Relations and Functions

2-1

Relations and FunctionsRelations and FunctionsALGEBRA 2 LESSON 2-1ALGEBRA 2 LESSON 2-1

Graph each ordered pair on the coordinate plane.

1. (–4, –8) 2. (3, 6) 3. (0, 0) 4. (–1, 3) 5. (–6, 5)

Evaluate each expression for x = –1, 0, 2, and 5.

6. x + 2 7. –2x + 3 8. 2x2 + 1 9. |x – 3|

(For help, go to Skills Handbook page 848 and Lesson 1-2.)


Solutions

1. 2. 3.

4. 5.

2-1


Solutions (continued)

6. x + 2 for x = –1, 0, 2, and 5:–1 + 2 = 1; 0 + 2 = 2; 2 + 2 = 4; 5 + 2 = 7

7. –2x + 3 for x = –1, 0, 2, and 5:–2(–1) + 3 = 2 + 3 = 5; –2(0) + 3 = 0 + 3 = 3;–2(2) + 3 = –4 + 3 = –1; –2(5) + 3 = –10 + 3 = –7

8. 2x2 + 1 for x = –1, 0, 2, and 5:2 • (–1)2 + 1 = 2 • 1 + 1 = 2 + 1 = 3;2 • 02 + 1 = 2 • 0 + 1 = 0 + 1 = 1;2 • 22 + 1 = 2 • 4 + 1 = 8 + 1 = 9;2 • 52 + 1 = 2 • 25 + 1 = 50 + 1 = 51

9. |x – 3| for x = –1, 0, 2, and 5:|–1 – 3| = |–4| = 4; |0 – 3| = |–3| = 3;|2 – 3| = |–1| = 1; |5 – 3| = |2| = 2

2-1

Graph the relation {(–3, 3), (2, 2), (–2, –2), (0, 4), (1, –2)}.

ALGEBRA 2 LESSON 2-1ALGEBRA 2 LESSON 2-1

Relations and FunctionsRelations and Functions

Graph and label each ordered pair.

2-1



Write the ordered pairs for the relation. Find the domain

and range.

{(–4, 4), (–3, –2), (–2, 4), (2, –4), (3, 2)}

The domain is {–4, –3, –2, 2, 3}.

The range is {–4, –2, 2, 4}.

2-1



Make a mapping diagram for the relation {(–1, 7), (1, 3),

(1, 7), (–1, 3)}.

Pair the domain elements with the range elements.

2-1



Determine whether the relation is a function.

The element –3 of the domain is paired with both 4 and 5 of the range.

The relation is not a function.

2-1



Use the vertical-line test to determine whether the graph

represents a function.

If you move an edge of a ruler from left to right across the graph, keeping the edge vertical as you do so, you see that the edge of the ruler never intersects the graph in more than one point in any position.

Therefore, the graph does represent a function.

2-1



Find ƒ(2) for each function.

a. ƒ(x) = –x2 + 1

ƒ(2) = –22 + 1 = –4 + 1 = –3

b. ƒ(x) = |3x|

ƒ(2) = |3 • 2| = |6| = 6

c. ƒ(x) = 9

1 – x

ƒ(2) = = = –99

1 – 29

–1

2-1


Pages 59–61 Exercises

1.

2.

3.

4.

5. (–2, –2), (–1, 1), (1, 1), (1, 0), (3, 3), (3, –2); domain {–2, –1, 1, 3}, range {–2, 0, 1, 3}

6. (–2, 3), (0, 1), (2, –1), (3, –2); domain {–2, 0, 2, 3}, range {–2, –1, 1, 3}

7. (–2, 0), (–1, 2), (0, 3), (1, 2), (2, 0); domain {–2, –1, 0, 1, 2}, range {0, 2, 3}

8.

2-1


9.

10.

11.

12. not a function

13. function

14. function

15. function

16. function

17. not a function

18. function

19. function

20. not a function

21. function

22. –7, –3, 4, 11

23. 13, 7, –3.5, –14

24. 4.5, 6.5, 10, 13.525. –2, –4, –7.5, –11

26. 21, 13, –1, –1527. –13, –9, –2, 5

28. 29 , 17 , –3 , –24

29. – , – , , 13

13

23

23

236

136

34

113

2-1


30. – , – , – , 0

31. y = , where y

is the number of meters and x is the number of inches; 1.50 m

32.

domain {2, 3, 4, 5}, range {4, 5, 6, 7}

92

72

74

x39.37

33.

domain {–4, –3, –2, –1}, range {1, 2, 3, 4}

34.

domain { – , 0, 2},

range { –2, – , , 2}

12

12

12

35.

domain { – , , , }

range { – , }

36. domain {2, 4, 8}, range {4, 8, 16}; function

37. domain {–2, –1, 0, 9}, range {2, 5, 7}; not a function

32

12

32

52

12

12

2-1


38. domain {all real numbers}, range {y 0}; function

39. domain {–3.2 x 3.2}, range {–1 y 1}; not a function

40. A

41. B

42. C

43. yes

44. no

45. no

46. v(s) = s3; 2460.375 cm3

47. v(r) = r 3; about 4849 cm3

48. No; since 2 in the domain maps to 1 and 4, it is not a function.

49. Check students’ work.

50. –3

51. –

52. –4x – 14

53. 7

54.

55. Yes; each x is paired with a unique y.

56. No; each positive x is paired with two y values.

34

23

2-1

>–

<– <–<– <–


57. Yes; each x is paired with a unique y.

58. Function; if the domain and range are interchanged, it is not a function because 6 would be paired with both 2.5 and 3.0.

59. Domain {all integers}, range {all even integers}; function, each integer pairs to a unique even integer.

60. Domain {all integers}, range {all integers}; function, each integer pairs with its opposite.

61. Domain {all integers}, range {all even integers}; not a function, each nonzero integer pairs with a pos. and neg. even integer.

62. C

63. G

64. [2] ƒ(5); ƒ(5) = –2(5) + 3 = –7 and g(–2) =

4(–2) – 3 = –11, and –7 > –11.

[1] only includes answer ƒ(5) or g(–2)

65. [4] Yes; S(c) = 6c2; S(2.5) = 6(2.5)2 = 37.5; the surface area is 37.5 cm2.

[3] minor computational error

[2] includes only function

[1] no work shown

2-1


2-1

66.

67.

68.

69. 0

70.

71.

72.

111

211

411

611

711

73. –6, 14

74. –2 b 2

75. x 6

311

76. 11.1% increase

77. 50% increase

78. 25% decrease

=/

<– <–



1. Write the ordered pairs for the relation. Find the domain and range.

2. Determine whether the relation {(–2, 3), (–5, 0), (3, 0), (1, 1)} is a function.

3. Delete one ordered pair so that the relation {(–4, 2), (1, 6), (0, 0), (–4, 6)} is a function.

2-1



4. Determine whether the graph represents a function.

5. Find ƒ(–5) for each function.

a. ƒ(x) = 5x + 35

b. ƒ(x) = x2 – x

2-1



1. Write the ordered pairs for the relation. Find the domain and range.

2. Determine whether the relation {(–2, 3), (–5, 0), (3, 0), (1, 1)} is a function.

3. Delete one ordered pair so that the relation {(–4, 2), (1, 6), (0, 0), (–4, 6)} is a function.

{(–3, 2), (–2, 0), (–1, –1), (2, 1), (2, 3), (4, 1)};D = {–3, –2, –1, 2, 4], R = {–1, 0, 1, 2, 3}

function

(–4, 2) or (–4, 6)

2-1



4. Determine whether the graph represents a function.

5. Find ƒ(–5) for each function.

a. ƒ(x) = 5x + 35

b. ƒ(x) = x2 – x

not a function

10

30

2-1

2-2

Linear EquationsLinear EquationsALGEBRA 2 LESSON 2-2ALGEBRA 2 LESSON 2-2

(For help, go to Lesson 1-2.)

Evaluate each expression for x = –2, 0, 1, and 4.

1. x + 7 2. x – 2

3. 3x + 1 4. x – 8

23

35

12


1. x + 7 for x = –2, 0, 1, 4:

(–2) + 7 = – + = = or 5 ;

(0) + 7 = 0 + 7 = 7;

(1) + 7 = + 7 = 7 or ;

(4) + 7 = + = = or 9

Solutions

23

23

23

23

23

43

213

–4 + 213

173

23

23

23

233

83

213

8 + 213

293

23

2-2

2. x – 2 for x = –2, 0, 1, 4:

(–2) – 2 = – + – = = – or –3 ;

(0) – 2 = 0 – 2 = –2;

(1) – 2 = + – = = – = – or –1 ;

(4) – 2 = + – = =



35

35

35

35

35

65

105

165

15

35

105

75

75

25

125

105

25

3. 3x + 1 for x = –2, 0, 1, 4:3(–2) + 1 = –6 + 1 = –5; 3(0) + 1 = 0 + 1 = 1;3(1) + 1 = 3 + 1 = 4; 3(4) + 1 = 12 + 1 = 13

2-2

–6 + (–10)5

3 + (–10)5

12 + (–10)5



4. x – 8 for x = –2, 0, 1, 4:

(–2) – 8 = –1 – 8 = –9;

(0) – 8 = 0 – 8 = –8;

(1) – 8 = – 8 = –7 or – ;

(4) – 8 = 2 – 8 = –6

12

12

12

12

12

12

12

152

2-2

Plot the points (0, 2) and (3, –2)

and then draw the line through these two points.


Linear EquationsLinear Equations

Graph the equation y = – x + 2.43

If x = 0, then y = 2.

If x = 3, then y = –2.

2-2



The equation 10x + 5y = 40 models how you can give $.40

change if you have only dimes and nickels. The variable x is the

number of dimes, and y is the number of nickels. Graph the

equation. Describe the domain and the range. Explain what the x-

and y-intercepts represent.

10x + 5y = 40 10x + 5y = 4010x + 5(0) = 40 10(0) + 5y = 40

10x = 40 5y = 40x = 4 y = 8

Set x or y equal to zero to find each intercept.

2-2

The x-intercept is (4, 0), which means that the change can be given using 4 dimes and 0 nickels.

The y-intercept is (0, 8), which means that the change can be given using 0 dimes and 8 nickels.



(continued)

Use the intercepts to graph the equation.

The number of dimes and the number of nickels must each be a whole number.

The possible solutions for this situation are limited to those points on the line segment connecting (0, 8) and (4, 0) whose x- and y-coordinates are whole numbers.

Therefore, the domain is {0, 1, 2, 3, 4} and the range is {0, 2, 4, 6, 8}.

2-2



Find the slope of the line through the points (–2, 7) and (8, –6).

Slope = Use the slope formula.y2 –y1

x2 – x1

= Substitute (–2, 7) for (x1, y1) and (8, –6) for (x2, y2).

–6 – 78 – (–2)

2-2

= – Simplify.1310

The slope of the line is – .1310



Write in standard form an equation of the line with slope 3

through the point (–1, 5).

y – y1 = m(x – x1) Use the point-slope equation.

y – 5 = 3[x – (–1)] Substitute 3 for m, 5 for y1, and –1 for x1.

y – 5 = 3[(x + 1)] Simplify.

y – 5 = 3x + 3 Distributive Property

3x – y = –8 Write in standard form.

2-2



Write in point-slope form an equation of the line through

(4, –3) and (5, –1).

y – y1 = m(x – x1) Write the point-slope equation.

y + 3 = 2(x – 4) Simplify.

You can also use (5, –1) for (x1, y1) and (4, –3) for (x2, y2).

This gives the equation y + 1 = 2(x – 5). Both equations define the same line.

y – y1 = (x – x1) Substitute the slope formula for m.y2 – y1

x2 – x1

y – (–3) = (x – 4) Substitute: x1 = 4, y1 = –3, x2 = 5, y2 = –1.–1 – (–3)

5 – 4

2-2



Find the slope of –7x + 2y = 8.

–7x + 2y = 8 Add 7x to both sides.

2-2

y = x + 4 Write in the slope-intercept form.72

The slope of the line is .72



Write an equation of the line through (5, –3) and perpendicular

to y = 4x + 1. Graph both lines.

2-2

m = – Find the negative reciprocal of 4.14

y = mx + b Use slope-intercept form.

y = – x + b Slope is – .14

14

14–3 = – (5) + b Substitute (5, –3) for (x, y).

–3 = – + b Simplify.54

– = b Solve for b.74

y = – x – Write the equation.14

74



1.

2.

3.

4.

5.

6.

2-2


7.

8.

9. a.

y = 0.23xdomain {x | x

0}range {y | y

0}

10. a.

b. The x-intercept is the point that represents selling

1000 caps and 0 sweatshirts in

order to raise the $4500. The

y-intercept represents

selling 0 caps and 360 sweatshirts to

raise the $4500.2-2

>–>–

b. x-intercept (0, 0), y-intercept (0, 0); when no miles have been driven, there is no cost.c. 0.23 represents a cost of $.23 per mile driven.


11. –1

12. –2

13. 3

14.

15. –

16. 1

17. undefined

18. 0

19.

411

15

175

20. 3x – y = –2

21. x – y =

22. x + y = –

23. y = –2

24. x + y = 2

25. 5x – y = –2

26. y – 3 = –1(x + 10)27. y – 0 = (x – 1)

28. y – 10 = – (x + 4)

56

193

35

125

54

52

29. y + 1 = – (x – 0)

30. y – 11 = 1(x – 7)31. y – 9 = – (x – 1)

32. –5

33.

34. –

35. –

36.

37. 0

43

75

32

12

AB

AB

2-2

40. y = 10

41. x = 1


38. y = –3x – 5

39. y = x +52

132

42.

43.

44.

2-2


45.

46.

47.

48.

49.

50.

51.

52.

13

23

53. –

54. , (0, 4), (–6, 0)55. –1, (0, 1000),

(1000, 0)

56. , 0, – , , 0

57. 5, (0, –1), , 0

58. undefined slope, no y-intercept, (–3, 0)

59. 0, (0, 0), all pts. onx-axis

14

23

RS

TS

TR

15

2-2


60. – , 0, – , (–5, 0)

61. –0.8, (0, 0.4), (0.5, 0)

62. – , 0, , , 0

63. –

64.

65. –

66. y = x + 3

67. y = 3x + 2

AB

CB

CA

12

52

513

75

710

34

68. y = – x –

69. a. I, II; III; graphs I and II show

constant rate of change.

b. I and III

c. II

70. Vertical lines cannot be graphed by this method.

71. y = –1

72. y = 2x + 1

73. y = x + 56

103

2-2

32

12


74. y = – x – 1

75. 3x – 2y = 2

76. 9x + 3y = 2

77. 3x + 12y = –4

78. a–e. Check students’ work. The polygon is a rectangle.

32

79. yes

80. no

81. a.

b. y = 3x + 6

c. y = – x +

d. They are perpendicular.

13

83

82. The equation of the line connecting (1, 3) to (–2, 6) is y = –x + 4. The equation of the line connecting (1, 3) to (3, 5) is y = x + 2. The slopes are negative reciprocals so the lines are perpendicular. Therefore by def. of a right triangle it is a right triangle.

2-2


83. The slope of the line connecting (2, 5) to

(4, 8) is , (2, 5) to

(5, 3) is – , (4, 8) to

(7, 6) is – , and

(5, 3) to (7, 6) is .

Since the adjacent sides’ slopes are negative reciprocals they are perpendicular. By the def. of a rectangle, it is a rectangle.

32

2323

32

84. p: y = 4x + 16

q: y = – x +

r : y = 4

85. A

86. G

87. C

88. B

14

134

89. C

90. B

91. domain {–2, 1, 2, 3, 4}, range {–2, –1, 2, 3}; not a function

92. domain {all reals}, range {all reals}; function

93. domain {–3, 0, 1, 7}, range {–10, –5, –1, 3}; not a function

2-2


94. Commutative Prop. of Add.

95. multiplicative inverses

96. Distributive Prop.

97. additive inverses, additive identity

98. studio: $6.26; designer: $9.39

2-2



1. Find the slope of the line through the points (–5, –1) and (2, 3).

2. Write an equation in standard form for the line with slope 3 through (9, –4).

3. Write in point-slope form an equation of the line through the points (–3, 8) and (7, 6). Use (–3, 8) as the point for the equation.

4. Write the equation 3x – 12y = 6 in slope-intercept form.

5. What is the slope of a line perpendicular to y = x – 7? What is the slope of a line parallel to y = x – 7?

3x – y = 31

232

3

2-2

47

y – 8 = – (x + 3)15

y = x –14

12

32

23;–

Direct VariationDirect Variation

(For help, go to Lesson 1-3 and Skills Handbook page 844.)


Solve each equation for y.

1. 12y = 3x 2. 12y = 5x 3. y = 15

4. 0.9y = 27x 5. 5y = 35

6. 7. 8. 9. 14

28

25

1236

2024

3036

615

924

Tell whether each equation is true.

2-3

34

Direct VariationDirect VariationALGEBRA 2 LESSON 2-3ALGEBRA 2 LESSON 2-3

Solutions

1. 12y = 3x

y = = =3x12

3 • x3 • 4

x4

2. 12y = 5x

y = = x5x12

512

5. 5y = 35

y = 7

3. y = 15

y = 15 = = 20

34

43

603

4. 0.9y = 27x

y = = 30x27x0.9

6.

1(8) 4(2)

8 = 8

true

14

28

2-3


2-3


7.

2(15) 5(6)

30 = 30

true

25

615 8.

9(36) 24(12)

324 288

false

1236

924

9.

20(36) 24(30)

720 = 720

true

2024

3036

=/


For each function, determine whether y varies directly with x.

If so, find the constant of variation and write the equation.

x –1 2 5y 3 –6 15

a.

Since the three ratios are not all equal, y does not vary directly with x.

x 7 9 –4y 14 18 –8

b.

The constant of variation is 2.

The equation is y = 2x.

2-3

= and are both equal to –3, but = 3. yx

3–1

–62

155

= 2, so y does vary directly with x.yx

147

189

–8–4= = =


For each function, tell whether y varies directly with x. If so,

find the constant of variation.

a. 3y = 7x + 7

Since you cannot write the equation in the form y = kx, y does not vary directly with x.

b. 5x = –2y

2-3

5x = –2y is equivalent to y = – x, so y varies directly with x.52

The constant of variation is – . 52


The perimeter of a square varies directly as the length of a

side of the square. The formula P = 4s relates the perimeter to the

length of a side.

a. Find the constant of variation.

The equation P = 4s has the form of a direct variation equation with k = 4.

b. Find how long a side of the square must be for the perimeter to be 64 cm.

P = 4s Use the direct variation.

64 = 4s Substitute 64 for P.

16 = s Solve for s.

The sides of the square must have length 16 cm.

2-3


Suppose y varies directly with x, and y = 15 when x = 27.

Find y when x = 18. Let (x1, y1) = (27, 15) and let (x2, y2) = (18, y).

15(18) = 27(y) Write the cross products.

y = 10 Simplify.

Write a proportion.y1

x1

y2

x2

=

Substitute.=1527

y18

2-3

y = Solve for y.15 • 18

27



1. yes; k = 2, y = 2x2. yes; k = –3, y = –3x

3. no

4. yes; k = , y = x

5. yes; k = 7, y = 7x6. no

7. yes; k = –2, y = –2x

8. no

9. yes; k = 12

10. yes; k = 6

11. yes; k = –2

12. no

13. no

14. yes; k = –5

15. yes; k = 6

16. no

13

13

17. k = ; –

18. k = – ;

19. k = –1; 5

20. k = 2; –10

21. k = – ; 21

22. k = – ; 1

23. a. k =

b. s = h

c. 23 ft 1 in.

27


107

53

253

174

14

14

14

13361336

2-3


24. –3

25. 4

26. 10.5

27.

28. 681.8 mi/h

29. yes; k = , y = x

30. no

31. no

32. yes; k = 1.3, y = 1.3x


53

23

23

33. y = 2x

34. y = x

35. y = – x

36. y = –500x

37. y = x

38. y = – x

39. y = x

40. y = – x

41. 9

73

92

35

19

27

143

42. 6

43. 90

44. –140

45. 1.2

46. No; y = 1.7x does not contain the point (9, –9).

47. Yes; y = – x contains

the point (15, –12 ).

48. Yes; y = x contains the

point (6 , 22 ).

56 1

2

72

12

34

2-3


49. Answers may vary. Sample: y = 0.5x

50. Answers may vary. Sample: y = 3.2x

51. Answers may vary. Sample: y = – x


34

52. a. y = 28xb. 103.6 mic. 417.9 gald. $0.056/mi

53. Answers may vary. Sample: No; y = 0 passes through the origin, but is not a direct variation.

54. Answers may vary. Sample: If y varies directly with x2, and y = 2 when x = 4,

then y = 10 when

x = 9.

18

55. y is doubled.

56. y is halved.

57. y is divided by 7.58. y is multiplied by 10.

59. a.

b. 32

c. z = kxy, x = k1w, so z = kk1wy,

z varies jointly with w and y.

60. B

12

2-3


61. G

62. D

63. I

64. [2] No; the ratio is not constant.

[1] correct answer with no

explanation


65.

66.

67.

68.

69.

70.

domain {4, 7}, range {0, –1}

2-3

yx

72.

domain {1, 2, 3, 4}, range {7, 8, 9, 10}


71.

domain {1, 2, 4, 5}, range {–2, –1, 1, 2}


73. 72%

2-3

1. For each function, tell whether y varies directly as x. If so, find the constant of variation and write the equation.

a. b.

2. Determine whether y varies directly as x. If so, find the constant of variation.

a. y = 7x + 4

b. y = x

c. 2y = –12x


x y–4 16–2 8 8 –32

x y2 103 156 25

53

2-3


3. Assume y varies directly as x. If y = 8 when x = 42, find y when x = 126.

4. How can you tell from the graph of a linear function whether the function is a direct variation?

2-3

1. For each function, tell whether y varies directly as x. If so, find the constant of variation and write the equation.

a. b.

2. Determine whether y varies directly as x. If so, find the constant of variation.

a. y = 7x + 4

b. y = x

c. 2y = –12x


x y–4 16–2 8 8 –32

x y2 103 156 25

yes; –4; y = –4x

no

53

no

yes; –6

2-3

yes;53


If the graph is a non-horizontal line through the origin, then the linear function is a direct variation.

3. Assume y varies directly as x. If y = 8 when x = 42, find y when x = 126.

4. How can you tell from the graph of a linear function whether the function is a direct variation?

24

2-3

Using Linear ModelsUsing Linear Models

(For help, go to Lessons 2-1 and 2-2.)


1. (–0.2, 9) and (3.4, 7.3) 2. (10, 17) and (11.5, 13.5)

3. (0, ) and (–1, )

4. ƒ(x) = x – 2 for x = –3, 0,

5. g(x) = 3(2 – x) for x = 0, , 1

310

25

43

12

16

Find the change in x and the change in y between each pair of points.

Evaluate each function for the given values.

2-4

Using Linear ModelsUsing Linear ModelsALGEBRA 2 LESSON 2-4ALGEBRA 2 LESSON 2-4

1. (– 0.2, 9) and (3.4, 7.3)change in x: 3.4 – (–0.2) = 3.4 + 0.2 = 3.6change in y: 7.3 – 9 = –1.7

2. (10, 17) and (11.5, 13.5)change in x: 11.5 – 10 = 1.5change in y: 13.5 – 17 = –3.5

3. 0, and –1,

change in x: –1 – 0 = –1

change in y: – = – =

310

25

310

25

410

310

110

Solutions

2-4


4. ƒ(x) = x – 2 for x = –3, 0, :

ƒ(–3) = (–3) – 2 = –4 – 2 = –6;

ƒ(0) = (0) – 2 = 0 – 2 = –2;

ƒ = – 2 = – = = – = – or –1

5. g(x) = 3(2 – x) for x = 0, , 1:

g(0) = 3(2 – 0) = 3(2) = 6;

g = 3 2 – = 3 = 3 = 3 = = or 5 ;

g(1) = 3(2 – 1) = 3(1) = 3

43

12

43

43

12

12

43

46

126

4 – 126

86

43

13

16

16

16

16

126 –

12 – 16

116

336

112

12


2-4


Suppose an airplane descends at a rate of 300 ft/min from

an elevation of 8000 ft.

Write and graph an equation to model the plane’s elevation as a function of the time it has been descending. Interpret the intercept at which the graph intersects the vertical axis.

Relate: plane’s elevation = rate • time + starting elevation .

Define: Let t = time (in minutes) since the plane began its descent.

Let d = the plane’s elevation.

Write: d = –300 • t + 8000

2-4


(continued)

An equation that models the plane’s elevation is d = –300t + 8000.

The d-intercept is (0, 8000).

This tells you that the elevation of the plane was 8000 ft at the moment it began its descent.

2-4


A spring has a length of 8 cm when a 20-g mass is hanging

at the bottom end. Each additional gram stretches the spring

another 0.15 cm. Write an equation for the length y of the spring as

a function of the mass x of the attached weight. Graph the equation.

Interpret the y-intercept.

Step 1: Identify the two points as (x1, y1) and (x2, y2).

Adding another 20 g of mass at the end of the spring will give a total mass of 40 g and a length of 8 + 0.15(20) = 11 cm. Use the points (x1, y1) = (20, 8) and (x2, y2) = (40, 11) to find the linear equation.

2-4


(continued)

Step 2: Find the slope of the line.

m = Use the slope formula.

m = Substitute.

m = , or 0.15 Simplify.

y2 – y1

x2 – x1

11 – 840 – 20

320

Step 3: Use one of the points and the point-slope form to write an equation for the line.

y – y1 = m(x – x1) Use the point-slope form.

y – 8 = 0.15(x – 20) Substitute.

y = 0.15x + 5 Solve for y.

2-4


(continued)

An equation of the line that models the length of the spring is y = 0.15x + 5.

The y-intercept is (0, 5). So, when no weight is attached to the spring, the length of the spring is 5 cm.

2-4


Use the equation from Additional Example 2. What mass

would be needed to stretch the spring to a length of 9.5 cm?

y = 0.15x + 5 Write the equation.

9.5 = 0.15x + 5 Substitute 9.5 for y.

30 = x Simplify.

The mass should be 30 g.

2-4

= x Solve for x.9.5 – 5

0.15


An art expert visited a gallery and jotted down her guesses

for the selling price of five different paintings. Then, she checked the

actual prices. The data points (guess, actual) show the results, where

each number is in thousands of dollars. Graph the data points.

Decide whether a linear model is reasonable. If so, draw a trend line

and write its equation.

{(12, 11), (7, 8.5), (10, 12), (5, 3.8), (9, 10)}

A linear model seems reasonable since the points fall close to a line.

Trend lines and equations may vary.

2-4


(continued)

A possible trend line is the line through (6, 6) and (10.5, 11). Using

these two points to write an equation in slope-intercept form gives

y = x – .109

23

2-4



1. d = 62.5h + 15 2. a. y = –50t + 1000

b.

The y-intercept (0, 1000) represents the filled pool, and the t-intercept (20, 0) represents the time needed to empty the pool.

3. h = 8x + 60

4. y = x – 1; 5 leaves

5. y = 0.5x + 0.75; 3.25 lb

6. y = 58.3x – 3.3; 172 blades of grass

2-4


7. y = 1.75x + 1.75; $8.40

8.

Linear model is reasonable; models may vary. Sample: y = –1.3x + 11

9.

Linear model is reasonable; models may vary. Sample: y = 2.6x – 0.6

10.

Linear model is reasonable; models may vary. Sample: y = –0.75x – 3.7

11.

not reasonable

2-4


12. a. Linear model is reasonable.

b. 40

c. Answers may vary. Sample: After drawing a trend line

on the graph, locate the European size on the y-axis. Then find the corresponding U.S. size on the

x-axis.

13. a.

Answers may vary. Sample: y = 125x + 975

b. 2975 cal c. Answers may vary. Sample: No;

adults need fewer Calories, not more.

2-4


14. a.

b–e. Answers may vary. Samples are given.

b. c = 0.04A

c. The model fits the data very closely.

d. No; the area of the tarp is 150 ft2 so the price should be $6.00.

e. 6 8 ft; $0.07

15. y = –4x + 10

16. y = –3x – 6

17. y = –7.5x – 2.5

18. y = 1.4375x – 7.33125

2-4


19. a. y = 29.95

b. y = 2.95x; slope = 2.95, y-intercept = 0

c. Answers may vary. Sample:

Either way, you will average the same costs over the long run.

20. a. Answers may vary. Sample:

y = 0.0714x – 9.3

b. 14.3 g

c. 200 Cal; a 200 Cal hamburger is closer to 5 g.

2-4


21. a. population

b–c.

d. 2 million

e. Answers may vary. Sample: Strong; the points fall close to a straight line.

22.104.5

23.85.8

24.6.5

25.13

26.882

2-4


27. a.

b.about $3900

c. about $1400

d.Answers may vary. Sample: No, expenditure would be predicted to be about $5000.

28. a. y = – x + 14.27

b. y = –2x + 10

c. Answers may vary. Sample: neither; y = –1.5x + 12

29. y = x – 2

30. 13.37

31. 1.10

32. 6.10

33. 8.09

1711

52

2-4


34. a. y = – x + 72.6

b. 35.3 tons

35. a. Answers may vary. Sample:

y = x +

b. 70

36. –2.7; 13.5

37. –3; 15

38. – ; 6

39. ; –42

163

163

23

65

425

40. {–7, –3, –1, 2, 7}

41. {1, 3.5, 5, 6, 8}

42. {3, 4, 5.25, 12, 19}

43. { }

44. {–299, –99, 1, 151, 401}

45. {1, 6, 9, 11, 15}

46. a. 20.91 ft/s; 2091 ft

b. d = 20.91t

c. 14.25 mi/h

– , – , –2, – , 072

52

54

2-4


1. A family built a house in a beach resort area. In 1990, the house was 430 ft from the water. With erosion, the house was 400 ft from the water in 2000.

a. Assuming a linear relationship between the number of years x since 1990 and the distance y of the house from the water, write an equation for y as a function of x.

b. In what year will the house be only 250 ft from the water?

2. Graph the set of data. Decide whether a linear model is reasonable. If so, draw a trend line and write its equation.{(1, 2), (3, 3), (3, 3.75), (4, 4), (5, 3.25), (6, 4.5)}

y = –3x + 430

2050

A linear model seems reasonable; Answers may vary.Sample: y = 0.4x +2

2-4

2-5

Absolute Value Functions and GraphsAbsolute Value Functions and GraphsALGEBRA 2 LESSON 2-5ALGEBRA 2 LESSON 2-5

(For help, go to Lesson 2-2 and Skills Handbook page 851.)

1. y = x for real numbers x and y 0

2. y = 2x – 4 for real numbers x and y 0

3. y = –x + 6 for real numbers x and y 3

Graph each equation for the given domain and range.

>–

<–

>–


1. y = x x = real numbersy 0

2. y = 2x – 4 x = real numbersy 0

3. y = –x + 6 x = real numbersy 3

Solutions

2-5

>–

<–

>–


Graph y = |2x – 1| by using a table of values.

Graph the function.

x –1.5 –1 –0.5 0 0.5 1 1.5 2 2.5y 4 3 2 1 0 1 2 3 4

Make a table of values.

Evaluate the equation for several values of x.

2-5


Use the absolute value key. Graph the equation

Y1 = abs(X – 1) – 1

Graph y = |x – 1| – 1 on a graphing calculator.

2-5


Use the definition of absolute value to graph y = |3x + 6| – 2.

Step 1: Isolate the absolute value.

y = |3x + 6| – 2

y + 2 = |3x + 6|

when 3x + 6 < 0y + 2 = –(3x + 6) y = –3x – 8

2-5

Step 2: Use the definition of absolute value. Write one equation for 3x + 6 0 and a second equation for 3x + 6 < 0.>–

when 3x + 6 0y + 2 = 3x + 6 y = 3x + 4

>–


(continued)

2-5

Step 3: Graph each equation for the appropriate domain.

When 3x + 6 0, or x –2, y = 3x + 4.

When 3x + 6 < 0, or x < –2, y = –3x – 8.

>– >–


A train traveling on a straight track at 50 mi/h passes a

certain crossing halfway through its journey each day. Sketch a graph

of its trip based on its distance and time from the crossing.

The equation d = |50t| models the train’s distance from the crossing.

2-5

Absolute Value Functions and GraphsAbsolute Value Functions and Graphs


1–9. Tables may vary. Samples are given.

1.


2.

3.

4.

5.

2-5


6.

7.


8.

9.

10.

11.

2-5


12.

13.

14.


15.

16.

17.

18.

19.

20.

2-5


21.

22.

23.


24.

25.

26.

27.

2-5


28.

29. B

30. C

31. A

32. D


33.

34.

35.

36.

37.

38.

2-5


39.

40.

41.


42.

43.

44.

45.

46.

47.

2-5


48.

49.

50.


51. a. time and distance before and after the roadside

stand

b. A

52. a.

b. Answers may vary. Sample: Same shape and size, different vertices,

and one points down.

53.

54.

2-5


55.

56.

57.


58.

59. Answers may vary. Samples:

a. y = |x|, y = –|x|

b. y = |x|, y = |x – 1| + 1

60. B

61. I

62. A

63. I

64. C

65. [2] The vertex of y = |3x – 6|

would be where 3x – 6 = –(3x – 6) because that would be where the graphs of both lines meet. In this case it is at x = 2.

[1] only includes solution x = 2

2-5

[1] y = –|5x + 1| is not interpreted correctly but graph is correct

67.

Answers may vary. Sample: y = 8.7x – 10.9


66.

[4] y = –5x – 1, y = 5x + 1[3] does not label vertex

[2] does not label vertex and does not include both

equations


2-5


68.

Answers may vary.

Sample: y = x + 6

69.

not reasonable

23

23

70.

Answers may vary. Sample: y = –0.5x + 7

71. –3

72. 4

73. –

74. 4

75. –

76. 5

77. y = 18h; linear78. 11

79. 9

80. 13

81. 4

82. 719

32

2-5


1. Graph y = –|x + 3|.

2. Graph y = |2x – 6| – 2.

2-5

2-6

Vertical and Horizontal TranslationsVertical and Horizontal TranslationsALGEBRA 2 LESSON 2-6ALGEBRA 2 LESSON 2-6

1. y = x, y = x + 4 2. y = –2x, y = –2x – 3

3. y = |x|, y = |x| – 2 4. y = –|2x|, y = –|2x| ± 1

5. y = |x|, y = |x – 1| 6. y = – x , y = – x + 2


12

12

Graph each pair of functions on the same coordinate plane.


1. y = xy = x + 4

3. y = |x|y = |x| – 2

5. y = |x|y = |x – 1|

2. y = –2xy = –2x – 3

4. y = – |2x|y = – |2x| + 1

6. y = – x

y = – x + 2

1212

Solutions

2-6


Graph y = |2x| and y = |2x| + 3 on the same coordinate

plane. Describe how the graphs are related. Make a table of values

and graph the equation.

x y = |2x| y = |2x| + 3

–2 4 7

–1 2 5

0 0 3

1 2 5

2 4 7

For each value of x, y = |2x| + 3 is 3 more than the value of y = |2x|.

The graph of y = |2x| + 3 is the graph of y = |2x| shifted 3 units up.

2-6

The parent is y = 2x, and k = –3.


Identify the parent function for y = 2x – 3 and the value of k.

Then, graph the function by translating the parent function.

Translate the graph of y = 2x down 3 units.

2-6


Write an equation to translate the graph of y = |4x|,

5 units down.

The graph of y = |4x|, shifted 5 units down, means k = –5.

An equation is y = |4x| – 5.

2-6


Identify the parent function and the value of h for y = | x – 5|.

Graph both functions.

The parent function is y = |x|, and h = 5.

The minus sign means to move to the right. Translate the graph of y = |x| right 5 units.

2-6


The graph is a translation of y = –|x|. Write an equation for

the graph.

The given graph can be obtained by translating the graph of y = –|x|, 3 units left.

So an equation for the given graph is y = –| x + 3|.

2-6


Describe a possible translation of Figures M and N in the

design shown below.

Translate Figure M 2 units down.

For Figure N, there are two possible translations: 4 units down, or 1 unit right and 2 units down.

2-6

The parent function is y = |x|, h = 1, and k = 2.


Graph y = | x + 1| + 2.

The first plus sign means move 1 unit to the left.

The second plus sign means move k units up.

Place the vertex at (–1, 2) and draw the graph opening upward.

2-6


Write an equation for each translation of y = |x|.

a. 3 units up, 7 units right

An equation is y = | x – 7| + 3.

b. 5 units down, 1 unit left

An equation is y = | x + 1| – 5.

7 units right h = 7; minus sign

3 units up k = 3; plus sign

1 unit left h = 1; plus sign

5 units down k = 5; minus sign

2-6

Vertical and Horizontal TranslationsVertical and Horizontal Translations


1. y = –|x| + 3 is y = –|x| 3 units up.

2. f(x) = |x| – 1 is f(x) = |x| one unit down.

3. g(x) = |3x| + 2 is g(x) = |3x| 2 units up.

4. is

units down.

y = |x| –12

23

y = |x|12

23

5. y = x, k = 3

6. y = –x, k = 4


7. y = |x|, k = 2

8. y = –|x|, k = 6

2-6


9. y = x –

10. y = |x| + 4

11. y = –|x| + 2

12. y = |x|, h = 4


23

13. y = |x|, h = 5

14. y = –|x|, h = 1

15. y = –|x|, h = 2

16. y = |x + 2|

17. y = |x – 3|

18. y = –|x – 4|

19. Answers may vary. Sample: 1 unit left and 1 unit up

20. Answers may vary.

Sample: units left

and 1 unit up

65

2-6


21.

22.

23.


24.

25.

26.

27. y = |x – 3| – 1

28. y = –|x + 1| + 2

29. y = – |x – | +

30. y = |x – 7| + 3

31. vertical

32. diagonal

52

12

2-6


33. horizontal

y = |x + 1|

34. vertical

35. vertical

36. vertical

37. diagonal

38. diagonal

39. diagonal

40. y = |x| – 3

41. y = x + 2

42. y = |x – 3| – 2

43. h = 1, k = 0

44. h = 0, k = 3

2-6


45. h = 2, k = 2

46. Horizontal; the graph is shifted left.

47. Answers may vary. Sample: Since a vert. translation affects only y-values and a horiz. translation affects only x-values, order is irrelevant.

48. a.

b. y = 3x + 5

c. y = 3x – 1

d. E

e. y = m(x – h) +

k


50. y = |x – 7| – 2

51. y = –|x| – 3

52. y = –|x + 5|

53. y = x + 7

54. y = |x – a| + b

55. y = x + (q – p)

56.

6 right

2-6

57.

4 right, 3 down

58.

3 left, 1 up

59.

3 right, 3 up


60. a.

b. Answers may vary. Sample: It is a

function because

each t is paired with a unique height.

c. vertical translation

5 ft up

d. C

61. a.

b.

c.

62. C

63. H

64. A

2-6


65. [4] The graph of y = |x + 3| – 2 is y = |x| first translated 3 units to the left for

y = |x + 3| and then 2 units down for y = |x + 3| – 2.

[3] includes equations and graph with no

explanation

[2] includes either the translation equations OR graph

[1] attempts to interpret the translation for y = |x + 3| – 2 and graphs it but makes an

error and does not translate it 3 units left and 2 units down

66. Answers may vary. Sample: ƒ(0) = 5; ƒ(1) = 4; ƒ(2) = 3; ƒ(3) = 2; ƒ(–1) = 6

2-6


67. Answers may vary. Sample: ƒ(0) = –2; ƒ(1) = 0; ƒ(2) = 2; ƒ(–2) = 0; ƒ(–1) = –2

y = |2x + 1| – 3

68. Answers may vary. Samples: ƒ(0) = 1;

ƒ(1) = ; ƒ(3) = ;

ƒ(–1) = ; ƒ(–3) =

69. x –10

70. a > 4.5

89

23

109

43

71. b > –1.5

72. 150 • 75 • x = 281,250; 25 ft

2-6

<–


1. Graph y = |x| and y = |x + 4| + 3 on the same coordinate plane. Describe the translation that takes the graph of the parent function to the graph of the other function.

2. Write equations for the graphs obtained by translating y = |x| and y = –|x| as described.

a. 10 units right

b. 4 units down

c. 7 units left, 6 units up

Translate the parent graph 4 units left and 3 units up.

y = |x – 10|; y = –|x – 10|

y = |x| – 4; y = –|x| – 4

y = |x + 7| + 6; y = –|x + 7| +6

2-6

2-7

Two-Variable InequalitiesTwo-Variable InequalitiesALGEBRA 2 LESSON 2-7ALGEBRA 2 LESSON 2-7


Solve each inequality. Graph the solution on a number line.

Solve and graph each absolute value equation or inequality.

1. 12p 15 2. 4 + t > 17 3. 5 – 2t 11

4. |4c| = 18 5. |5 – b| = 3 6. |2h| 7

>–<–

>–


1. 12p 15

p

p 1.25

3. 5 – 2t 11 –2t 6 t –3

5. | 5 – b | = 35 – b = 3 or 5 – b = –3 –b = –2 or –b = – 8 b = 2 or b = 8

2. 4 + t > 17 t > 13

4. |4c| = 18 4c = 18 or 4c = –18 c = 4.5 or c = –4.5

6. |2h| 7

2h 7 or 2h –7

h or h –

h 3.5 or h –3.5

1512

72

72

Solutions

2-7

>–

<–

<–<–

>–<–

>–>–>–>–

<–<–<–

Step 1: Graph the boundary line y = x + 1. Since the inequality is greater than, not greater than or equal to, use a dashed boundary line.

32


Graph y > x + 1.32

Step 2: Since the inequality is greater than, y-values must be more than those on the boundary line. Shade the region above the boundary line.

2-7


A restaurant has only 15 eggs until more are delivered. An

order of scrambled eggs requires 2 eggs. An omelet requires 3

eggs. Write an inequality to model all possible combinations of

orders of scrambled eggs and omelets the restaurant can fill till

more eggs arrive. Graph the inequality.

2-7

Relate:plus

is less than or equal to

number of eggs needed for x orders of scrambled eggs

number of eggs needed for y orders

of omelets

15

Define: Let x = the number of orders for scrambled eggs.

Let y = the number of orders for omelets.

Write: 2 x + 3 y 1515<–


(continued)

Step 1: Find two points on the boundary line. Use the points to graph the boundary line.

when x = 0, 2(0) + 3y = 15

3y = 15

y = 5

when y = 0, 2x + 3(0) = 15

2x = 15

x = 152

Graph the points ( , 0) and (0, 5). Since the

inequality is less than or equal to, use a solid

boundary line.

152

2-7


(continued)

Step 2: Since the inequality is less than, y-values must be less than those on the boundary line. Shade the region below the boundary line.

All ordered pairs with whole-number coordinates in the shaded area and on the boundary line represent a combination of x orders of scrambled eggs and y orders of omelets that the restaurant could fill.

2-7


Graph y |2x| – 3.

Since the inequality is greater than or equal to, the boundary is solid and the shaded region is above the boundary.

2-7

>–


Write an inequality for each graph. The boundary line is given.

a. Boundary: y = | x – 2| – 1

The boundary line is dashed. The shaded region is above the boundary. This is the graph of y > |x – 2| – 1.

b. Boundary: y = – x + 312

2-7

The boundary is solid. The shaded region is below

the boundary. This is the graph of y – x + 3.12

<–



1.

2.

3.

4.

5.

6.

7.

8.

9.

2-7


10. a. y 20x if x 6,

y 15x if x > 6

b. Answers may vary. Sample:

11.

12.

13.

14.

15.

16.

2-7

>– <–>–


17.

18.

19.

20. y < –x – 2

21. 5x + 3y 9

22. 2y |2x + 6|

23.

24.

y > x + 225

25.

26.

27.

2-7

>–

<–


28.

29.

30.

31.

32.

33.

34.

35.

2-7

40. y –2x + 4

41. y |x + 2|

42. y < –|x – 4|

43. y > |x + 1| – 1

44. Answers may vary. Sample:

y – x +


36.

37.

38. x > –3

39. y x + 232

53

953

45. a. c + s

b. c + s 200

c.

Yes, (20, 50) is a solution.

d. 140 sundaes

46. Answers may vary. Sample: when it lies on the boundary line

2-7

>–

<–

<–

<–

<–

50.

51. A

52. I

53. D

54. F

55. C


47.

48.

49.

2-7


56. [2] y 300x, where y is the number of tornadoes that could occur in the next x years. The domain is all whole numbers greater than 0, since years are whole numbers and can’t be negative. The range is whole numbers 300,

since you cannot have a fraction of a tornado and in the first year you will have at least 300.

[1] includes only inequality with no explanation of domain and range

57.

58.

59.

60.

61.

2-7

>–

>–


62.

63. no

64. yes; 100

65. yes; –5

66. no

67. yes; 3

68. no

69. yes; –10

70. no

71. $7800

72.

73.

74.

2-7


2. –y + 1 > – | x + 3 |

1. y – x + 314

Graph each inequality.

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>–

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Functions, Equations and GraphsFunctions, Equations and GraphsALGEBRA 2 CHAPTER 2ALGEBRA 2 CHAPTER 2

1. domain {0, 1, 2, 3, 4}, range {–16, –9, –4, –1, 0}

2. domain {3, 4, 5, 6, 7}, range {2, 3, 4, 5, 6}

3. domain {–2, 1, 3, 4 }

range {1, 2, 3, 4, 5}

12

4. domain {– , 0.3, 5}

range {–0.01, 0, 2.5}

5. 1

6. 4

7. 1

8. 12

9. –17

12


10. 22


12. 2

13. –

14.

15. 3x + y = 0

16. 2x – 5y = –23

17. 4x – y = –3

18. x + 2y = 12


y = – x

20. Answers may vary. Sample: y + 6 = –16(x + 1)


y = (x – 3)

22. Answers may vary. Sample: y – 2 = 3(x – 8)

43

12

12


a. y = –x

b. y = x + 15

c. y = x

d. y = –x – 5e. rectangle

24. 8; –4

25. ; –

26. 4.2 min

74

23

13

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27.

28.

29.

30.

31. a–b.

Answers may vary. Sample y = 222x + 1125, where 1990 corr. to x = 0

c. Answers may vary. Sample: $4011

d. Check students’ work.

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32. vertical

33. diagonal

34. diagonal

35. vertical

36. vertical

37. diagonal

38.

39.

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40.

41.

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