A6© 2014 College Board. All rights reserved. SpringBoard Algebra 1, Unit 2 Practice

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 1

31

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

A7© 2014 College Board. All rights reserved. SpringBoard Algebra 1, Unit 2 Practice

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 x12

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