Appendix A — Answer Key, Chapter 16: Quadratic Functions

SOL A.F.2 (b, c, d, g) · Covers textbook Chapter 16 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 126 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below: an intercept is a point; a zero is a number. The vertex is a minimum when $a > 0$ and a maximum when $a < 0$. Domains and ranges use words or inequalities (no interval notation required). Transformations are limited to $f(x) + k$ and $kf(x)$ with rational $k$. When solving $f(x) = c$ in context, reject a physically meaningless input with a stated reason.

The figures used repeatedly in the chapter, for reference:

• Figure 1 is $y = x^2 - 4x - 5$ with vertex $(2, -9)$ and axis $x = 2$
• Figure 2 is $y = x^2 - 2x - 3$ with zeros $-1$ and $3$, $y$-intercept $(0, -3)$, vertex $(1, -4)$, range $y \ge -4$
• Figure 3 pairs $y = x^2 - 4$ (minimum $(0, -4)$) with $y = -x^2 + 4$ (maximum $(0, 4)$)
• Figure 4 is $h = -16t^2 + 48t$ with context domain $0 \le t \le 3$ and range $0 \le h \le 36$
• Figure 5 is the parent $y = x^2$
• Figure 6 shows vertical shifts $y = x^2 + 3$ and $y = x^2 - 4$ of the parent
• Figure 7 shows $y = 2x^2$, $y = \tfrac12 x^2$, and $y = -x^2$ as $kf(x)$ of the parent
• Figure 8 is $y = x^2 - 6x + 5 = (x - 1)(x - 5)$
• Figure 9 is $f(x) = x^2 - 4x + 3$ with $f(5) = 8$ and $f(x) = 3$ at $x = 0$ and $x = 4$
• Figure 10 is four blank practice grids
• Figure 11 is the standard-vs-factored feature reference

Lesson 16.1 — Vertex, Axis, and Opening

Guided practice

1. $y = x^2 - 4x - 5$; vertex $(2, -9)$; axis $x = 2$.
2. Opens up, because $a = 1 > 0$.
3. Minimum; $(0, -4)$.
4. Maximum; $(0, 4)$.
5. Vertex $(0, 0)$; axis $x = 0$; domain all real numbers; range $y \ge 0$.
6. The parabola is symmetric left-right about its turning point, so the mirror line is the vertical line through the vertex.

Independent practice

7. a) Opens up; minimum. b) Opens down; maximum. c) Opens up; minimum. d) Opens down; maximum.
8. $(2, 3)$. The axis is $x = 4$; $6$ is $2$ units right of $4$, so the mirror is $2$ units left at $x = 2$, same output $3$.
9. U-shaped curve through $(0, 0)$, $(\pm 1, 1)$, $(\pm 2, 4)$; vertex marked at origin; axis the $y$-axis.
10. The defining equation is $y = ax^2 + bx + c$, which assigns one output to each input. A sideways parabola would fail the vertical-line test and would not be a function of $x$.
11. The student confused "vertex" with "minimum." Because $a < 0$, the vertex is a maximum of $4$.
12. $f(0) = -5$ and $f(4) = -5$, so $(0, -5)$ and $(4, -5)$ are mirrors across $x = 2$.
13. Maximum, because $a = -16 < 0$ (opens down).
14. $y = -x^2$ is the parent reflected across the $x$-axis. Both have vertex $(0, 0)$; one is a min, the other a max of the same height $0$.
15. Any window containing $(2, -9)$ works (e.g. $-6 \le x \le 8$, $-12 \le y \le 10$). The min/vertex feature should report $(2, -9)$.
16. If $a = 0$, there is no $x^2$ term and the graph is a line (or constant), not a parabola.

Exit ticket 16.1

17. Opens up; minimum.
18. Axis $x = -2$; $5$ is a maximum value.
19. Vertex $(0, 0)$; axis $x = 0$.
20. It is the vertical mirror line: points the same distance left and right of the axis share the same output.

Lesson 16.2 — Intercepts, Zeros, and the Vertex Formula

Guided practice

21. Zeros $-1$ and $3$; $y$-intercept $(0, -3)$; vertex $(1, -4)$; domain all real numbers; range $y \ge -4$.
22. $h = -\dfrac{-2}{2(1)} = 1$; $k = 1^2 - 2(1) - 3 = -4$.
23. $(0, -8)$.
24. Zeros $2$ and $3$; $x$-intercepts $(2, 0)$ and $(3, 0)$.
25. $h = 3$, $k = 9 - 18 + 5 = -4$; vertex $(3, -4)$, a minimum.
26. A zero is the number $r$ with $f(r) = 0$; an $x$-intercept is the point $(r, 0)$. For this function the zeros are $-1$ and $3$, and the $x$-intercepts are $(-1, 0)$ and $(3, 0)$.

Independent practice

27. a) $(0, -2)$ b) $(0, 11)$ c) $(0, 0)$
28. a) Zeros $-3$, $3$; intercepts $(-3, 0)$, $(3, 0)$. b) Zeros $-5$, $3$; intercepts $(-5, 0)$, $(3, 0)$. c) Zeros $4$, $-1$; intercepts $(4, 0)$, $(-1, 0)$.
29. a) Axis $x = 2$; vertex $(2, -3)$; minimum. (Zeros $2 \pm \sqrt{3}$ if asked.) b) Axis $x = 2$; vertex $(2, 5)$; maximum. c) Axis $x = -3$; vertex $(-3, 1)$; minimum. No real zeros (discriminant $36 - 40 = -4 < 0$).
30. $(x - 3)(x + 1) = 0$ gives zeros $-1$, $3$; $f(0) = -3$; $h = 1$, $k = -4$. All match the figure.
31. Axis $x = \dfrac{-2 + 6}{2} = 2$. The axis is the midpoint of the zeros.
32. The student used $h = \dfrac{b}{2a}$ (or dropped the minus on $-b$). Correct: $h = -\dfrac{-4}{2} = 2$.
33. $y$-intercept $(0, 0)$: selling $0$ items produces $\$0$ revenue. Vertex $(20, 400)$: maximum revenue is $\$400$ when $20$ items are sold.
34. Zeros $-2$, $2$; $y$-intercept $(0, 4)$; vertex $(0, 4)$ (maximum); domain all reals; range $y \le 4$.
35. Zero feature: $x = -1$ and $x = 3$. Minimum feature: $(1, -4)$.
36. Yes. $h = -\dfrac{-4}{2} = 2$, $k = 4 - 8 - 5 = -9$; $f(0) = -5$. Both claims hold.

Exit ticket 16.2

37. $y$-intercept $(0, -6)$; zeros $-2$ and $3$.
38. $h = \tfrac12$, $k = -\tfrac{25}{4}$; vertex $\left(\tfrac12, -\tfrac{25}{4}\right)$, a minimum.
39. $x = 1$ (midpoint of $-3$ and $5$).
40. The number $h = -\dfrac{b}{2a}$ is the $x$-coordinate of the vertex, so the axis is the vertical line $x = h$.

Lesson 16.3 — Domain and Range, Including Context

Guided practice

41. Domain: all real numbers. Range: $y \ge -4$.
42. Domain: all real numbers. Range: $y \le 4$.
43. Domain: $0 \le t \le 3$ seconds. Range: $0 \le h \le 36$ feet.
44. $(0, 0)$: thrown from ground at $t = 0$. $(1.5, 36)$: maximum height $36$ ft at $1.5$ s. $(3, 0)$: lands at $t = 3$ s.
45. The algebraic rule accepts every real $t$; the story only models the flight from throw to landing, so $t$ is restricted to $0$ through $3$.
46. Domain: all real numbers. Range: $y \ge 0$.

Independent practice

47. a) Domain all reals; range $y \ge -4$ (vertex $(-3, -4)$). b) Domain all reals; range $y \le 8$ (vertex $(2, 8)$). c) Domain all reals; range $y \ge -9$ (vertex $(0, -9)$). d) Domain all reals; range $y \le -1$ (vertex $(0, -1)$).
48. Domain $0 \le t \le 3$ seconds is the flight time. Range $0 \le h \le 36$ feet is every height the ball reaches. The vertex $(1.5, 36)$ is the highest point of the flight.
49. Vertex $(3, 9)$. Domain in context: $0 \le x \le 6$ meters. Range in context: $0 \le y \le 9$ meters.
50. An upward parabola has a lowest output $k$ at the vertex; every output below $k$ is impossible, so the range cannot be all real numbers.
51. The function is not a height model; its vertex output is $-4$, so the range is $y \ge -4$, which correctly includes negative values.
52. Vertex $(20, 400)$. Domain in context: $0 \le x \le 40$ items. Range in context: $0 \le R \le 400$ dollars.
53. Unrestricted: domain all reals; range $y \ge -4$. Restricted to $-2 \le x \le 2$: domain $-2 \le x \le 2$; range $-4 \le y \le 0$ (minimum $-4$ at $x = 0$; value $0$ at the endpoints).
54. Window showing $0 \le t \le 3$ and $0 \le h \le 40$; maximum feature near $(1.5, 36)$.
55. Downward parabola through $(\pm 3, 0)$ and $(0, 9)$; shade $y \le 9$ (or mark the range along the $y$-axis).
56. The equation is unchanged, but the inputs the story allows are a subset of the real numbers — times after a throw, widths of a garden, and so on — so the domain (and often the range) shrinks to match the story.

Exit ticket 16.3

57. Domain all reals; range $y \ge -4$ (vertex $(4, -4)$).
58. Domain all reals; range $y \le 5$ (vertex $(0, 5)$).
59. $0 \le t \le 3$: $t = 0$ is the throw; $t = 3$ is the landing.
60. For every real input, $ax^2 + bx + c$ produces a real output; nothing in the formula forbids any real $x$.

Lesson 16.4 — Transformations $f(x) + k$ and $kf(x)$

Guided practice

61. $y = x^2 + 3$: shift up $3$; vertex $(0, 3)$. $y = x^2 - 4$: shift down $4$; vertex $(0, -4)$.
62. $2x^2$: stretch. $\tfrac12 x^2$: shrink. $-x^2$: reflect.
63. Reflect across the $x$-axis and stretch by $2$ (multiply outputs by $-2$).
64. Shrink by $\tfrac12$, then shift down $3$.
65. Both share axis $x = 0$; vertices at $(0, 2)$ (min) and $(0, 2)$ (max) respectively — upward and downward U's meeting at height $2$ on the $y$-axis.
66. $y = (x - 3)^2$ moves the axis to $x = 3$. A.F.2c only names $f(x) + k$ and $kf(x)$, which keep the axis at $x = 0$. Horizontal shifts are outside this chapter's transformation list.

Independent practice

67. a) Shift down $7$; vertex $(0, -7)$. b) Stretch by $4$; vertex $(0, 0)$. c) Reflect, then shift up $1$; vertex $(0, 1)$, maximum. d) Shrink by $\tfrac13$, then shift up $2$; vertex $(0, 2)$.
68. Sketches: (a) parent up $4$; (b) narrow downward through origin; (c) wide upward with vertex $(0, -2)$.
69. Compare $|k|$ to $1$: $|k| > 1$ is narrower (stretch); $0 < |k| < 1$ is wider (shrink).
70. $y = x^2 - 4x$ is not a vertical shift of the parent; the axis is not $x = 0$. Correct vertex: $h = 2$, $k = -4$, so $(2, -4)$.
71. Wider (shrink by $\tfrac14$). Vertex $(0, 0)$.
72. Vertex $(2, -9)$; zeros $-1$ and $5$; $y$-intercept $(0, -5)$ — matches figure 1.
73. Reflect, then shift up $4$. Range $y \le 4$.
74. $2x^2$ is narrower than the parent; $\tfrac12 x^2$ is wider; all three share the vertex $(0, 0)$.
75. $y = -x^2 - 6$.
76. Vertex $(0, -3)$; $y$-intercept $(0, -3)$; narrower than the parent, opens up.

Exit ticket 16.4

77. Shift down $5$.
78. Multiply outputs by $\tfrac12$ (shrink); wider.
79. $(0, 9)$; maximum.
80. Those two moves keep the axis at $x = 0$, but $y = x^2 - 6x + 5$ has axis $x = 3$.

Lesson 16.5 — Standard Form, Factored Form, and the Graph

Guided practice

81. Standard $y = x^2 - 6x + 5$; factored $y = (x - 1)(x - 5)$; zeros $1$ and $5$; $y$-intercept $(0, 5)$.
82. $x^2 - 6x + 5$.
83. Standard: $y$-intercept $(0, c)$ (and opening from $a$). Factored: zeros $r$ and $s$ (and opening from $a$).
84. Zeros $-2$ and $4$; axis $x = 1$; $y$-intercept $(0, -8)$; vertex $(1, -9)$.
85. $y = (x + 5)(x - 3)$; zeros $-5$ and $3$.
86. Expanding turns factored form into standard form; factoring turns standard form into factored form. Both name the same outputs for every input, so they name the same graph.

Independent practice

87. a) Zeros $2$, $6$; axis $x = 4$; $y$-int $(0, 12)$; vertex $(4, -4)$. b) Zeros $-3$, $-1$; axis $x = -2$; $y$-int $(0, 3)$; vertex $(-2, -1)$. c) Zeros $1$, $5$; axis $x = 3$; $y$-int $(0, -5)$; vertex $(3, 4)$ (maximum).
88. a) $(x - 2)(x - 3)$ b) $(x - 3)(x + 3)$ c) $(x + 4)(x - 3)$
89. With $a = 1$: $y = (x + 4)(x - 2) = x^2 + 2x - 8$, and $f(0) = -8$. Matches; no adjustment needed. (Accept $y = (x + 4)(x - 2)$.)
90. Expand: $(x - 6)(x + 5) = x^2 - x - 30$, not $x^2 - 6x + 5$. Correct: $(x - 1)(x - 5)$.
91. $A(x) = -x^2 + 10x$. Vertex $(5, 25)$: maximum area is $25$ square meters when the width is $5$ meters (a square).
92. $y = (x - 3)(x + 1)$; zeros $-1$ and $3$ — yes, matches figure 2.
93. Same zeros $1$ and $5$. Vertex $y$-coordinate is doubled: $k = 2(3 - 1)(3 - 5) = 2(-4) = -8$, so vertex $(3, -8)$.
94. The two graphs coincide for every visible $x$.
95. $y = x^2 - 3x - 4$ (or $y = (x + 1)(x - 4)$).
96. Zeros $(-2, 0)$, $(4, 0)$; $y$-int $(0, -8)$; vertex $(1, -9)$.

Exit ticket 16.5

97. $y = (x - 1)(x - 5)$.
98. Zeros $1$ and $5$; axis $x = 3$.
99. $(3, -4)$.
100. Standard form; factored form.

Lesson 16.6 — Evaluating $f(x)$ and Finding $x$

Guided practice

101. $f(5) = 8$ from the graph; $25 - 20 + 3 = 8$ by substitution.
102. $x = 0$ or $x = 4$.
103. $f(2) = -1$; $f(1) = 0$.
104. $x = 1$ or $x = 3$. Those are exactly the $x$-coordinates of the $x$-intercepts.
105. $h(1.5) = 36$. At $1.5$ seconds, the ball is at its maximum height of $36$ feet.
106. A horizontal line $y = c$ can cross a parabola at two points (except at the vertex, where it touches once, or below/above the vertex, where it misses).

Independent practice

107. a) $f(0) = 5$; $f(3) = -4$; $f(7) = 12$. b) $x = 0$ or $x = 6$. c) $x = 3$ (only the vertex).
108. a) $g(0) = 4$; $g(1) = 3$. b) $x = -2$ or $x = 2$. c) $x = -1$ or $x = 1$.
109. $h(0.5) = 20$: at half a second, the height is $20$ feet. $h(t) = 0$ at $t = 0$ and $t = 3$: thrown and landing times.
110. $R(10) = 300$: selling $10$ items produces $\$300$ revenue. $R(x) = 300$ at $x = 10$ and $x = 30$: two sales levels give the same revenue.
111. The minimum value is $f(2) = -1$, so the range is $y \ge -1$. The number $-5$ is not in the range.
112. The student found only one of the two inputs on the horizontal line $y = 3$. The complete solution is $x = 0$ or $x = 4$.
113. $x = -1$ or $x = 5$. (Figure 9 confirms $f(5) = 8$.)
114. Table/trace should show $f(5) = 8$ and outputs of $3$ at $x = 0$ and $x = 4$.
115. $h(0) = 48$: thrown from $48$ feet. $h(t) = 48$ gives $t = 0$ or $t = 2$. Keep both as times when height is $48$ ft (start, and again on the way down), or note $t = 0$ is the launch — neither is rejected for being negative.
116. Parabola through $(1, 0)$, $(3, 0)$, $(2, -1)$; horizontal line $y = 3$ meeting at $(0, 3)$ and $(4, 3)$.

Exit ticket 16.6

117. $f(6) = 15$.
118. $x = 0$ or $x = 4$.
119. After $1$ second, the ball's height is $32$ feet.
120. A non-horizontal line meets a horizontal line $y = c$ at most once; a parabola can meet it twice because it turns around at the vertex.

Chapter 16 Review

Part A — A.F.2b

121. Zeros $-1$, $3$; $y$-intercept $(0, -3)$; vertex $(1, -4)$ (minimum); domain all real numbers; range $y \ge -4$.
122. Domain $0 \le t \le 3$ seconds; range $0 \le h \le 36$ feet. Vertex $(1.5, 36)$: maximum height $36$ feet at $1.5$ seconds after the throw.

Part B — A.F.2c

123. Reflect and stretch by $2$, then shift up $3$; vertex $(0, 3)$, a maximum.
124. $\tfrac12 x^2 - 4$: shrink by $\tfrac12$, then shift down $4$ (transformations). $x^2 - 4x - 5$: vertex $(2, -9)$ and intercepts (characteristics) — transformations alone cannot move the axis off $x = 0$.

Part C — A.F.2d

125. $y = (x - 1)(x - 5)$. Zeros $1$ and $5$ from either form; $y$-intercept $(0, 5)$ from either form — matches figure 8.

Part D — A.F.2g

126. $f(5) = 8$; $f(x) = 3$ at $x = 0$ or $x = 4$. For $h(t) = 32$: $t = 1$ s (on the way up) and $t = 2$ s (on the way down) are the two times the ball is $32$ feet high.