40p6zu91z1c3x7lz71846qd1-wpengine.netdna-ssl.com40p6zu91z1c3x7lz71846qd1-wpengine.netdna-ssl.com/... · Chapter 2 Functions and Linear Functions 44 9 9 fx x() 3 2.1 Check Points 1

Chapter 2 Functions and Linear Functions

44

99

( ) 3f x x

2.1 Check Points

1. The domain is {0, 10, 20, 30, 34}. The range is {9.1, 6.7, 10.7, 13.2, 15.5}.

2. a. The relation is not a function because an element, 5, in the domain corresponds to two elements in the range.

b. The relation is a function.

3. a. ( ) 4 5(6) 4(6) 5(6) 29

f x xff

= += +=

b. 2

2

( ) 3 10

( 5) 3( 5) 10( 5) 65

g x x

gg

= −

− = − −− =

c. 2

2

( ) 7 2

( 4) ( 4) 7( 4) 2( 4) 46

h r r r

hh

= − +

− = − − − +− =

d. ( ) 6 9( ) 6( )( ) 6 6

F x xF a h a hF a h a h

= ++ = + ++ = + +

4. a. Every element in the domain corresponds to exactly one element in the range.

b. The domain is {0, 1, 2, 3, 4}. The range is {3, 0, 1, 2}.

c. (1) 0g =

d. (3) 2g =

e. and 0x = 4.x =

2.1 Exercise Set

2. The relation is a function. The domain is {4,6,8}. The range is {5,7,8}.

4. The relation is not a function. The domain is {5, 6}. The range is {6, 7}.

6. The relation is a function. The domain is {–7, –5, –3, 0}. The range is {–7, –5, –3, 0}.

8. The relation is a function. The domain is {4, 5, 6}. The range is {1}.

10. = +

(0) 0 3 3f

a. = + =

(5) 5 3 8f

b. = + =

( 8) 8 3 5f

c. − = − + = −

(2 ) 2 3f a a

d. + =

( )( 2) 2 32 3 5

f a aa a

+ = + + e. = + + = +

( ) 4 3

12. g x x − =

a. ( ) ( )0 4 0 3 0 3 3g − = − = − =

b. ( ) ( )5 4 5 3 20 3 23g = − − = − − = − −

c. 3 34 3 3 3 04 4

⎛ ⎞ ⎛ ⎞g = − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

d. ( ) ( )5 4 5 3 20 3b bg b − = − =

( e. )( 5) 4 5 34 20 3 4 17

bb b

+ = + −g b+ − = +

=

Intermediate Algebra for College Students 5E Section 2.1

14. 2( ) 2 4h x x= −

a. ( ) ( ) ( )20 2 0 4 2 0 40 4 4

h = − = −

= − = −

b. ( ) ( ) ( )21 2 1 4 2 1 42 4 2

h − = − − = −

= − = −

c. ( ) ( ) ( )25 2 5 4 2 25 450 4 46

h = − = −

= − =

d. ( ) ( ) ( )23 2 3 4 2 9 418 4 14

h − = − − = −

= − =

e. ( ) ( ) ( )2 2

2

5 2 5 4 2 25 4

50 4

h b b b

b

= − =

= −

−

2

2−

−

16. 2( ) 3 4 2f x x x= + −

a. ( ) ( ) ( )( )

20 3 0 4 0 2

3 0 0 20 0 2 2

f = + −

= + −

= + − = −

b. ( ) ( ) ( )( )

23 3 3 4 3 2

3 9 12 227 12 2 37

f = + −

= + −

= + − =

c. ( ) ( ) ( )( )

25 3 5 4 5 2

3 25 20 275 20 2 53

f − = − + − −

= − −

= − − =

d. ( ) ( ) ( )2

2

3 4

3 4 2

f b b b

b b

= + −

= + −

e. ( ) ( ) ( )

( )2

2

2

5 3 5 4 5

3 25 20 2

75 20 2

f a a a

a a

a a

= +

= +

= + −

18. 3 1( )5

xf xx−

=−

a. ( ) ( )( )3 0 1 0 1 1 10

0 5 5 5 5f

− − −= = = =

− − −

b. ( ) ( )3 3 1 9 1 83 43 5 2 2

f− −

= = = = −− − −

c. ( ) ( )3 3 1 9 133 5 8

10 58 4

f− − − −

− = =− − −

−= =

−

d. ( ) ( )3 10 1 30 1 291010 5 5 5

f− −

= = =−

e. ( ) ( )3 15

3 3 15

a hf a h

a ha ha h

+ −+ =

+ −+ −

=+ −

f. Five must be excluded from the domain, because 5 would make the denominator zero. Division by zero is undefined.

20. a. ( 3) 8f − =

b. (3) 16f =

c. 0x =

22. a. ( 2) 2h − = −

b. (1) 1h = −

c. 1x = − and 1x =

24. ( ) ( )( )( ) ( ) ( ) ( )2

1 3 1 5 3 5 8

1 8 8 8

64 8 4 76

g

f g f

− = − − = − − = −

4− = − = − − − +

= + + =

45

Chapter 2: Functions and Linear Functions

26. ( ) ( )24 1 3 3 3

4 1 9 3 3 6

3 9 1 63 9 6 6 6 0

− − − − − + − ÷ ⋅−

= − + − + − ÷ ⋅−

= − − + − ⋅−

= − + = − + =

6

7

3

1

28. ( ) ( )( ) ( ) ( )2 2

2 2

3 7 3

3 7 3 76

f x f x

x x x x

x x x xx

− −

= − − − + − − +

= + + − + −=

30. a. ( ) ( )3 6 3 1 18 1 19f − = − − = − − = −

b. ( ) ( )0 7 0 3 0 3f = + = + =

c. ( ) ( )4 7 4 3 28 3 3f = + = + =

d. ( ) ( )( ) ( )100 100

6 100 1 7 100 3600 1 700 3

100 2 102

f f− +

= − − + +

= − − + += + =

32. a. {(Nicholson, 12), (Olivier, 10), (Newman, 9), (Tracy, 9), (Brando, 8), (Lemmon, 8), (Pacino, 8)}

b. Yes, the relation is a function because each actor in the domain corresponds to exactly one number of nominations in the range.

c. {(12, Nicholson), (10, Olivier), (9, Newman), (9, Tracy), (8, Brando), (8, Lemmon), (8, Pacino)}

d. No, the relation is not a function because 9 nominations in the domain corresponds to two actors in the range, Newman and Tracy, and 8 nominations in the domain corresponds to three actors in the range, Brando, Lemmon, and Pacino.

34. – 36. Answers will vary.

38. makes sense

40. does not make sense; Explanations will vary. Sample explanation: The range is the chance of divorce.

42. false; Changes to make the statement true will vary. A sample change is: Functions can have ordered pairs with the same second component. It is the first component that cannot be duplicated.

44. true

46. false; ( 4) ( 4)( 1) ( 1)

2

g f− + −= − + −= −

48. Answers will vary. An example is {(1, 1), (2, 1)}.

50. ( )

( )

( )( )

( )

2

2

2

24 4 2 5 2 6

24 4 2 3 6

24 4 1 6

24 4 1 6

6 1 6 6 6 0

÷ − − −⎡ ⎤⎣ ⎦

= ÷ − −⎡ ⎤⎣ ⎦

= ÷ − −

= ÷ −

= − = − =

51. 2 2 22 2 2 5 10

3 5 23 3

3 9 4x y x y

y y x

− −−⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

yx

52. 3 43 5x x= +

( )

( )

315 15 43 5

315 15 15 43 55 3 3 605 9 60

5 9 9 9 604 604 604 4

15

x x

x x

x xx x

x x x xxx

x

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= +

= +− = − +− =−

=− −

= −

The solution set is { }15− .

46


53. x ( ) 2f x x= ( ),x y −2 ( )2 2( 2) 4f − = − = − ( )2, 4− − −1 ( )1 2( ) 21f − = = −− ( )1, 2− − 0 ( )0 2(0) 0f = = ( )0,0 1 ( )1 2(1) 2f = = ( )1,2 2 ( )2 2(2) 4f = = ( )2, 4

54. x ( ) 2 4f x x= + ( ),x y

−2 ( )2 2( 2) 4 0f − = − + =

( )2,0−

−1

( )1 2( ) 4 21f − = + =−

( )1, 2−

0 ( )0 2(0) 4 4f = + = ( )0, 4 1 ( )1 2(1) 4 6f = + = ( )1,6 2 ( )2 2(2) 4 8f = + = ( )2,8

55. a. When the x-coordinate is 2, the y-coordinate is

3.

b. When the y-coordinate is 4, the x-coordinates are –3 and 3.

c. The x-coordinates are all real numbers.

d. The y-coordinates are all real numbers greater than or equal to 1.

2.2 Check Points

1. x ( ) 2f x x= ( ),x y

−2 ( )2 2( 2)f 4− = − = − ( )2, 4− −

−1 ( )1 2( ) 21f − = =− − ( )1, 2− −

0 ( )0 2(0) 0f = = ( )0,0

1 ( )1 2(1)f 2= = ( )1, 2

2 ( )2 2(2) 4f = = ( )2,4

x ( ) 2 3g x x= − ( ),x y −2 ( ) 2( ) 32 2g − 7− = − = − ( )2, 7− − −1 ( ) 2( ) 3 51 1g − = −− = − ( )1, 5− − 0 ( ) 2(0) 3 30g − = −= ( )0, 3− 1 ( ) 2(1) 3 11g − = −= ( )1, 1− 2 ( ) 2(2) 3 12g − == ( )2,1

The graph of g is the graph of f shifted down by 3 units.

2. a. The graph represents a function. It passes the

vertical line test.

b. The graph represents a function. It passes the vertical line test.

c. The graph does not represent a function. It fails the vertical line test.

3. a. (5) 400f =

b. When x is 9, the function’s value is 100. i.e. (9) 100f =

c. The minimum T cell count during the asymptomatic stage is approximately 425.

47


4. a. The domain is { } .

The range is { } .

| 2 1x x− ≤ ≤

| 0 3y y≤ ≤

b. The domain is { } .

The range is { } .

| 2 1x x− < ≤

| 1 2y y− ≤ <

c. The domain is { } .

The range is { } .

| 3 0x x− ≤ <

| 3, 2, 1y y = − − −

2.2 Exercise Set

2. x ( )f x x= ( ),x y

−2 ( )2 2f − = − ( )2, 2− −

−1 ( )1 1f − = − ( )1, 1− −

0 ( )0 0f = ( )0,0

1 ( )1 1f = ( )1,1

2 ( )2 2f = ( )2, 2

x ( ) 4g x x= − ( ),x y 2− ( )2 2 4g − = − − = −6 ( )2, 6− − 1− ( )1 1 4g − = − − = −5 ( )1, 5− − 0 ( )0 0 4 4g = − = − ( )0, 4− 1 ( )1 1 4 3g = − = − ( )1, 3− 2 ( )2 2 4 2g = − = − ( )2, 2−

The graph of g is the graph of f shifted down 4 units.

4. x ( ) 2f x x= − ( ),x y −2 ( ) ( )2 2 2 4f − = − − = ( )2, 4− −1 ( ) ( )1 2 1 2f − = − − = ( )1, 2− 0 ( ) ( )0 2 0 0f = − = ( )0,0 1 ( ) ( )1 2 1 2f = − = − ( )1, 2− 2 ( ) ( )2 2 2 4f = − = − ( )2, 4− x ( ) 2 3g x x= − + ( ),x y 2− ( ) ( )2 2 32g 7− = − + =− ( )2,7− 1− ( ) ( )1 2 31g 5− = − + =− ( )1,5− 0 ( ) ( )0 2 30g 3= − + = ( )0,3 1 ( ) ( )1 2 31g 1= − + = ( )1,1 2 ( ) ( )2 2 2 3g 1= − + = − ( )2, 1−

The graph of g is the graph of f shifted up 3 units.

48


6. x ( ) 2f x x= ( ),x y -2 ( ) ( )22 2 4f − = − = ( )2, 4−

-1 ( ) ( )21 1 1f − = − = ( )1,1−

0 ( ) ( )20 0 0f = = ( )0,0

1 ( ) ( )21 1 1f = = ( )1,1

2 ( ) ( )22 2 4f = = ( )2, 4

x ( ) 2 2g x x= − ( ),x y 2− ( ) ( )22 2 2g − = − − = 2 ( )2, 2−

1− ( ) ( )21 1 2g − = − − = −1 ( )1, 1− −

0 ( ) ( )20 0 2g = − = −2 ( )0, 2−

1 ( ) ( )21 1 2g = − = −1 ( )1, 1−

2 ( ) ( )22 2 2g = − = 2 ( )2, 2

The graph of g is the graph of f shifted down 2 units.

8. x ( )f x x= ( ),x y 2− ( )2 2f − = − = 2 ( )2, 2− 1− ( )1 1f − = − =1 ( )1,1− 0 ( )0 0f = = 0 ( )0,0 1 ( )1 1f = =1 ( )1,1 2 ( )2 2f = = 2 ( )2, 2 x ( ) 1g x x= + ( ),x y 2− ( )2 2 1g − = − + = 3 ( )2,3− 1− ( )1 1 1g − = − + = 2 ( )1, 2− 0 ( )0 0 1g = + =1 ( )0,1 1 ( )1 1 1g = + = 2 ( )1,2 2 ( )2 2 1g = + = 3 ( )2,3

The graph of g is the graph of f shifted up 1 unit.

49


10. x ( ) 3f x x= ( ),x y

2− ( ) ( )32 2f − = − = −8 ( )2, 8− −

1− ( ) ( )31 1f − = − = −1 ( )1, 1− −

0 ( ) ( )30 0f = = 0 ( )0,0

1 ( ) ( )31 1f = =1 ( )1,1

2 ( ) ( )32 2f = = 8 ( )2,8

x ( ) 3 1g x x= − ( ),x y

2− ( ) ( )32 2 1g − = − − = −9 ( )2, 9− −

1− ( ) ( )31 1 1g − = − − = −2 ( )1, 2− −

0 ( ) ( )30 0 1g = − = −1 ( )0, 1−

1 ( ) ( )31 1 1g = − = 0 ( )1,0

2 ( ) ( )32 2 1g = − = 7 ( )2,7

The graph of g is the graph of f shifted down 1 unit.

12. The graph represents a function. It passes the vertical line test.

14. The graph does not represent a function. It fails the vertical line test.

16. The graph does not represent a function. It fails the vertical line test.

18. The graph represents a function. It passes the vertical line test.

20. (2) 4f = −

22. ( 4) 4f − =

24. ( 1) 0f − =

26. ( )2 2g = −

28. (10) 2g = −

30. When 1, ( ) 1.x g x= = −

32. The domain is { | 5 0x x− < ≤ }. The range is { | 3 3y y− ≤ < }.

34. The domain is { }. The range is { }.

| 1x x ≥ −| 0y y ≥

36. The domain is { | 3 2x x− ≤ ≤ }. The range is { | 5 5y y− ≤ ≤ }.

38. The domain is { |x x is a real number}. The range is { }. | 0y y ≥

40. The domain is { }. The range is {

| 5, 2,0,1,x x = − − 4| 2y y = − }.

42. a. The domain is { }| 6x x ≤ .

b. The range is { }| 1y y ≤ .

c. ( 4) 1f − = −

d. –6 and 6; i.e. ( 6) (6) 3f f− = = −

e. f crosses the x-axis at ( 3,0) and (3,0).−

f. f crosses the y-axis at (0,1).

g. for ( ) 0f x > { }| 3 3 .x x− < <

h. ( 2)f − is positive.

44. a. (20) 3.5(20) 472 542.M = + = Approximately 542,000 bachelor’s degrees were awarded to men in 2000. This is represented as (20, 542) on the graph.

b. overestimates the actual data shown by the bar graph by 2 thousand.

(20)M

50


62. The cost to mail a letter weighing 1.8 ounces is $0.58.

46. a.

Approximately 19,500 more bachelor’s degrees were awarded to women than to men in 1985. The points on the graph with first coordinate 5 are 19.5 units apart.

[ ] [ ](5) (5)13.2(5) 443 3.5(5) 472 509 489.5

19.5

W M−= + − + = −

= 64. – 66. Answers will vary.

68.

The number of physician’s visits per year based on age first decreases and then increases over a person’s lifetime.

These are the approximate coordinates of the point (20.3, 4.0). The means that the minimum number of physician’s visits per year is approximately 4. This occurs around age 20.

b. underestimates the actual difference of 20 thousand, as shown by the data in the bar graph, by 0.5 thousand.

(5) (5)W M−

48.

Fifty-year-old drivers have 200 accidents per 50 million miles driven. This is represented on the graph by point (50,200).

( ) ( )( )

2(50) 0.4 50 36 50 1000

0.4 2500 1800 10001000 1800 1000 200

f = − +

= − +

= − + =

50. Answers will vary. One possible answer is age 16 and age 74.

Both 16-year-olds and 74-year-olds have approximately 526.4 accidents per 50 million miles driven.

( ) ( )( )

2(16) 0.4 16 36 16 1000

0.4 256 576 1000102.4 576 1000 526.4

f = − +

= − +

= − + =

( ) ( )( )

2(74) 0.4 74 36 74 1000

0.4 5476 2664 10002190.4 2664 1000 526.4

f = − +

= − +

= − + =

70. makes sense

72. does not make sense; Explanations will vary. Sample explanation: The domain is the set of the various ages of the people.

74. true

76. false; Changes to make the statement true will vary. A sample change is: The range of f is { }2 2y y− ≤ <

52. In 2000, 2.2% of the U.S. population was made up of Jewish Americans.

(100) 2.2f ≈

.

54. For and , This means that in 1914 and 1990, 2.5% of the U.S. population was made up of Jewish Americans.

14x ≈ 90x ≈ ( ) 2.5.f x = 78. false; Changes to make the statement true will vary. A sample change is: (0) 0.6f =

80. 2

2

( 2.5) (1.9) [ ( )] ( 3) (1) ( )

2 ( 2) [3] 2 ( 2)( 4)

4 9 ( 1)( 4)2 9 4

3

f f f f f fπ π− − − − + − ÷ ⋅

= − − − + ÷ − −

= − + − −= − += −

56. The percentage of Jewish Americans in the U.S. population was at a minimum in 1900. Approximately 1.4% of the U.S. population was made up of Jewish Americans.

58. The percentage of Jewish Americans in the U.S. population increased from 1900 to 1940 and then decreased from 1940 to 2000. 81. The relation is a function. Every element in the

domain corresponds to exactly one element in the range. 60.

The cost of mailing a first-class letter weighing 3.5 ounces is $0.92.

(3.5) 0.92f =

51


82.

32

12 2(3 1) 4 512 6 2 4 5

10 6 4 56 4 5 1

10 1510 1510 10

x xx x

x xx x

xx

x

− + = −− − = −

− = −− − = − −

− = −− −

=− −

=

0

The solution set is { }32 .

83. Let x = the width of the rectangle. Let 3 length of the rectangle. 8x + =

2 2624 2(3 8) 2624 6 16 2624 8 168 608

763 8 236

P l wx x

x xx

xx

x

= += + += + += +

− = −=

+ =

The dimensions of the rectangle are 76 yards by 236 yards.

84. 3 must be excluded from the domain of f because it would cause the denominator, to be equal to zero. Division by 0 is undefined.

3,x −

85.

(4) (4)2(4) (4) (4 4) (4 5)

20 ( 1)19

f g

f g+ = + + −= + −=

86. ( )2 2

2 2

2

7.4 15 4046 3.5 20 2405

7.4 15 4046 3.5 20 2405

10.9 35 1641

x x x x

x x x x

x x

− + − − + +

= − + + − −

= − +

2.3 Check Points

1. a. Domain of { }= is a real numberf x x .

b. Domain of {= is a real number and 5g x x x ≠ −

2. a. ( ) ( ) ( )2

2

( ) 3 4 1 2 7

3 6 6

f g x x x x

x x

+ = + − + +

= + +

b. ( ) 2(4) 3(4) 6(4) 678

f g+ = + +

=

3. a. ( ) 5 7( )8

f g xx x

− = −−

b. Domain of { }= is a real number and 0 and 8f g x x x x− ≠ ≠

4. a. ( ) ( ) ( ) ( )

2

5 5 5

[5 2 5] [5 3] 23

f g f g+ = +

= − ⋅ + + =

b. ( ) ( ) ( ) ( )2 2[ 2 ] [ 3] 3

f g x f x g x

x x x x x

− = −

3= − − + = − −

( )( ) 21 ( 1) 3( 1) 3f g 1− − = − − − − =

c. ( ) ( )( )

2 23

f xf xxg g x x

⎛ ⎞ −= =⎜ ⎟ +⎝ ⎠

x

( )2(7) 2(7) 35 77

(7) 3 10 2fg

⎛ ⎞ −= = =⎜ ⎟ +⎝ ⎠

d. ( ) ( ) ( ) ( )

( ) ( )2

4 4 4

( 4) 2( 4) ( 4) 3 24

fg f g− = − ⋅ −

= − − − − + = −

5. a. ( ) ( )( ) ( )

2 2

2 2

2

(7.4 15 4046) ( 3.5 20 2405)

7.4 15 4046 3.5 20 2405

3.9 5 6451

B D x

B x D x

x x x x

x x x x

x x

+

= +

= − + + − + +

= − + − + +

= + +

} .

b. ( )

The number of births and deaths in the U.S. in 2005 is 6573.5 thousand.

( )( ) ( )

2

2

3.9 5 6451

5 3.9(5) 5(5) 64516573.5

B D x x x

B D

+ = + +

+ = + +

=

c. ( )( )B D x+ underestimates the actual number of births and deaths in 2005 by 1.5 thousand.

52


2.3 Exercise Set

2. Domain of { }= is a real numberf x x .

4. Domain of { }= is a real number and 5g x x x ≠ − .

6. Domain of { }= is a real number and 2f x x x ≠ .

8. Domain of { }= is a real number and 6g x x x ≠ .

10. Domain of { }= is a real number and 8 and 10f x x x x≠ − ≠ .

12. ( ) ( ) (( ) 4 2 2 94 2 2 96 7

f g x x xx xx

+ = + + −= + + −= −

)

( ) ( ) ( )5 6 5 7 30 7 23f g+ = − = − =

14. ( ) ( ) ( )2

2

2

( ) 6 2

6 22 6

f g x x x

x xx x

+ = − +

= − +

= + −

( )( ) ( )( )

25 2 5 5 62 25 5 650 5 6 49

f g+ = + −= + −= + − =

16. ( ) ( ) ( )2

2

2

( ) 4 3 1

4 3 1

4 2

f g x x x x

x x x

x

+ = − − + +

= − − + +

= −

( ) ( ) ( ) ( )25 4 5 2 4 25 2100 2 98

f g+ = − = −

= − =

18. Domain of { }= is a real number .f g x x+

20. Domain of { } = is a real number and 6 .f g x x x+ ≠

22. Domain of { } = is a real number and 0 and 6 .f g x x x x+ ≠ ≠

24. Domain of { } = is a real number and 8 and 4 .f g x x x x+ ≠

26. Domain of { }= is a real number and 4 .f g x x x+ ≠

28. Domain of { }= is a real number .f g x x+

30. 2

2

( )( ) ( ) ( )

4 2

3 2

f g x f x g x

x x x

x x

+ = +

= + + −

= + +

( ) ( ) ( )2( ) 4 4 3 4 2 3f g+ = + + = 0

3 32. ( ) ( ) ( ) ( )( ) ( )( )23 3 3 4 3 2

3 5 2

f g− + − = − + − + − −

= − + =

34.

( ) ( )2

2

2

( )( ) ( ) ( )

4 2

4 2

5 2

f g x f x g x

x x x

x x x

x x

− = −

= + − −

= + − +

= + −

( ) ( ) ( )2( ) 6 6 5 636 30 2 64

f g 2− = + −

= + − =

36. ( ) ( ) ( ) ( )( ) ( )( )23 3 3 4 3 2

3 5 8

f g 3− − − = − + − − − −

= − − = −

38. ( ) ( ) ( ) ( )

( ) ( )( ) ( )( )( )

2

3 3 3

3 4 3 2 3

3 5 15

fg f g− = − ⋅ −

= − + − ⋅ − −

= − = −

40. ( ) ( ) ( ) ( )

( ) ( )( ) ( )( )( )

2

6 6 6

6 4 6 2 6

60 4 240

fg f g= ⋅

= + ⋅ −

= − = −

42. ( ) ( )( )

2 42

f xf x xxg g x x

⎛ ⎞ += =⎜ ⎟ −⎝ ⎠

( ) ( ) ( )23 4 3 9 12 213 22 3 1 1

fg

+⎛ ⎞ + 1= = = = −⎜ ⎟ − − −⎝ ⎠

− ≠

44. ( ) ( )( )

2 42

f xf x xxg g x x

⎛ ⎞ += =⎜ ⎟ −⎝ ⎠

( ) ( ) ( )20 4 0 0 0 00 02 0 2 2

fg

+⎛ ⎞ += = = =⎜ ⎟ −⎝ ⎠

53


46. Domain of { is a real numberf g x x− = } . 64. First, find ( )( )R C x− . ( )( ) 65 (600,000 45 )

65 600,000 4520 600,000

R C x x xx xx

− = − += − −= −

( )( )(20,000)

20 20,000 600,000400,000 600,000 200,000

R C−= −

= − = −

This means that if the company produces and sells 20,000 radios, it will lose $200,000.

48. ( )( ) ( ) ( ) ( )( )2 4 2fg x f x g x x x x= ⋅ = + −

Domain of { } is a real numberfg x x= .

50. ( )( ) ( ) ( )2 2 2 2 3g f g f− − = − − − = − = −1

52. ( ) ( )( )3 03 03 3

ggf f

⎛ ⎞= = =⎜ ⎟ −⎝ ⎠

( )( )(30,000)

20 30,000 600,000600,000 600,000 0

R C−= −

= − =

If the company produces and sells 30,000 radios, it will break even with its costs equal to its revenue.

54. The domain of fg

is { | . 4 3}x x− < <

56. The graph of f g−

( )( )(40,000)

20 40,000 600,000800,000 600,000 200,000

R C−= −

= − =

This means that if the company produces and sells 40,000 radios, it will make a profit of $200,000.

66. – 68. Answers will vary. 58. ( ) ( ) ( )( )( ) ( ) ( ) ( )( ) ( )

( )

1 0

1 1 0 0

3 2 4 2

3 2 4 2 3 2 6

f g g f

f g g f

+ − − −

= − + − − −⎡ ⎤⎣ ⎦= + − − − −⎡ ⎤⎣ ⎦= + − − + = + − − = −

70. 1 2

3 1 2

4 2y x yy y y

x= − == −

5

60. ( )( ) ( )

( ) ( ) ( )( )

( ) ( )

2

2

22

2 0

02 2

0

40 1 0 22

0 4 4

gfgf

gf g

f

⎡ ⎤⎛ ⎞− ⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

⎡ ⎤= − ⎢ ⎥

⎢ ⎥⎣ ⎦

⎛ ⎞= − = − −⎜ ⎟−⎝ ⎠= − = −

72. 21 2

13

2

2 y x x y xy

yy

= − =

=

62. a. ( )( ) ( ) ( )

(1.48 120.6) (1.58 114.4)0.1 6.2

F M x F x M xx x

x

− = −= + − += − +

74. makes sense

76. makes sense

78. true b.

In 2005 there were 4.2 million more women than men.

( )( ) 0.1 6.2( )(20) 0.1(20) 6.2 4.2

F M x xF M

− = − +− = − + = 80. true

c. The result in part (b) overestimates the actual difference by 0.2 million.

54

Intermediate Algebra for College Students 6E Mid-Chaspter Check Point

82. ( )33 3

3 33 or

3 3

R a bR a b

R a bR a Rb b

= +

= +− =

−= =

87. 5 3 123 5 13 5 13 3

5 43

x yy xy x

y x

223

+ = −= − −−

= −

= − −

a−

)

83.

18

( ) (3 6 3 2 418 3 3 2 818 3 11 2

x xx xx x

− = − −

− = − +− = −

117

xx

= +=

The solution set is { } 7 .

Mid-Chapter Check Point – Chapter 2

1. The relation is not a function. The domain is {1 The range is { 6

,2}., 4,6}.−

84. ( ) ( )2 6 2 46 12 4 6 8

f b bb b

+ = + −

= + − = + 2. The relation is a function.

The domain is {0 The range is {1

, 2,3}., 4}.

85. a. 4 3 64 3(0) 6

4 632

x yx

x

x

− =− =

=

=

3. The relation is a function.

The domain is { | The range is { |

2 2}.x x− ≤ <0 3}.y y≤ ≤

4. The relation is not a function. The domain is { | The range is { |

3 4}.x x− < ≤1 2}.y y− ≤ ≤

b. 4 3 64(0) 3 6

3 62

x yyyy

− =− =− =

= − 5. The relation is not a function.

The domain is { 2, 1,0,1, 2}.− − The range is { 2, 1,1,3}.− −

86. a. ( ),2 4y x= + x y x 6. The relation is a function.

The domain is { | The range is { |

1}.x x ≤1}.y y ≥ −

( )3− 2( 3) 4 2− + = − 3, 2− −

( )2− 2( 2) 4 0− + = 2,0−

( )1, 2− 1− 2( 1) 4 2− + = 7. The graph of f represents the graph of a function

because every element in the domain is corresponds to exactly one element in the range. It passes the vertical line test.

( ) 2(0) 4 4+ = 0,40

( )2(1) 4 6+ = 1,61

8. ( )4 3f − =

9. The function ( ) 4f x = when 2.x = −

10. The function ( ) 0f x = when and 2x = 6.x = −

11. The domain of f is { | is a real number}.x x b. The graph crosses the x-axis at the point ( 2,0).−

12. The range of f is { | 4}.y y ≤ c. The graph crosses the y-axis at the point (0 , 4).

13. The domain is { | is a real number}.x x

55


14. The domain of g is { | is a real number and 2 and 2}.x x x x≠ − ≠

15. ( ) ( )( ) ( )( ) ( )

20 0 3 0 8 8

10 2 10 5 20 5 15

0 10 8 15 23

f

g

f g

= − + =

− = − − − = − =

+ − = + =

16. ( ) ( ) ( )( ) ( )( ) ( ) ( )

21 1 3 1 8 1 3 8 1

3 2 3 5 6 5 11

1 3 12 11 12 11 23

f

g

f g

− = − − − + = + + =

= − − = − − = −

− − = − − = + =

2

11

5−

3

)

3

17. ( )( ) ( )

( ) ( )

2

2

2

3 8

3 2 3 5 2 6 5 2 11

3 3 8 2

5 3

f a a a

g a aa a

f a g a a a a

a a

= − +

+ = − + −

= − − − = − −

+ + = − + + − −

= − −

18. ( )

( ) 2

2

3 8 2

5 3

f g x x x x

x x

+ = − + + −

= − +

( )( ) ( ) ( )22 2 5 24 10 3 17

f g+ − = − − − +

= + + =

19. ( )

( ) (2

2

2

3 8 2 5

3 8 2 5

13

f g x x x x

x x x

x x

− = − + − − −

= − + + +

= − +

( )( ) ( )25 5 5 1325 5 13 33

f g− = − +

= − + =

20. ( ) ( ) ( )

( ) ( )( )( ) ( )

21 1 3 1 81 3 8 12

1 2 1 5 2 5

1 12 3 36

f

g

fg

− = − − − +

= + + =

− = − − − = − = −

− = − = −

21. ( )

( ) ( ) ( )( )

2

2

3 82 5

4 3 4 84

2 4 516 12 8 36 12

8 5 3

f x xxg x

fg

⎛ ⎞ − +=⎜ ⎟ − −⎝ ⎠

− − − +⎛ ⎞− =⎜ ⎟ − − −⎝ ⎠

+ += =

−=

22. The domain of fg

is 5| .2

x x⎧ ⎫≠ −⎨ ⎬⎩ ⎭

2.4 Check Points

1. 3 2 6x y− = Find the x–intercept by setting y = 0.

3 2 63 2(0) 6

3 62

x yx

xx

− =− =

==

Find the y–intercept by setting x = 0. 3 2 6

3(0) 2 62 6

3

x yyyy

− =− =− =

= −

2. a. 2 1

2 1

2 4 6 64 ( 3) 1

y ym

x x− − − −

= = =− − − − −

=

b. 2 1

2 1

5 ( 2) 7 71 4 5 5

y ym

x x− − −

= = = =− − − −

−

3. 8 4 204 8 24 8 24 4

2 5

x yy xy x

y x

00

− =− = − +− − +

=− −

= −

The slope is 2 and the y-intercept is –5.

4. Begin by plotting the y-intercept of –3. Then use the slope of 4 to plot more points.

56


5. Begin by plotting the y-intercept of 0. Then use the

slope of 23− to plot more points.

6. is a horizontal line. 3y =

7. is a vertical line. 3x = −

8. 2 1

2 1

4.4 2.7 1.7 0.572005 2002 3

y ym

x x− −

= = = ≈− −

For the 50-59 age group, the percentage reporting using illegal drugs in the previous month increased by about 0.57% each year.

9. 2 1

2 1

0.05 0.03 0.02 0.013 1 2

y ym

x x− −

= = = ≈− −

The average rate of change between 1 hour and 3 hours is 0.01. This means that the drug’s concentration is increasing at an average rate of 0.01 milligram per 100 milliliters per hour.

10. a. We will use the line segment passing through and to obtain a model. We

need values for m, the slope, and b, the y-intercept.

(13,25.3) (0, 23.9)

2 1

2 1

25.3 23.9 1.4 0.1113 0 13

y ym

x x− −

= = = ≈− −

The point gives us the y-intercept of 23.9. Thus,

(0, 23.9)

( )( ) 0.11 23.9

A x mx bA x x

= += +

b. ( ) 0.11 23.9(40) 0.11(40) 23.9

28.3

A x xA

= += +=

The model predicts the median age of first marriage for U.S. women will be 28.3 years in 2030.

2.4 Exercise Set

2. 2x y+ = Find the x–intercept by setting y = 0.

0 22

xx

+ ==

Find the y–intercept by setting x = 0. 0 2

2yy

+ ==

57


4. 2 4 Find the x–intercept by setting y = 0.

Find the y–intercept by setting x = 0.

x y+ =

2 02 4

2

xxx

+ ===

4

( )2 0 44

yy

+ =

=



18x y− =

( )6 9 0 16 1

3

xxx

− =

==

8. 4 8x y− = Find the x–intercept by setting y = 0.

( )4 0 80 8

8

xx

x

+ =

+ ==

Find the y–intercept by setting x = 0. 0 4 8

4 82

yyy

− =− =

= −

10. 2 5x y− = Find the x–intercept by setting y = 0. 2 0 5

2 552

xx

x

− ==

=

Find the y–intercept by setting x = 0. ( )2 0 5

55

yyy

− =

− == −

88

( )6 0 9 180 9 18

9 182

yyyy

− =

− =− =

= −

58




1x y= − 5

( )3 5 0 153 0 153 15

5

xxxx

= −

= −= −= −

( )3 0 5 150 5 15

15 53

yyy

y

= −

= −==


12x y− =

( )8 2 0 128 12

12832

xx

x

x

− =

=

=

=


( )8 0 2 122 12

6

yyy

− =

− == −

16. 4 1 35 3 2

m −= =

−

The line rises.

18. ( )( )

5 2 5 2 72 3 2 3

m− − +

= =− − +

20. ( )( )

3 3 3 3 0 04 6 4 6 10

m− − − − +

= = = =− − +

The line is horizontal.

22. ( )3 1 3 1 4 16 2 8 8 2

m− − +

= = = = −− − − −

The line falls.

24. ( )1 16 6

6 6 undefined3 3 02 2

m− − +

= = =−

undefined slope; The line is vertical.

26. passes through 1L ( ) ( )1,0 and 0, 1 .−

1 0 1 10 1 1

m − − −= = =

− −

passes through 2L ( ) (4, 2 and 1, 4 .− − )( )4 2 4 2 2 2

1 4 3 3 3m

− − − − + −= = = =

− − −

passes through 3L ( ) (4, 2 and 3,0 .− − )

( )0 2 2 2 2

3 4 3 4 1m − − −= = = =− − − − +

−

28. 3 2y x= + 3 intercept 2m y= − =

30. ( ) 3 2f x x= − + 3 intercept 2m y= − − =

5=

The line rises.

59


32. ( ) 3 34

f x x= −

3 intercept 34

m y= − = −

34. ( ) 2 65

f x x= − +

2 intercept 65

m y= − − =

36. 13

y x= −

1 intercept 03

m y= − − =

38. 13

y = −

10 intercept3

m y= − = −

40. a. 3 03

x yy x

+ == −

b. 3 intercept 0m y= − − =

c.

42. a. 4 3

34

y x

y x

=

=

b. 3 intercept 04

m y= − =

c.

60


44. a. 2 42 4

x yy x

+ == − +

b. 2 intercept 4m y= − − =

c.

46. a. 7 2 14

2 7 17 72

x yy x

y x

+ == − +

= − +

4

b. 7 intercept 72

m y= − − =

c.

48. 5y =

50. which is the same as ( ) 4f x = − 4y = −

52. 5 306

yy= −= −

54. ( ) 1 or 1f x y= =

56. 4x =

58. 4 123

xx= −= −

60. 0y = This is the equation of the x–axis.

62. ( )

00

b bma a

− − − ba

= = = −− −

The line falls.

61


64. ( )

( )a c c am

a a b b+ −

= =− −

The line rises.

66. Ax By CAx C By

A Cx yB B

= −+ =

+ =

The slope is AB

and the intercept is y − .CB

68. ( )

1 43 4 2

y− −=

− −

( )

1 43 4 21 43 66 3 46 12 318 3

6

y

y

yy

yy

− −=

+− −

=

= − −

= − −= −

− =

70. ( )( )

( )

6 5 20

5 66 45

x f x

f x x

f x x

− =

− = − +

= −

20

72. ( )36 22

6 33

b

bb

− = − +

− = − +− =

74. Decreasing order: 2 1 4 3, , ,b b b b

76. The slope is 2. This means the amount spent by the drug industry to market drugs to doctors is increasing by $2 billion each year.

78. The slope is –0.28. This means the percentage of U.S. taxpayers audited by the IRS is decreasing by 0.28% each year.

80. a. 15% of marriages in which the wife is over age 25 when she marries end in divorce within the first five years.

b. 25% of marriages in which the wife is over age 25 when she marries end in divorce within the first ten years.

c. 2 1

2 1

25 15 10 210 5 5

y ym

x x− −

= = =− −

=

There is an average increase of 2% of marriages ending in divorce per year.

82. a. 2 1

2 1

1489 449 1040 2604 0 4

y ym

x x− −

= = = =− −

The number of U.S. college students enrolled exclusively in online education is increasing by 260 thousand per year.

b. ( ) 260 449L x x= +

c. (8) 260(8) 449 2529L = + = The function predicts that the number of U.S. college students enrolled exclusively in online education in 2010 will be 2529 thousand.

84. ( ) 1.3 23P x x= +


104. 3 6y x= − +

Two points found using [TRACE] are (0, 6) and (2, 0). Based on these points, the slope is:

6 0 6 3.0 2 2

m −= = = −

− −

This is the same as the coefficient of x in the line’s equation.

62


106. 3 24

y x= −

Two points found using [TRACE] are ( )0, 2− and

. Based on these points, the slope is: (1, 1.25− )2 ( 1.25) 0.75 0.75.

0 1 1m − − − −= = =

− −

This is equivalent to the coefficient of x in the line’s equation.

108. does not make sense; Explanations will vary. Sample explanation: Since this value is increasing, it will have a positive slope.

110. makes sense

112. false; Changes to make the statement true will vary. A sample change is: Slope-intercept form is

. Vertical lines have equations of the form y mx b= +

x a= . Equations of this form have undefined slope and cannot be written in slope-intercept form.

114. false; Changes to make the statement true will vary. A sample change is: The graph of is a vertical line that passes through the point ( )

7x =7,0 .

116. We are given that the and the

slope is

intercept is 6y − −

1 .2

So the equation of the line is 1 6.2

y x= −

We can put this equation in the form ax by c+ = to find the missing coefficients.

( )

1 621 62

12 22

2 122 12

y x

y x

y x

y xx y

= −

− = −

⎛ ⎞− = −⎜ ⎟⎝ ⎠− = −

− =

118. ( ) ( ) (22 2 22 3 2 2 3

3

4 6

4 4 4

16

x )2x y xy

x y

−

⎛ ⎞= =⎜ ⎟⎜ ⎟

⎝ ⎠

=

y

119. ( )( )( )

7 3 4

1 4

3

8 10 4 10 32 10

3.2 10 10

3.2 10

− −

−

−

× × = ×

= × ×

= ×

120. ( ) [ ]5 3 4 6 5 3 12 6

5 3 12 63 17

x x x x

x xx

− − − = − − −⎡ ⎤⎣ ⎦= − + += +

121. 5 7( 4)5 7 28

7 33

y xy x

y x

− = +− = +

= +

122. 73 ( 1)37 733 37 73 3 33 37 23 3

y x

y x

y x

y x

+ = − −

+ = − +

+ − = − + −

= − −

123. a. 4 8 04 84 84 4

1 24

x yy xy x

y x

+ − == − +− +

=

= − +

The slope is 1 .4

−

b. 2

2

2

1 141 14

4

m

m

m

− ⋅ = −

⋅ =

=

The slope of the second line is 4. 6

Therefore, the coefficient of x is 1 and the coefficient of is y 2.−

63


2.5 Check Points

1. S

Slope-Intercept Form

lope 2, passing through (4, 3)Point-Slope Form

= − −

( )( )( )

1 1

( 3) 2 4

3 2 4

y y m x x

y x

y x

− = −

− − = − −

+ = − −

( )

( )

3 2 43 2 8

2 52 5

y xy x

y xf x x

+ = − −

+ = − += − +

= − +

2. a. Passing through First, find the slope.

(6, 3) and (2,5)−

5 ( 3) 8 22 6 4

m − −= = =

− −−

Then use the slope and one of the points to write the equation in point-slope form.

or

( )( )

1 1

5 2 2

y y m x x

y x

− = −

− = − −

( )( )( )

1 1

( 3) 2 6

3 2 6

y y m x x

y x

y x

− = −

− − = − −

+ = − −

b. Slope-Intercept Form

( )

( )

5 2 25 2 4

2 92 9

y xy x

y xf x x

− = − −

− = − += − +

= − +

3. First, find the slope. 79.7 74.7 5 0.17

40 10 30m −= = ≈

−

Then use the slope and one of the points to write the equation in point-slope form. Using the point (10, 74.7):

( )( )

1 1

74.7 0.17 100.17 73

( ) 0.17 73

y y m x x

y xy x

f x x

− = −

− = −

= += +

Thus, (60) 0.17(60) 73 83.2.f = + = This means that the life expectancy of American women in 2020 is predicted to be 83.2 years. Answers vary due to rounding and choice of point. If point (40, 79.7) is chosen, and the life expectancy of American women in 2020 is predicted to be 83.1 years.

( ) 0.17 72.9f x x= +

4. Since the line is parallel to we know it will have slope

3 1y x= + ,3.m = We are given that it passes

through ( 2,5).− We use the slope and point to write the equation in point-slope form.

( )( )( )

( )

1 1

5 3 2

5 3 2

y y m x x

y x

y x

− = −

− = − −

− = +

Solve for y to obtain slope-intercept form. ( )5 3 2

5 3 63 11

( ) 3 11

y xy x

y xf x x

− = +

− = += += +

5. a. Solve the given equation for y to obtain slope-intercept form.

3 123 1

1 43

x yy x

y x

2+ =

= − +

= − +

Since the slope of the given line is 1 ,3

− the

slope of any line perpendicular to the given line is 3.

b. We use the slope of 3 and the point ( 2, 6)− − to write the equation in point-slope form.

( )( ) ( )( )

( )

1 1

6 3 2

6 3 2

y y m x x

y x

y x

− = −

− − = − −

+ = +


6 3 63

( ) 3

y xy x

y xf x x

+ = +

+ = +==

64


2.5 Exercise Set

2. S


lope 4, passing through (3,1)Point-Slope Form

=

( )1 4 3y x− = −

( )1 4 31 4 12

y xy x− = −

− = −

( )4 114 11

y xf x x

= −

= −

4. S


lope 8, passing through ( 4,1)Point-Slope Form

= −

( )( )( )

1 8 4

1 8 4

y x

y x

− = − −

− = +

( )

1 8 328 338 33

y xy x

f x x

− = += +

= +

6. S


lope 6, passing through ( 2, 4)Point-Slope Form

= − − −

( ) ( )( )( )

4 6 2

4 6 2

y x

y x

− − = − − −

+ = − +

( )

4 6 16 166 16

y xy x

f x x

+ = − −= − −

= − −

2

8. S


lope 4, passing through (0, 3)Point-Slope Form

= − −

( ) ( )( )

3 4 0

3 4 0

y x

y x

− − = − −

+ = − −

( )

3 44 34 3

y xy x

f x x

+ = −= − −

= − −

10. ( )14Slope 1, passing through , 4

Point-Slope Form

= − − −

( ) ( )( )( )

14

14

4 1

4 1

y x

y x

− − = − − −

+ = − +


( )

1414174

174

4

4

y x

y x

y x

f x x

+ = − −

= − − −

= − −

= − −

12. ( )15Slope , passing through 0,0

Point-Slope Form

=

( )150 0y x− = −


( )

1515

y x

f x x

=

=

14. 25Slope , passing through (15, 4)

Point-Slope Form

= − −

( ) ( )( )

2525

4 1

4 1

y x

y x

− − = − −

+ = − −

5

5


( )

252525

4 6

2

2

y x

y x

f x x

+ = − +

= − +

= − +

65


16. P First, find the slope.

assing through (1,3) and (2, 4)

4 3 1 12 1 1

m −= = =

−


or Slope-Intercept Form

(4 1 2y x− = − ) ( )3 1 1y x− = −

( )

4 1 222

y xy x

f x x

− = −= +

= +


assing through (2,0) and (0, 1)−

1 0 1 10 2 2 2

m − − −= = =

− −


( ) ( ) ( )

( )

1 10 2 or 12 2

11 02

y x y x

y x

− = − − − = −

+ = −

0


( )

1 121 12

y x

f x x

= −

= −


assing through ( 3,2) and (2, 8)− −

( )8 2 10 10 2

2 3 2 3 5m − − − −= = = =

− − +

22. Passing through (4, 8) and (8, 3)− − First, find the slope.

( )3 8 3 8 58 4 4 4

m− − − − +

= = =−


( ) ( )

( )

( ) ( )

( )

53 8453 84

or58 4458 44

y x

y x

y x

y x

− − = −

+ = −

− − = −

+ = −


( )

53 145 1345 134

y x

y x

f x x

0+ = −

= −

= −

24. Passing through ( 1, 4) and (3, 4)− − − First, find the slope.

( )( )

4 4 0 03 1 4

m− − −

= = =− −


( ) ( )( )( )

( ) ( )( )

4 0 14 0 1

or4 0 34 0 3

y xy x

y xy x

− − = − −+ = +

− − = −+ = −


( )

4 044

yy

f x

+ == −= −

−

)



( ) ( )( )

( )( )

8 2 2

8 2 2or

2 2 ( 3)

2 2 3

y x

y x

y x

y x

− − = − −

+ = − −

− = − − −

− = − +

( ) (

( )

8 2 28 2 4

2 42 4

y xy x

y xf x x

− − = − −

+ = − += − −

= − −

66


26. P If the line has a it passes through

First, find the slope.

assing through ( 4,5) and with -intercept 3y− = −-intercept 3,y = −

( )0, 3 .−

( )3 5 8 8 2

0 4 0 4 4m − − − −= = = =

− − +−



( ) ( )( )

( )( )

3 2 0

3 2 0or

5 2 ( 4)

5 2 4

y x

y x

y x

y x

− − = − −

+ = − −

− = − − −

− = − +

( )

3 22 32 3

y xy x

f x x

+ = −= − −

= − −

28. and If the line has an it passes through the point If the line has a

-intercept 2x = − -intercept 4y =-intercept 2,x = −

( 2,0 .− ) -intercept 4,y =

it passes through First, find the slope.

( )0,4 .

( )4 0 4 2

0 2 2m −= =

− −=



( )

( )( )

4 2 0or

0 2 ( 2)

0 2 2

y x

y x

y x

− = −

− = − −

− = +

( )

4 22 42 4

y xy x

f x x

− == +

= +

30. a. Parallel: 3m =

b. Perpendicular: 13

m = −

32. a. Parallel: 9m = −


m =

34. a. Parallel: 14

m =

b. Perpendicular: 4m = −

36. a. Parallel: 37

m = −


m =

38. 8 118 11

x yy x

+ == − +

a. Parallel: 8m = −


m =

40. 3 2 62 3 6

3 32

x yy x

y x

+ == − +

= − +

a. Parallel: 32

m = −


m =

42. 3 4 74 3

3 74 4

x yy x

y x

7− = −− = − −

= +

a. Parallel: 34

m =


m = −

44. 9y = is a horizontal line with slope 0.m =

a. Parallel: 0m =

b. Perpendicular: is undefinedm

67


46. L will have slope . Using the point and the slope, we have

Solve for y to

obtain slope-intercept form.

2m = −(4 2 3y x− = − − ).

( )

4 2 62 102 10

y xy x

f x x

− = − += − +

= − +

48. L will have slope 1 .2

m = The line passes through

(–1, 2). Use the slope and point to write the equation in point-slope form.

( )( )

( )

12 1212 12

y x

y x

− = − −

− = +

Solve for y to obtain slope-intercept form. 1 122 21 1 22 2

y x

y x

− = +

= + +

( )

1 52 21 52 2

y x

f x x

= +

= +

50. L will have slope The line passes through (–2, –7). Use the slope and point to write the equation in point-slope form.

Solve for y to obtain slope-intercept form.

5.m = −

( ) ( )( )( )

7 5 2

7 5 2

y x

y x

− − = − − −

+ = − +

( )

7 5 15 175 17

y xy x

f x x

+ = − −= − −

= − −

0

2

52. L will have slope The line passes through (–4, 2 ). We use the slope and point to write the equation in point-slope form.


3.m = −

( )( )( )

2 3 4

2 3 4

y x

y x

− = − − −

− = − +

( )

2 3 13 103 10

y xy x

f x x

− = − −= − −

= − −

54. Find the slope. 3 2 5

2 33 52 2

x yy x

y x

5− =− = − +

= −

Since the lines are parallel, it will have slope 3 .2

m = The line passes through (–1, 3 ). Use the

slope and point to write the equation in point-slope form.

( )( )

( )

33 1233 12

y x

y x

− = − −

− = +


( )

3 332 23 3 32 23 92 23 92 2

y x

y x

y x

f x x

− = +

= + +

= +

= +

56. Find the slope. 7 127 12

1 127 7

x yy x

y x

+ == − +

= − +

Since the lines are perpendicular, the slope is 7.m = The line passes through (5, –9). Use the

slope and point to write the equation in point-slope form.

( ) ( )( )

9 7 5

9 7 5

y x

y x

− − = −

+ = −

Solve for y to obtain slope-intercept form. ( )

( )

9 7 59 7 35

7 447 44

y xy x

y xf x x

+ = −

+ = −= −

= −

58. Since the line is perpendicular to which is a vertical line, we know the graph of

4x = −f is a horizontal

line with 0 slope. The graph of f passes through

( )2,6− , so the equation of f is ( ) 6.f x =

68


60. First we need to find the slope of the line with x − intercept of 3 and intercept of This line will pass through and We use these points to find the slope.

y − 9.−

(3,0) )(0, 9 .−

9 0 9 30 3 3

m − − −= =

− −=

Since the graph of f is perpendicular to this line, it

will have slope 1 .3

m = −

Use the point and the slope ( 5,6− ) 13

− to find the

equation of the line. ( )

( )( )

( )

( )

1 1

16 5316 531 563 31 133 31 133 3

y y m x x

y x

y x

y x

y x

f x x

− = −

− = − − −

− = − +

− = − −

= − +

= − +

62. First put the equation in slope-intercept form.

The equation of will have slope

4x y− = 6

4 64 6

4 6

x yy xy x

− =− = − +

= −

f 14

− since it is

perpendicular to the line above and the same intercept

So the equation of is

y − 6.−

f ( ) 1 6.4

f x x= − −

64. The graph of is just the graph of f g shifted up 3 units. So add 3 to the equation of ( )g x to obtain

the equation of ( ).f x

( ) ( ) 3 2 5 3 2 2f x g x x x= + = − + = −

66. From exercise 65, we know the slope of the line is

.AB

− So the slope of the line that is perpendicular

would be .BA

68. a. First, find the slope using and ( )20,51.7

( )10,45.2 .

51.7 45.2 6.5 0.6520 10 10

m −= = =

−


( )( )

( )

1 1

45.2 0.65 10or

51.7 0.65 20

y y m x x

y x

y x

− = −

− = −

− = −

b. ( )

( )

45.2 0.65 1045.2 0.65 6.5

0.65 38.70.65 38.7

y xy x

y xf x x

− = −

− = −= +

= +

c. ( )35 0.65(35) 38.7 61.45f = + = The linear function predicts the percentage of never married American males, ages 25 – 29, to be 61.45% in 2015.

70. a.

69


b.

Use the two points and

to find the slope. ( )10,317

(50,367)367 317 50 1.25

50 10 40m −= = =

−



( )( )

( )

1 1

317 1.25 10or

367 1.25 50

y y m x x

y x

y x

− = −

− = −

− = −

( )

( )

317 1.25 10317 1.25 12.5

1.25 304.51.25 304.5

y xy x

y xf x x

− = −

− = −= +

= +

c. ( )100 1.25(100) 304.5 429.5f = + = The function predicts that the average atmospheric concentration of carbon dioxide will be 429.5 parts per million in 2050.

72. a. 970 582 388 43.12016 2007 9

m −= =

−≈

The cost of Social Security is projected to increase at a rate of approximately $43.1 billion per year.

b. 392 195 197 21.92016 2007 9

m −= =

−≈

The cost of Medicaid is projected to increase at a rate of approximately $21.9 billion per year.

c. No, the slopes are not the same. This means that the cost of Social Security is projected to increase at a faster rate than the cost of Medicaid.


80. 1 133 2

y x

y x

= +

= − −

a.

The lines do not appear to be perpendicular.

b.

The lines appear to be perpendicular.

Explanations will vary. Sample explanation: The calculator screen is rectangular and does not have the same width and height. This causes the scale of the x–axis to differ from the scale on the y–axis despite using the same scale in the window settings. In part (a), this causes the lines not to appear perpendicular when indeed they are. The zoom square feature compensates for this and in part (b), the lines appear to be perpendicular.

70


82. a.

b.

c.

84. makes sense

86. does not make sense; Explanations will vary. Sample explanation: If we know the slope and the y-intercept, it may be easier to write the slope-intercept form of the equation.

88. true

90. true

92. The slope of the line containing ( and

94. First find the slope. 0

0b bm

a a− b

a= = = −

− −

Use the slope and point to write the equation in point-slope form.

( )0by b xa

− = − −

Solve this equation for to obtain slope-intercept form.

y

by b xaby xa

− = −

b= − +

Divide both sides by .b

1

1

y xb a

x ya b

= − +

+ =

This is called intercept form because the variable x is being divided by the x − intercept, and the variable is being divided by the

,ay y − intercept, . b

95. ( ) ( ) ( )( )

22 3 2 8 2

3 4 16 512 16 5 33

f 5− = − − − +

= + +

= + + =

96. ( ) ( ) ( )21 1 3 1 4 1 3 4f 8− = − − − + = + + =

( ) ( )1 2 1 5 2 5 7g − = − − = − − = −

( )( ) ( ) ( ) ( )( )1 1 1 8( 7)fg f g

)1, 3− ( )2, 4− has slope

( )4 3 4 3 7 72 1 3 3 3

m− − +

= = = =− − − −

56− = − ⋅ − = − = − − .

Solve for to obtain slope-intercept form.

So the slope of this line is

2Ax y+ = y

22

Ax yy Ax

+ == − +

.A− This line is perpendicular to the line above so its

slope is 3 .7

Therefore, 37

A− = so 3 .7

A = −

97. Let x = the measure of the smallest angle. x + 20 = the measure of the second angle. 2x = the measure of the third angle.

( )20 2 18020 2 180

x x xx x x+ + + =

+ + + =

4 20 180

4 16040

xxx

+ ===

Find the other angles. x + 20 = 40 + 20 = 60 2x = 2(40) = 80 The angles are , , and 80 . 40° 60° °

71


98. a.

The point satisfies the equation.

2 42( 5) ( 6) 4

10 6 44 4, tru

x y− = −− − − = −− + = −

− = − e

b.

The point satisfies the equation.

3 5 153( 5) 5( 6) 15

15 30 1515 15, true

x y− =− − − =− + =

=

99. The graphs intersect at (3, 4).−

100.

The solution set is {1}.

7 2( 2 4) 37 4 8 3

11 8 311 11

1

x xx x

xxx

− − + =+ − =

− ===

Chapter 2 Review

1. The relation is a function. Domain {3, 4, 5} Range {10}

2. The relation is a function. Domain {1, 2, 3, 4} Range {–6, π , 12, 100}

3. The relation is not a function. Domain {13, 15} Range {14, 16, 17}

4. a. ( )(0) 7 0 5 0 5 5f = − = − = −

b. ( )(3) 7 3 5 21 5 16f = − = − =

c. ( )( 10) 7 10 5 75f − = − − = −

d. ( )(2 ) 7 2 5 14 5f a a a= − = −

e. ( )( 2) 7 2 57 14 5 7 9

f a aa a

+ = + −

= + − = +

5. a. ( ) ( )2(0) 3 0 5 0 2 2g = − + =

b. ( ) ( )( )

2(5) 3 5 5 5 2

3 25 25 275 25 2 52

g = − +

= − +

= − + =

c. ( ) ( ) ( )24 3 4 5 4 2 70g − = − − − + =

d. ( ) ( ) ( )2

2

3 5

3 5 2

g b b b

b b

2= − +

= − +

e. ( ) ( ) ( )

( )2

2

2

4 3 4 5 4

3 16 20 2

48 20 2

g a a a

a a

a a

2= − +

= − +

= − +

72

Intermediate Algebra for College Students 6E Chapter 2 Review

6. g shifts the graph of f down one unit

7. g shifts the graph of f up two units.

8. The vertical line test shows that this is not the graph of a function.

9. The vertical line test shows that this is the graph of a function.





14. ( 2) 3f − = −

15. (0) 2f = −

16. When 3, ( ) 5.x f x= = −

17. The domain of f is { | 3 5}.x x− ≤ <

18. The range of f is { | 5 0}.y y− ≤ ≤

19. a. The eagle’s height is a function of its time in flight because every time, t, is associated with at most one height.

b. ( )15 0f = At time t = 15 seconds, the eagle is at height zero. This means that after 15 seconds, the eagle is on the ground.

c. The eagle’s maximum height is 45 meters.

d. For x = 7 and 22, This means that at times 7 seconds and 22 seconds, the eagle is at a height of 20 meters.

( ) 20.f x =

e. The eagle began the flight at 45 meters and remained there for approximately 3 seconds. At that time, the eagle descended for 9 seconds. It landed on the ground and stayed there for 5 seconds. The eagle then began to climb back up to a height of 44 meters.

20. The domain of f is { is a real number}.x x

21. The domain of {is is a real number and 8}.f x x x ≠ −

22. The domain of { }is is a real number and 5 .f x x x ≠

23. a. ( ) ( ) ( )( ) 4 5 2 14 5 2 16 4

f g x x xx xx

+ = − + +

= − + += −

b. ( ) ( )(3) 6 3 418 4 14

f g+ = −

= − =

24. a. ( )

( ) ( )2

2 2

( )

5 4 3

5 4 3 5

f g x

x x x

1x x x x

+

= − + + −

= − + + − = +

b. ( ) ( ) ( )2(3) 5 3 1 5 9 145 1 46

f g+ = + =

= + =

+

25. The domain of f g+ is

{ }is a real number and 4x x x ≠

26. The domain of f g+ is { | is a real number and 6 and 1}.x x x x≠ − ≠ −

73


27. 2( ) 2 , ( ) 5f x x x g x x= − = −

( ) ( ) ( )2

2

2

( ) 2 5

2 5

5

f g x x x x

x x x

x x

+ = − + −

= − + −

= − −

( ) ( ) ( )2 ( 2) 2 2 54 2 5 1

f g+ − = − − − −

= + − =

28. From Exercise 27 we know We can use this to find

( ) 2( ) 5.f g x x x+ = − −

( ) ( )3 3f g+ .

5

( ) ( ) ( )( )( ) ( )2

3 3 3

3 39 3 5 1

f g f g+ = +

= − −

= − − =

29. 2( ) 2 , ( ) 5f x x x g x x= − = −

( ) ( ) ( )2

2

2

( ) 2 5

2 5

3 5

f g x x x x

x x x

x x

− = − − −

= − − +

= − +

( )( ) ( ) ( )

2

2

( ) 3 5

(1) 1 3 1 51 3 5 3

f g x x x

f g

− = − +

− = − +

= − + =

30. From Exercise 29 we know We can use this to find

( ) 2( ) 3 5.f g x x x− = − +

( ) ( )4 4f g− .

5

( ) ( ) ( )( )( ) ( )2

4 4 4

4 3 416 12 5 9

f g f g− = −

= − +

= − + =

31. Since ( )( ) ( ) ( )3 3fg f g− = − ⋅ −

32. 2( ) 2 , ( ) 5f x x x g x x= − = − 2 2( )

5f x xxg x

⎛ ⎞ −=⎜ ⎟ −⎝ ⎠

( ) ( )24 2 4 16 8(4)4 5 1

8 81

fg

−⎛ ⎞ −= =⎜ ⎟ − −⎝ ⎠

= = −−

33. ( ) 2( ) 3 5f g x x x− = − +

The domain of { } is is a real number .f g x x−

34. 2 2( )

5f x xxg x

⎛ ⎞ −=⎜ ⎟ −⎝ ⎠

The domain of

is { is a real number and 5}.f x x xg

≠

35. 2 4x y+ = Find the x–intercept by setting y = 0 and the y–intercept by setting x = 0.

( )2 0 40 4

4

xx

x

+ =

+ ==

Choose another point to use as a check. Let x = 1.

0 2 42 4

2

yyy

+ ===

1 2 42 3

32

yy

y

+ ==

=

3 , find ( )3f − and

first.

( )3g −

( ) ( )2( 3) 3 2 39 6 15

f − = − − −

= + =( 3) 3 5 8g − = − − = −

( )( ) ( ) ( )( )

3 3

15 8 120

fg f g− = − ⋅ −

= − = −

3

74


36. 2 3 Find the x–intercept by setting y = 0 and the y–intercept by setting x = 0.


12x y− =

( )2 3 0 122 0 12

2 16

xx

xx

− =

+ ===

2

( )2 0 3 120 3 12

3 124

yyyy

− =

− =− =

= −

( )2 1 3 122 3 12

3 10103

yyy

y

− =

− =− =

= −

37. 4 8 2x y= − Find the x–intercept by setting y = 0 and the y–intercept by setting x = 0.


( )4 8 2 04 8 04 8

2

xxxx

= −

= −==

( )4 0 8 20 8 2

2 84

yy

yy

= −

= −==

( )4 1 8 24 8 24 22

yy

yy

= −

= −− = −

=

38. ( )2 4 6 25 2 3

m− −

= =−

39. ( )3 3 6 22 7 9 3

m− −

= = = −− − −

The line through the points falls.

40. ( )2 1 33 3 0

m− −

= =−

m is undefined. The line through the points is vertical.

41. ( )

4 4 0 03 1 2

m −= = =− − − −

The line through the points is horizontal.

42. 2 1y x= − 2 intercept 1m y= − = −

43. ( ) 1 42

f x x= − +

1 intercept 42

m y= − − =

44. 23

y x=

2 intercept 03

m y= − =

=

The line through the points rises.

75


45. To rewrite the equation in slope-intercept form, solve for y.

2 42 4

x yy x

+ == − +

2 intercept 4m y= − − =

46. 3 553

y x

y x

− =

= −

5 intercept 03

m y= − − =

47. 5 3 63 5 6

5 23

x yy x

y x

+ == − +

= − +

5 intercept 23

m y= − − =

48. 2y =

49. 7 2

50. ( ) 44

f xy= −

= −

51. 3x =

52. 2 105

xx= −= −

53. In ( ) 0.27 70.45,f t t= − + the slope is 0.27− . A slope of 0.27− indicates that the record time for the women’s 400-meter has been decreasing by 0.27 seconds per year since 1900.

13

yy= −= −

54. a. 1163 617 546 1371998 1994 4

m −= =

−≈

The was an average increase of approximately 137 discharges per year.

b. 668 1273 605 2022004 2001 3

m − −= = ≈

−−

The was an average decrease of approximately 202 discharges per year.

76


55. a. Find the slope of the line by using the two points (0, 32) and (100,212).

212 32 180 9100 0 100 5

m −= =

−=

We use the slope and one of the points to write the equation in point-slope form.

( )( )

1 1932 0593259 3259 325

y y m x x

y x

y x

y x

F C

− = −

− = −

− =

= +

= +

b. Let . 30C =

( )9 30 32 54 32 865

F = + = + =

The Fahrenheit temperature is 86 when the Celsius temperature is 30 .

°°

56. S


lope 6, passing through ( 3, 2)Point-Slope Form

= − −

( )( )( )

( )

1 1

2 6 3

2 6 3

y y m x x

y x

y x

− = −

− = − − −

− = − +

( )2 6 32 6 18

6 16( ) 6 16

y xy x

y xf x x

− = − +

− = − −= − −= − −


assing through (1,6) and ( 1, 2)−

( )6 2 4 2

1 1 2m −= = =

− −


( )( )

1 16 2 1

y y m x xy x− = −− = −

or ( )

( )( )( )

1 1

2 2 12 2 1

y y m x x

y xy x

− = −

− = − −− = +

Slope-Intercept Form ( )6 2 1

6 2 22 4

( ) 2 4

y xy x

y xf x x

− = −− = −

= += +

58. Rewrite 3 9x y 0+ − = in slope-intercept form. 3 9 0

3 9x y

y x+ − =

= − +

Since the line we are concerned with is parallel to this line, we know it will have slope 3.m = − We are given that it passes through (4, –7). We use the slope and point to write the equation in point-slope form.

( )( ) ( )

( )

1 17 3 47 3 4

y y m x xy x

y x

− = −− − = − −

+ = − −


7 3 123 5

y xy x

y x

+ = − −+ = − +

= − +

In function notation, the equation of the line is ( ) 3 5f x x .= − +

77


59. The line is perpendicular to 1 4,3

y x= + so the

slope is –3. We are given that it passes through (–2, 6). We use the slope and point to write the equation in point-slope form.


In function notation, the equation of the line is

( )( )( )

( )

1 16 3 26 3 2

y y m x xy xy x

− = −− = − − −− = − +

( )6 3 26 3 6

3

y xy x

y x

− = − +− = − −

= −

( ) 3 .f x x= −

60. a. First, find the slope using the points (2, 34.6) and (5, 37.0).

37.0 34.6 2.4 0.85 2 3

m −= =

−=


( )( )

( )

1 1

34.6 0.8 2or

37.0 0.8 5

y y m x x

y x

y x

− = −

− = −

− = −

b. Solve for y to obtain slope-intercept form.

( )34.6 0.8 234.6 0.8 1.6

0.8 33( ) 0.8 33

y xy x

y xf x x

− = −− = −

= += +

c. ( The linear function predicts that 41 million Americans will be living below poverty level in 2010.

10) 0.8(10) 33 41f = + =

Chapter 2 Test

1. The relation is a function. Domain {1, 3, 5, 6} Range {2, 4, 6}

2. The relation is not a function. Domain {2, 4, 6} Range {1, 3, 5, 6}

3. ( ) ( )4 3 4 23 12 2 3 10

f a aa a

+ = + −

= + − = +

4. ( ) ( ) ( )( )

22 4 2 3 2 6

4 4 6 6 16 6 6 28

f − = − − − +

= + + = + + =

5. g shifts the graph of f up 2 units.



8. ( )6 3f = −

9. ( ) 0f x = when 2x = − and 3.x =

10. The domain of f is { is a real number}x x .

11. The range of f is { | 3}.y y ≤

12. The domain of f is { | is a real number and 10}.x x x ≠

13. ( ) ( )2 4 and 2f x x x g x x= + = +

( )( ) ( ) ( )

( ) ( )2

2

2

4 2

4 2

5 2

f g x f x g x

x x x

x x x

x x

+ = +

= + + +

= + + +

= + +

( )( ) ( ) ( )23 3 5 39 15 2 26

f g 2+ = + +

= + + =

14. ( ) ( )2 4 and 2f x x x g x x= + = +

( ) ( ) ( ) ( )

( ) ( )2

2

2

4 2

4 2

3 2

f g x f x g x

x x x

x x x

x x

− = −

= + − +

= + − −

= + −

( ) ( ) ( ) ( )21 1 3 11 3 2 4

f g 2− − = − + − −

= − − = −

78

Intermediate Algebra for College Students 6E Chapter 2 Test

20. ( ) 44

f xy=

=

An equation of the form y = b is a horizontal line.

15. We know that ( )( ) ( ) ( )fg x f x g x= ⋅ . So, to find

we use and .

( )( )5 ,fg − ( )5f − ( )5g −

( ) ( ) ( )25 5 4 5 25 20 5f − = − + − = − =

( )5 5 2g − = − + = −3

5( )( ) ( ) ( )( )

5 5

5 3 15

fg f g− = − ⋅ −

= − = −

16. ( ) ( )2 4 and 2f x x x g x x= + = +2 4( )

2f x xxg x

⎛ ⎞ +=⎜ ⎟ +⎝ ⎠

( ) ( )22 4 2 4 8 12(2) 32 2 4 4

fg

+⎛ ⎞ += = =⎜ ⎟ +⎝ ⎠

21. 4 2 2 11 5 4 2

m −= = = −

− −

The line through the points falls. =

22. 5 ( 5) 10

4 4 0is undefined

m

m

− −= =

−

The line through the points is vertical. 17. Domain of f

g is { is a real number and 2}.x x x ≠ −

23. ( )(10) 3.6 10 14036 140 176

V = +

= + =

In the year 2005, there were 176 million Super Bowl viewers.



12x y− =

( )4 3 0 124 1

3

xxx

− =

==

2

( )4 0 3 123 12

4

yyy

− =

− == −

24. The slope is 3.6. This means the number of Super Bowl viewers is increasing at a rate of 3.6 million per year.


assing through ( 1, 3) and (4, 2)− −

( )( )

2 3 5 14 1 5

m− −

= = =− −


( )( ) ( )( )

( )

1 1

3 1 1

3 1 1

y y m x x

y x

y x

− = −

− − = − −

+ = +

or ( )2 1 4

2 4y xy x− = −

− = −

Slope-Intercept Form 2 4

2y x

y x− = −

= −

In function notation, the equation of the line is ( ) 2.f x x= −

19. ( ) 1 23

1 intercept 23

f x x

m y

= − +

= − − =

79


26. The line is perpendicular to 1 4,2

y x= − − so the

slope is 2. We are given that it passes through . We use the slope and point to write the

equation in point-slope form.



( 2,3− )

.

.

( )( )( )

( )

1 1

3 2 2

3 2 2

y y m x x

y x

y x

− = −

− = − −

− = +

( )3 2 23 2 4

2 7

y xy x

y x

− = +

− = += +

( ) 2 7f x x= +

27. The line is parallel to Put this equation in slope-intercept form by solving for

2 5x y+ =

.y2 5

2 51 52 2

x yy x

y x

+ == − +

= − +

Therefore the slopes are the same; 1 .2

m = −

We are given that it passes through . We use the slope and point to write the equation in point-slope form.

( )6, 4−

( )

( ) ( )

( )

1 1

14 6214 62

y y m x x

y x

y x

− = −

− − = − −

+ = − −


( )14 6214 321 12

y x

y x

y x

+ = − −

+ = − +

= − −


( ) 1 1.2

f x x= − −

28. a. First, find the slope using the points . (2,476) and (4, 486)

486 476 10 54 2 2

m −= = =

−

Then use the slope and a point to write the equation in point-slope form.

( )( )

( )

1 1

486 5 4or

476 5 2

y y m x x

y x

y x

− = −

− = −

− = −

b. ( )

( )

486 5 4486 5 20

5 4665 466

y xy x

y xf x x

− = −

− = −= +

= +

c. ( )10 5(10) 466 516f = + = The function predicts that in 2010 the number of sentenced inmates in the U.S. will be 516 per 100,000 residents.

Cumulative Review Exercises

1. {0, 1, 2, 3}

2. False. π is an irrational number.

3. ( )

2

28 3 9

5 5 18 6

− ÷

− − − ÷⎡ ⎤⎣ ⎦

( ) [ ]2 28 9 9 8 1

5 25 5 3

7 7 75 4 1

− ÷ −= =

−− −⎡ ⎤⎣ ⎦

= = =−

4. ( )( )

0 2

0

4 2 9 3 1 3

4 7 9 1 3 4 1 9 1 34 1 9 3 3 9 3 15

− − + ÷ +

= − − + ÷ + = − + ÷ +

= − + + = + + =

5. ( )[ ] [

3 2 2 5

3 2 4 5 3 3 43 3 4 3 7

x x

x x x ]x x

− − −⎡ ⎤⎣ ⎦= − − − = − − −

= + + = +

80

Intermediate Algebra for College Students 6E Cumulative Review

6.

The solution set is { 4 .

( )2 3 4 2 33 2 2 6

x xx x

+ − = −

− = −2 6

4x

x− = −

= −}−

7. ( )4 12 8 6 2 212 4 6 12 212 4 4 12

12 120 0

x x x xx x xx x

+ − = − − +

− = − + +− = − +

==

The solution set is { } is a real numberx x or The equation is an identity.

.

8.

( ) (

2 2 64 3

4 2 6 3 28 24 3 65 24 6

x x

x xx xx

− +=

+ = −

+ = −+ = −

)

0

The solution set is { } .

5 36

xx= −= −

6−

9. Let x = the price before reduction

The price of the computer before the reduction was $2250.

0.20 18000.80 1800

2250

x xxx

− ===

10. A p prtA p prtA p t

pr

= +− =−

=

11. ( )2 24 524 5

5 433

3 9

10

8x y yx yy x

−−− ⎛ ⎞ ⎛ ⎞

= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ x

12. 2 22 4 2 3

3 2 2 4

25 1

6 1

3 3

3 9

x y x xx y y y

13. ( )( )( )( )( ) ( )

8 2

8 2 6

6 6

5

7 10 3 10

7 3 10 10 21 10

2.1 10 10 2.1 10 10

2.1 10

−

− −

− −

−

× ×

= × × = ×

= × × = ×

= ×

14. The relation is a function. Domain {1, 2, 3, 4, 6} Range {5}

15. g shifts the graph of f up three units.

16. The domain of f is { is a real number and 15}.x x x ≠

17. ( )

( ) (2 2

2 2

2

( )

3 4 2 5 3

3 4 2 5 3

2 5

f g x

x x x x

x x x x

)

x x

−

= − + − − −

= − + − + +

= + +

( ) ( ) ( )( )

2( 1) 2 1 1 5

2 1 1 5 2 1 5 6

f g− − = − + − +

= − + = − + =

18. ( ) 2 42 4

f x xy x= − += − +

2 intercept 4m y= − − =

0

2x xy y

−

−

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠

81


19. Rewrite the equation of the line in slope intercept form.

2 6x y− =

2 62 6

x yy x

− =− = − +

1 32

y x= −

1 intercept 32

m y= − = −

20. The line is parallel to so the slope is 4. We are given that it passes through (3, –5). We use the slope and point to write the equation in point-slope form.



4 7y x= + ,

.

( )( ) ( )

( )

1 1

5 4 3

5 4 3

y y m x x

y x

y x

− = −

− − = −

+ = −

( )5 4 35 4 12

4 17

y xy x

y x

+ = −

+ = −= −

( ) 4 17f x x= −

82

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40p6zu91z1c3x7lz71846qd1-wpengine.netdna-ssl.com40p6zu91z1c3x7lz71846qd1-wpengine.netdna-ssl.com/... · Chapter 2 Functions and Linear Functions 44 9 9 fx x() 3 2.1 Check Points 1