Answers to Algebra 1 Unit 2 Practice

A1© 2014 College Board. All rights reserved. SpringBoard Algebra 1, Unit 2 Practice

1. {(1, 0), (3, 0), (25, 2)}

2. A

3. The input 0 is paired with 2 outputs, 3 and 29.

4. Change one of the 0s in the table to a number other than 18 or 12. Then each input will be paired with exactly one output.

5. {(B, 2), (C, 4), (A, 5), (B, 6)}

6. a. Domain: {15, 3, 212, 21, 9}; Range: {2, 0, 21, 2}

b. Domain: {25, 1, 3}; Range: {22, 1}

7. The domain is whole numbers because you can purchase neither a negative number of bagels nor partial bagels; the range includes $0 and positive multiples of $0.90.

8. B

9. a. Answers may vary. Sample answer:

x

y

1

12345

25242322212122232425 2 3 4 50

b. The domain and range values can be paired in 6 different ways, so there are 6 possible functions:

{(21, 25), (0, 2), (3, 4)} {(21, 25), (0, 4), (3, 2)}

{(21, 2), (0, 25), (3, 4)} {(21, 2), (0, 4), (3, 25)}

{(21, 4), (0, 2), (3, 25)} {(21, 4), (0, 25), (3, 2)}

10. a. Answers will vary.

21

6

12

23

8

b. No; if there are fewer elements in the domain than in the range, then at least one domain value must be paired with two range values, which cannot happen in a function.

11. D

12. x represents the number of movies Manny rents; f(x) represents the total cost for the movies.

13. f(23) 5 21; f(0) 5 5; f12

5 6; f(10) 5 25

14. Hudson evaluated the function at x 5 17, instead of finding the value of x for which f(x) 5 17. The correct answer is 6.

15. Answers will vary. Students should write a sequence of numbers whose third term is

14

.

16. There are no breaks or gaps in the line.

17. B

18. a. Domain: {x | 0 # x # 60}; range: {y | 0 # y # 400}

b. The domain for the second balloon is {x | 0 # x # 65} because the ride lasted 65 minutes instead of 60; the ranges are the same because neither balloon ever went higher than the other.

19. a. The y-intercept is (0, 0); it represents the height at the start of the ride (when time 5 0).

b. No; all balloon rides must start from the ground, where height 5 0.

20. The point where the graph intersects the y-axis is the y-intercept.

21. D

Answers to Algebra 1 Unit 2 Practice


22. There are no relative or absolute minima.

23. Answers may vary. All values of x correspond to a point on the graph.

24. No; the domain must be restricted so that the denominator is not 0. The correct domain is {x | x 5 4}.

25. a. The independent variable is x, the number of hours. The dependent variable is the total cost.

b. The reasonable domain is {x | x $ 0} because a negative number of hours does not make sense; the reasonable range is {y | y $ 0} because a negative cost does not make sense.

26. C

27. The function is discrete, and the least value for x is 0 and the greatest is 5; therefore, all real numbers greater than 0 cannot be the domain.

28. Answers will vary. Jason had a box containing 5 granola bars. Each day he ate one bar.

29. C

30. d 5 2w

31. The reasonable domain is {w | w $ 0} because a weight cannot be negative; the reasonable range is {d | d $ 0} because a distance cannot be negative.

32.

x

y

5

1015

25

5

20

3035

Spr

ing

Str

etch

(cm

)

4045

5550

60

10 15 20 25Weight (oz)

30 35 40 45 6055500

y 5 2x

33. Kathleen is incorrect. To check using the equation, substitute 25 for w: d 5 2(25) 5 52 inches. To check using the graph, find the point on the graph whose x-coordinate is 25. The y-coordinate of this point, 50, is the correct length of the stretch.

34. Time (seconds) Height (ft)

0 12001 11842 11363 10564 9445 8006 6247 4168 1769 –96

The object reaches the ground when the height is 0; the table shows that this happens between 8 and 9 seconds. 35. B

36. y

x1

200300

100

400500600700

Hei

ght

(ft)

800900

100011001200

2 3 4 5Time (seconds)

6 7 8 9 100

y 5 1200216x2

Explanations may vary. The object falls for a length of time between 8 and 9 seconds, so viewing the x-axis from 0 to 10 is appropriate. The object starts at 1200 feet and falls to the ground (0 feet), so view the y-axis from 0 to 1200.

37. (0, 1200); the height of the object at time 5 0 (before it is dropped)


38. When the object reaches the ground, its height will be 0. Find the point on the graph that has y-coordinate 0; use the trace or intersect features of the calculator to find that the x-coordinate of this point is about 8.66. The object reaches the ground in about 8.66 seconds. To check using the equation, substitute 8.66 for t: 1200 2 16(8.66)2 5 0.704 ≈ 0.

39. Years Since

PurchaseValue of Computer

0 $5401 $3602 $2403 $160

40. B

41. Explanations may vary. One-half of any amount greater than 0 will always be greater than 0.

42. Graph A; explanations may vary. Graph B shows an amount that is increasing, not decaying, over time.

43. 10 grams; the amount is the y-coordinate when time 5 2s; the line x 5 2 intersects the graph at (10, 2).

44. translation 4 units down

45. a. the y-coordinate of the y-intercept of g(x)

b. g(2)

c. f(4)

46. B

47. No; the graph of g(x) 5 x2 1 3 would be 3 units above the graph of f(x) 5 x2.

48. a. b x x( ) 5.25 45 1

b. a x x( ) 5.25 55 1

c. The graph of a(x) is one unit above the graph of b(x), because a(x) 5 b(x) 1 1.

d. The graph will be translated down by the amount of the reduction.

49. The graph of g(x) is translated 1 unit to the left and 12 units up from the graph of f(x).

50. C

51. Answers may vary.

a. g(x) 5 x2 2 5

b. g(x) 5 (x 2 3)2

c. g(x) 5 (x 2 6)2 2 9

d. g(x) 5 (x 1 2)2 1 1

To check by graphing, graph the functions above and check that each vertex matches the vertex given in the problem.

52. (5, 4); the graph of g(x) will be translated 5 units to the right from the graph of f(x). The graph of f(x) has vertex (0, 4) so the graph of g(x) will have vertex (5, 4).

53. a. f(x) 5 5x 1 8.50

b. Substitute x 2 3 for x in the function from part a: f(x 2 3) 5 5(x 2 3) 1 8.50

c. The graph of the function in part b is translated 3 units to the right from the graph of the function in part a.

54. B

55. 3

56. Answers will vary. (0, 0) and (1, 2);

slope 5 2 01 0

2

2 5 2.

57. a. –2

b. Answers will vary; y 5 4. If y 5 4, then the

slope of line b is 4 0

3 ( 1)2

2 2 5 1 . 22.

58. Yes; the value of y increases by 4 as x increases by 1.

59. a. f(x) 5 75x 1 250

b. Tables may vary.

Months Total Amount

0 $2501 $3252 $4003 $475


y

x1

200300

100

400500600700

Tota

l Am

ount

($) 800

9001000

2 3 4 5Months

6 7 8 9 100

y 5 75x 1250

c. $75/mo

d. The rate of change is $75/mo, which is the amount that Carlos pays each month. It is the slope of the graph and the coefficient of x in the equation.

60. B

61. Answers will vary. An elevator descends at a rate of 15 feet per second.

62. a. Positive; the slope is 10.

b. Negative; the slope is 26.

c. Positive; the slope is 60.

63. B

64. Explanations may vary. The x-coordinates for every point on a vertical line are equal. This means that the denominator in the slope formula will be 0. Division by 0 is undefined, so the slope is undefined, not 0.

65. Answers will vary; (21, 5). The slope of the line

containing the points (22, 0) and (4, 10) is 10 04 ( 2)

106

53

2

2 25 5 . Find another point (x, y) for

which the slope is 53

. Using the slope formula and

the point (22, 0), y

x0

( 2)2

2 2 5

53

; y 5 5 and x 5 21

is one possible solution.

66. 6; use any two ordered pairs in the table to find

that the slope is 22. If y is the missing y-value,

substitute (23, 12) and (0, y) into the slope

formula: y 120 ( 3)

2

2 2 5 22; solve this equation to

find y = 6.

67. B

68. y 5 x32

, or y 5 1.5x

69. a. Yes; explanations may vary. The cost of 0 bags of soil is $0 and each bag of soil costs the same amount, so the equation can be written in the form y = kx, where y is the total cost and x is the number of bags purchased. Three bags cost

$3.75, so k 5 3.753

5 $1.25.

b. $8.75

70. a. Answers will vary. Students should write a linear equation that cannot be written in the form y 5 kx, such as y 5 3x 1 2.

b. Answers will vary. Students should graph a line that does not pass through (0, 0).

71. Explanations may vary. Although the equation is not in the form y = kx, it can be written in that form by solving for y: y 5 2x.

72. C

73. yx40

5

y

x1

10

20

30

Tim

e (h

ours

)

40

50

2 3 4 5Painters

6 7 8 9 100


74. Color Your World; with 7 painters, Color Your World could paint walls in about 5.7 hours, which is faster than the 6 hours Good Hues would require.

75. x = 12

76. a.

x y 5 x120

y 5 x160

y 5 x192

1 120 160 1922 60 80 964 30 40 488 15 20 24

b. When x is doubled, y is divided in half.

77. a. p(x) 5 15x 2 450

y

x

100

2100

2200

2300

2400

2500

200

300

Prof

it ($

)

400

500

4 8 12 16 20 24 28 32 36 40 44 480

Hours Worked

y 5 15x 2450

b. (0, 2450); it represents Erin’s profit (2$450) when she has worked 0 hours.

c. 20 jobs; use the equation or the graph to find that Erin must work 30 hours to break even. If a job takes 1.5 hours, then Erin must work 301.5

5 20 jobs to break even.

d. Answers may vary. Erin could double her hourly rate to $30. Or Erin could keep her hourly rate at $15 but shop around for less expensive supplies; if she can reduce her initial costs to $225, she will break even after 10 jobs.

78. f(x) 5 2x 1 25

79. B

80. a. Answers will vary. Students may recommend that Mariah not participate, because she would have to donate $2 for each item, and if each item sells for $3, Mariah will be donating most of what she earns for each item.

b. Answers will vary. Knowing the amount it costs Mariah to make each magnet would be helpful, because you could then determine whether she will lose money by donating $2 per magnet. If she still profits by making $1 per magnet, then she may want to participate in the fair. Students may also want to know whether Mariah wants to advertise her products; if so, she may want to participate regardless of profits.

81. B

82. No. Use the equation to show that there are y-values that correspond to more than one x-value; for example, both f(2) and f(22) are equal to 16. The graph fails the horizontal line test.

83. a. f 21 (x) 5 x

9.25 b. The total amount spent on tickets

c. 5

84. a. {(January, 31), (February, 28), (March, 31), (April, 30), (May, 31), (June, 30), (July, 31), (August, 31), (September, 30), (October, 31), (November, 30), (December, 31)}

b. No; the function is not one-to-one. For example, April and June are both paired with 30.

c. Answers will vary. A function that pairs each person with their Social Security Number is one-to-one, because each person has a unique Social Security Number.

85. No; if two functions are inverses, then they undo each other. Because f(0) 5 3, if g(x) were its inverse then g(3) would be 0. However, g(3) 5 21. Therefore the functions are not inverses.

86. B

87. No; in an arithmetic sequence, each term is found by adding, a common difference to the term before, not multiplying.


88. a. 26

b. 12

c. 10

89. Yes; the common difference is 0.

90. Sequence 1

91. Sequence 3

92. D

93. an 5 3 1 2(n 2 1), or 2n 1 1; an is the number of tiles in the nth stage and d 5 2 is the number of tiles added to each stage.

94. If a4 5 11 and a9 5 36, then 11 1 5d 5 36. Solve this equation to find d 5 5. Then a1 1 3(5) 5 11; solve this equation to find a1 5 24. The explicit formula is an 5 24 1 5(n 2 1), or 5n 2 9.

y

x1

105

15

25

35

45

25

20

30

40

50

2 3 4 5 6 7 8 9 100

95. f(n) 5 5n 2 8

96. f(2) 5 0; f(11) 5 18

97. C

98. a. Yes; a graph of the sequence includes the points (3, 212) and (7, 0). Luisa knows that the slope of the graph is the common difference of the sequence, so she is using slope to find the value of d.

b. f (n) 5 3n 2 21

99. a. f –1(n) 5 n 131

b. The inputs are terms in the sequence; the outputs are the corresponding term numbers.

c. Answers will vary. The value 56 corresponds to what term in the sequence? Substitute 56 into the inverse function:

f –1(56) 5 56 1

31

5 19. 56 is the value of the 19th term in the sequence.

100. D

101. a.

a

a a

12

34n n

1

1

5

5 12

;

f

f n f n

(1) 12

( ) ( 1) 34

5

5 2 1

b.

aa a

32n n

1

1

52

5 12

;

ff n f n(1) 3( ) ( 1) 2

52

5 2 1

c.

aa a

54n n

1

1

5

5 12

;

ff n f n(1) 5( ) ( 1) 4

5

5 2 1

d.

aa a

86n n

1

1

52

5 12

;

ff n f n(1) 8( ) ( 1) 6

52

5 2 1

102. No; explanations may vary. The first 4 terms of this sequence are 1, 8, 22, and 48. There is no common difference between consecutive terms, so the sequence is not arithmetic.

103. Answers may vary. The first term must be defined to determine where the sequence begins. For example, an 5 an21 2 3 indicates a sequence with a common difference of 23, but there are infinitely many such sequences; two examples are 0, 23, 26, 29, … and 4, 1, 22, 25, … . Without knowing the first term, it is impossible to determine the sequence.

104. D

105. y 5 24x 1 1

106. y x12

152 2


107. a. y 5 50x 1 200

b. 9 months; find the value of x for which f(x) 5 650. Because 50(9) 1 200 5 650, x 5 9.

c. The equations will have the same constant term, 200, and the graphs will have the same y-intercept, (0, 200), because both Laura and her brother began with $200. The equations will have different coefficients for x and the graphs will have different slopes because each person deposits a different amount of money each month.

108. From the given equation, the y-intercept is (0, 25). Use this point and (22, 3) to find the slope. The equation is y 5 24x 2 5.

109. y 1 1 5 15(x 2 3)

110. Answers may vary; y 2 1 5 0.5(x 2 3)

111. Infinitely many; any point on the line can be substituted into the point-slope form.

112. One; the line has only one slope and one y-intercept, so only one equation can be written in slope-intercept form.

113. C

114. You cannot enter an equation in point-slope form into a graphing calculator because the equation is not solved for y. To enter an equation in point-slope form into a calculator, you must first solve the equation for y. For example, the equation y 2 2 5 23(x 1 5) can be entered into a graphing calculator as y 5 23(x 1 5) 1 2, or y 5 23x 2 13.

115. x-intercept 5 (27, 0); y-intercept 5 (0, 24); slope 5�

47

116. C

117. Chase has found the equation of the vertical line through (3, 5), not the horizontal line. The correct equation is 0x 1 y 5 5, or y 5 5.

118. h, g, e, d, b, a, f, c

119. 4x 1 3.75y 5 50

120. B

121. a 5 4; explanations may vary. The graph

of 2x 1 3y 5 6 has slope �23

. A line that is

perpendicular will have slope 32

. The graph of

ay 5 6x 1 10 has slope a6

; for this to be equal to 32

,

a must equal 4.

122. a. Answers will vary; y 2 x195

b. Yes; any line with slope 195

(except y 5 x195

72 ,

the equation of the given line) is a possible

answer, and there are infinitely many such lines.

Another example is y 5 x195

21 .

123. a. y x1212( 1)2 52 1 , or y 5 x

12

232

2 1

b. No; a given point and slope define exactly one line.

124. C

125. The finishing times decrease as the months of training increase.

126. y

x1

20

40

60

80

100

2 3 4 5 6 7 80


127. Trend lines and equations may vary.y

x1

20

40

60

80

2 3 4 5 6 7 80

Using the points (1, 65) and (6, 40), the equation is y 5 25x 1 70.

128. Yes; the point (8, 30) lies near or on the trend line.

129. D

130. a. There is a positive correlation. As the temperature increases, the number of people at the pool increases.

b. Yes; it is reasonable to assume that higher temperatures cause more people to visit the pool.

131. a. y 5 11.05x 2 798.62

b. 30

132. Answers will vary; the number of people at the public pool and the number of frozen yogurt cups sold at a frozen yogurt shop across town. It is reasonable to expect that there might be a positive correlation between these two variables, but it is probably not the case that an increase in one variable causes an increase in the other. Rather, increases in these variables are both most likely due to an increase in temperature.

133. B

134. Yes; an increase in the age of the painting causes the value of the painting to increase.

135. a. y 5 0.21 x2 1 12.07x 1 660; $1788.50

b. y 5 653.44 (1.02)x ; $1758.79

Answers to Algebra 1 Unit 2 Practice

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