intercept is (0,3) and the slope is 2/3.

2 3 9

3 2 9

3 2 9

3 3 32

33

x y

y x

y x

y x


Example continued

Plot a point at (0,3) on the y-axis, then move up 2 units and to the right 3 units.


Horizontal Lines

Graph y = 3

y is always equal to 3, the value of y can never be 0.

The graph is parallel to the x-axis.


Vertical Lines

Graph x = 3

x always equals 3, the value of x can never be 0.

The graph is parallel to the y-axis.


6.8

Linear Inequalities in Two Variables


To Graph Inequalities in Two Variables

Mentally substitute the equal sign for the inequality sign and plot points as if you were graphing the equation.

If the inequality is < or >, draw a dashed line through the points. If the inequality is or , draw a solid line through the points.

Select a test point not on the line and substitute the x and y-coordinates into the inequality. If the substitution results in a true statement, shade the area on the same side of the line as the test point. If the test point results in a false statement, shade the area on the opposite side of the line as the test point.


Graph 3x + 4y > 12

Draw a dashed line. Select a test point.

Try (0, 0) 3x + 4y > 12

3(0) + 4(0) > 12

0 + 0 > 12

0 > 12 false

Shade the opposite half plane.


Graph 3x + y 6

Draw a solid line. Select a test point.

Try (0, 0) 3x + y 6 3(0) + (0) 6 0 + 0 6 0 6

false Shade the

opposite half plane.


Graph y > 2

Draw a dashed line. Select a test point.

Try (0, 0) 0 > 2 true Shade the half plane

containing (0, 0).


6.9

Solving Quadratic Equations by Using Factoring and by Using the Quadratic Formula


FOIL

A binomial is an expression that contains two terms.

To multiply two binomials, we use the FOIL method.

F = First O = Outer I = Inner L = Last

(a + b)(c + d) = ac + ad + bc + bd

F

I

O

L


Example

Multiply: (2x + 4)(x + 6)

2

2

(2 4)

F

( 6) 2

O

2 6 4 4 6

2 12 4 24

2 16 2

I

4

L

x x x x x x

x x x

x x


To Factor Trinomial Expressions of the Form x2 + bx + c Find two numbers whose product is c and

whose sum is b. Write the factors in the form

(x + ) (x + )

Check your answer by multiplying the factors using the FOIL method.

One number from step 1

Other number from step 1


Factoring Example

Factor x2 7x + 12. We need to find two numbers whose product is 12 and

whose sum is 7.

(x 3)(x 4)

3 + 4 = 7

2 + 6 = 8

1 + 12 = 13

Sum of Factors

(3)(4)

(2)(6)

1(12)

Factors of 12

3 + 4 = 7(3)(4)

2 + 6 = 8(2)(6)

1 + 12 = 131(12)

Sum of FactorsFactors of 12


Factoring Trinomials of the Form ax2 + bc + c, a 1. Write all pairs of factors of the coefficient of the squared

term, a. Write all pairs of the factors of the constant, c. Try various combinations of these factors until the sum

of the products of the outer and inner terms is bx. Check your answer by multiplying the factors using the

FOIL method.


Example: Factoring

Factor 3x2 + 14x + 8. (3x + )(x + )

Thus, 3x2 + 14x + 8 = (3x + 2)(x + 4).

14x Correct middle term(3x + 2)(x + 4)

10x(3x + 4)(x + 2)

11x(3x + 8)(x + 1)

25x(3x + 1)(x + 8)

Sum of Outer and Inner Terms

Possible Factors


Solving Quadratic Equations by Factoring Standard Form of a Quadratic Equation

ax2 + bx + c = 0, a 0

Zero-Factor Property If a • b = 0, then a = 0 or b = 0.


To Solve a Quadratic Equation by Factoring Use the addition or subtraction property to make

one side of the equation equal to 0. Factor the side of the equation not equal to 0. Use the zero-factor property to solve the

equation.


Example: Solve by Factoring

Solve 4x2 + 17x 15 = 0.

The solutions are 5 and ¾.

24 17 15 0

(4 3)( 5) 0

4 3 0 or 5 0

4 3 or 5

3

4

x x

x x

x x

x x

x


Quadratic Formula

For a quadratic equation in standard form, ax2 + bx + c = 0, a 0, the quadratic formula is

2 4

2

b b acx

a


Example: Using the Quadratic Formula

Solve the equation 3x2 + 2x 7 = 0.

a = 3, b = 2 and

c = 7

2

1

2 2 4(3)( 7)

2(3)

2 4 84

6

2 88

6

2 2 22 2

6

123

22

6

1 22 1 22 or

3 3

2 4

2

b b acx

a


6.10

Functions and Their Graphs


Function

A function is a special type of relation where each value of the independent variable corresponds to a unique value of the dependent variable.

Domain the set of values used for the independent variable.

Range The resulting set of values obtained for the dependent variable.


Vertical Line Test

If a vertical line can be drawn so that it intersects the graph at more than one point, then each x does not have a unique y.


Types of Functions

Linear: y = ax + b

Quadratic: y = ax2 + bx + c


Graphs of Quadratic Functions

2

0

y ax bx c

a

2

0

y ax bx c

a

axis of symmetry axis of

symmetry


Graphs of Quadratic Functions continued

Axis of Symmetry of a Parabola

2

bx

a


General Procedure to Sketch the Graph of a Quadratic Equation

Determine whether the parabola opens upward or downward.

Determine the equation of the axis of symmetry. Determine the vertex of the parabola. Determine the y-intercept by substituting x = 0 into the

equation. Determine the x-intercept (if they exist) by substituting

y = 0 into the equation and solving for x. Draw the graph, making use of the information gained

in steps 1 through 5. Remember the parabola will be symmetric with respect to the axis of symmetry.


Graph y = x2 + 2x 3.

Since a = 1, the parabola opens up Axis:

y-coordinate of vertex (1, 4)

y-intercept: (0, 3)

2 21

2 2(1) 2

bx

a

2

2

2 3

( 1) 2( 1) 3 1 2 3 4

y x x

y

2

2

2 3

(0) 2(0) 3 3

y x x

y


Graph y = x2 + 2x 3 continued

x-intercepts:

Plot the points and sketch.

2 2 3 0

( 1)( 3) 0

1 or 3

x x

x x

x x


Graphs of Exponential Functions

Graph y = 3x. Domain: all real numbers Range is y > 0

(3, 1/27)1/273

(2, 1/9)1/92

(1, 1/3)1/31

(3, 27)273

9

3

1

y = 3x

(2, 9)2

(1, 3)1

(0, 1)0

(x, y)x


Graph

Domain: all real numbers Range is y > 0

1.

3

x

y

(3, 1/27)1/273

(2, 1/9)1/92

(1, 1/3)1/31

(3, 27)273

9

3

1

(2, 9)2

(1, 3)1

(0, 1)0

(x, y)x 1

3

x

y

Slide 6-1 Copyright © 2005 Pearson Education, Inc. SEVENTH EDITION and EXPANDED SEVENTH EDITION.

Documents