Everything below prints as one AP AP Precalculus practice paper set: papers A & B with their answer keys, plus the full-length study package exam. Use the Download PDF / Print button (or Cmd/Ctrl+P) to save it.

Paper A

AP Precalculus — Practice Paper A

Original unofficial practice questions · paper A · answer key on the last page

Total time: see section headers · No guessing penalty

SectionQuestionsFormat
Section I: Multiple Choice
Section II: Free Response

Section I — Multiple Choice

1.

The unit circle coordinate of the point at angle π/2 is

A. (0, 1)B. (1, 0)C. (-1, 0)D. (0, -1)
Answer:
2.

If f(x) = 2x + 3, then f⁻¹(7) =

A. 2B. 17C. 5D. 7
Answer:
3.

The amplitude of y = 3 sin(2x) is

A. 3B. 2C. 6D. 1
Answer:
4.

log₂(8) =

A. 3B. 2C. 8D. 16
Answer:
5.

The slope of the line through (1,2) and (3,6) is

A. 2B. 1C. 4D. 0.5
Answer:
6.

A polynomial of degree 3 has at most

A. 3 rootsB. 2 rootsC. 1 rootD. infinite roots
Answer:
7.

The midline of y = cos(x) + 2 is

A. y = 2B. y = 1C. y = 0D. y = -2
Answer:
8.

The domain of f(x) = √x is

A. x ≥ 0B. x > 0C. all realsD. x ≠ 0
Answer:
9.

Which angle is coterminal with 30°?

A. 390°B. 150°C. 60°D. 210°
Answer:
10.

The vertex of y = (x - 3)² + 1 is

A. (3, 1)B. (-3, 1)C. (3, -1)D. (1, 3)
Answer:

Section II — Free Response

1.

A ball thrown vertically has height h(t) = -16t² + 64t. (a) Find the vertex. (b) Interpret its meaning.

5 points · rubric: Vertex math 3 pts, interpretation 2 pts.

2.

Sketch and describe the transformations that map y = x² to y = -½(x-1)² + 4.

4 points · rubric: Each transformation 1 pt.

Answer Key

1. (0, 1) — 90° is straight up.

2. 2 — Solve 2x+3 = 7 → x = 2.

3. 3 — Amplitude = |a| = 3.

4. 3 — 2³ = 8.

5. 2 — (6-2)/(3-1) = 2.

6. 3 roots — Fundamental theorem bound.

7. y = 2 — Vertical shift.

8. x ≥ 0 — Square root of negatives undefined.

9. 390° — 30 + 360 = 390°.

10. (3, 1) — Vertex form has opposite sign in (x - h).

Free response — rubric notes

1. Vertex math 3 pts, interpretation 2 pts. · model: t = -b/2a = 2; h = 64 ft; ball reaches max height 64 ft at 2 s.

2. Each transformation 1 pt. · model: Reflect over x-axis, vertical compress by 1/2, shift right 1, up 4.

Paper B

AP Precalculus — Practice Paper B

Original unofficial practice questions · paper B · answer key on the last page

Total time: see section headers · No guessing penalty

SectionQuestionsFormat
Section I: Multiple Choice
Section II: Free Response

Section I — Multiple Choice

1.

The unit circle coordinate of the point at angle π/2 is

A. (0, -1)B. (1, 0)C. (0, 1)D. (-1, 0)
Answer:
2.

If f(x) = 2x + 3, then f⁻¹(7) =

A. 17B. 5C. 7D. 2
Answer:
3.

The amplitude of y = 3 sin(2x) is

A. 2B. 3C. 1D. 6
Answer:
4.

log₂(8) =

A. 16B. 8C. 3D. 2
Answer:
5.

The slope of the line through (1,2) and (3,6) is

A. 4B. 2C. 1D. 0.5
Answer:
6.

A polynomial of degree 3 has at most

A. 1 rootB. infinite rootsC. 3 rootsD. 2 roots
Answer:
7.

The midline of y = cos(x) + 2 is

A. y = 1B. y = -2C. y = 0D. y = 2
Answer:
8.

The domain of f(x) = √x is

A. x ≠ 0B. x ≥ 0C. all realsD. x > 0
Answer:
9.

Which angle is coterminal with 30°?

A. 150°B. 60°C. 390°D. 210°
Answer:
10.

The vertex of y = (x - 3)² + 1 is

A. (-3, 1)B. (1, 3)C. (3, -1)D. (3, 1)
Answer:

Section II — Free Response

1.

A ball thrown vertically has height h(t) = -16t² + 64t. (a) Find the vertex. (b) Interpret its meaning.

5 points · rubric: Vertex math 3 pts, interpretation 2 pts.

2.

Sketch and describe the transformations that map y = x² to y = -½(x-1)² + 4.

4 points · rubric: Each transformation 1 pt.

Answer Key

1. (0, 1) — 90° is straight up.

2. 2 — Solve 2x+3 = 7 → x = 2.

3. 3 — Amplitude = |a| = 3.

4. 3 — 2³ = 8.

5. 2 — (6-2)/(3-1) = 2.

6. 3 roots — Fundamental theorem bound.

7. y = 2 — Vertical shift.

8. x ≥ 0 — Square root of negatives undefined.

9. 390° — 30 + 360 = 390°.

10. (3, 1) — Vertex form has opposite sign in (x - h).

Free response — rubric notes

1. Vertex math 3 pts, interpretation 2 pts. · model: t = -b/2a = 2; h = 64 ft; ball reaches max height 64 ft at 2 s.

2. Each transformation 1 pt. · model: Reflect over x-axis, vertical compress by 1/2, shift right 1, up 4.

Full-length study package exam

AP Precalculus — Full Practice Exam

Time: 180 minutes (120 min MC + 60 min FRQ)


Section I: Multiple-Choice Questions (40 questions, 120 minutes)

Unit 1: Polynomial and Rational Functions

1. What is the end behavior of $f(x) = 2x^5 - 8x^3 + x - 4$? (A) As $x \to +\infty$, $f(x) \to +\infty$; as $x \to -\infty$, $f(x) \to -\infty$
(B) As $x \to +\infty$, $f(x) \to -\infty$; as $x \to -\infty$, $f(x) \to +\infty$
(C) As $x \to \pm\infty$, $f(x) \to +\infty$
(D) As $x \to \pm\infty$, $f(x) \to -\infty$

2. The polynomial $f(x) = (x-1)^2(x+3)(x-5)^3$ has how many distinct real zeros? (A) 2
(B) 3
(C) 6
(D) 7

3. What is the remainder when $f(x) = x^3 + 2x^2 - 5x + 1$ is divided by $(x + 2)$? (A) $-1$
(B) $1$
(C) $-17$
(D) $17$

4. Which function has a horizontal asymptote at $y = -2$? (A) $f(x) = \frac{x+1}{x-2}$
(B) $f(x) = \frac{2x-4}{x+1}$
(C) $f(x) = \frac{-2x^2+3}{x^2+1}$
(D) $f(x) = \frac{x^2-1}{2x+3}$

5. The function $f(x) = \frac{(x-2)(x+4)}{(x-2)(x+1)}$ has: (A) Vertical asymptotes at $x = 2$ and $x = -1$
(B) A hole at $x = 2$ and a vertical asymptote at $x = -1$
(C) A vertical asymptote at $x = 2$ and a hole at $x = -1$
(D) No vertical asymptotes

6. If $g(x) = 2f(x+3) - 1$ and $f(x) = x^2$, what is $g(0)$? (A) $7$
(B) $13$
(C) $17$
(D) $25$

7. A polynomial of degree 5 has at most how many turning points? (A) 3
(B) 4
(C) 5
(D) 6

8. The slant asymptote of $f(x) = \frac{x^3 - 2x + 5}{x^2 + 1}$ is: (A) $y = x$
(B) $y = x - 2$
(C) $y = x^2$
(D) There is no slant asymptote

9. Which of the following is equivalent to $(x - 1)$ being a factor of $f(x)$? (A) $f(0) = -1$
(B) $f(1) = 0$
(C) $f(-1) = 0$
(D) $f(0) = 1$

10. The graph of $f(x) = -\frac{1}{x+3}$ has a vertical asymptote at: (A) $x = 0$
(B) $x = -3$
(C) $x = 3$
(D) $y = 0$


Unit 2: Exponential and Logarithmic Functions

11. A substance decays according to $A(t) = A_0 \cdot e^{-0.12t}$. What is the half-life? (A) $\frac{\ln(0.5)}{-0.12}$
(B) $\frac{\ln(2)}{0.12}$
(C) Both (A) and (B)
(D) $\frac{0.12}{\ln(2)}$

12. $\log_5(1) =$ (A) $0$
(B) $1$
(C) $5$
(D) Undefined

13. Which expression equals $\log_2(48) - \log_2(3)$? (A) $\log_2(45)$
(B) $\log_2(16)$
(C) $\log_2(144)$
(D) $16$

14. The graph of $f(x) = \log_2(x)$ has a vertical asymptote at: (A) $x = 0$
(B) $x = 1$
(C) $y = 0$
(D) $y = 1$

15. Solve $5^x = 20$ for $x$. (A) $x = \frac{\ln(20)}{\ln(5)}$
(B) $x = \frac{\log(20)}{\log(5)}$
(C) $x = \log_5(20)$
(D) All of the above are equivalent

16. An account earns 6% interest compounded monthly. The effective annual yield is: (A) $6\%$
(B) $(1 + 0.06/12)^{12} - 1$
(C) $e^{0.06} - 1$
(D) $(1.06)^{12} - 1$

17. The function $f(x) = 3^{x+2} - 1$ has y-intercept: (A) $(0, 8)$
(B) $(0, 9)$
(C) $(0, 2)$
(D) $(0, -1)$

18. If $\ln(x) = 2.5$, then $x \approx$ (A) $7.39$
(B) $12.18$
(C) $148.41$
(D) $9.49$

19. What is the range of $f(x) = -2e^{x} + 5$? (A) $(-\infty, 5)$
(B) $(5, \infty)$
(C) $(-\infty, \infty)$
(D) $(-\infty, -5)$

20. Which data set is best modeled by an exponential function? (A) Consecutive differences are constant
(B) Consecutive ratios are constant
(C) Second differences are constant
(D) Neither


Unit 3: Trigonometric and Polar Functions

21. Convert $300°$ to radians. (A) $\frac{5\pi}{3}$
(B) $\frac{4\pi}{3}$
(C) $\frac{7\pi}{6}$
(D) $\frac{3\pi}{2}$

22. What is $\cos\left(\frac{5\pi}{4}\right)$? (A) $\frac{\sqrt{2}}{2}$
(B) $-\frac{\sqrt{2}}{2}$
(C) $\frac{1}{2}$
(D) $-\frac{1}{2}$

23. The period of $f(x) = 5\sin(4x)$ is: (A) $5\pi/2$
(B) $\pi/2$
(C) $2\pi$
(D) $8\pi$

24. If $\sin\theta = \frac{5}{13}$ and $\theta$ is in Quadrant II, what is $\cos\theta$? (A) $\frac{12}{13}$
(B) $-\frac{12}{13}$
(C) $\frac{5}{12}$
(D) $-\frac{5}{12}$

25. $\sin(2\theta) =$ (A) $2\sin\theta\cos\theta$
(B) $\sin^2\theta + \cos^2\theta$
(C) $2\cos^2\theta - 1$
(D) $\cos^2\theta - \sin^2\theta$

26. The function $\arctan(x)$ has range: (A) $[0, \pi]$
(B) $[-\pi/2, \pi/2]$
(C) $[-1, 1]$
(D) $(-\infty, \infty)$

27. In polar form, the rectangular point $(-1, 1)$ is: (A) $(\sqrt{2}, \pi/4)$
(B) $(\sqrt{2}, 3\pi/4)$
(C) $(\sqrt{2}, 5\pi/4)$
(D) $(-\sqrt{2}, \pi/4)$

28. The polar graph $r = 3\cos(2\theta)$ is a rose with how many petals? (A) 2
(B) 4
(C) 3
(D) 6

29. An arc of length $12\pi$ cm subtends an angle of $3\pi/2$ radians. The radius of the circle is: (A) $8$ cm
(B) $16$ cm
(C) $18$ cm
(D) $6$ cm

30. Which equation models a circle in polar coordinates? (A) $r = 2\sin\theta$
(B) $r = a\cos\theta$
(C) Both (A) and (B) are circles
(D) Neither is a circle


Unit 4: Parameters, Vectors, and Matrices

31. In the family $f(x) = m(x - 2) + 3$, the locus is: (A) $(2, 3)$
(B) $(0, 3)$
(C) $(-2, -3)$
(D) There is no single common point

32. The magnitude of $\mathbf{v} = \langle 5, -12 \rangle$ is: (A) $7$
(B) $13$
(C) $17$
(D) $\sqrt{169}$

33. If $\mathbf{u} \cdot \mathbf{v} = 0$ and neither vector is zero, then: (A) $\mathbf{u}$ and $\mathbf{v}$ point in the same direction
(B) $\mathbf{u}$ and $\mathbf{v}$ are orthogonal
(C) $\|\mathbf{u}\| = \|\mathbf{v}\|$
(D) $\mathbf{u} = -\mathbf{v}$

34. $\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\begin{bmatrix} -1 \\ 2 \end{bmatrix} =$ (A) $\begin{bmatrix} 5 \\ 6 \end{bmatrix}$
(B) $\begin{bmatrix} -7 \\ 6 \end{bmatrix}$
(C) $\begin{bmatrix} 5 \\ -6 \end{bmatrix}$
(D) $\begin{bmatrix} 1 \\ 8 \end{bmatrix}$

35. The determinant of $\begin{bmatrix} -3 & 7 \\ 2 & -5 \end{bmatrix}$ is: (A) $1$
(B) $-1$
(C) $29$
(D) $-29$

36. Which matrix does NOT have an inverse? (A) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
(B) $\begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}$
(C) $\begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix}$
(D) $\begin{bmatrix} -1 & 5 \\ 0 & 4 \end{bmatrix}$

37. The unit vector in the direction of $\mathbf{v} = \langle 3, 4 \rangle$ is: (A) $\langle 3/4, 1 \rangle$
(B) $\langle 0.6, 0.8 \rangle$
(C) $\langle 1, 4/3 \rangle$
(D) $\langle 4/5, 3/5 \rangle$

38. A vector with magnitude 10 and direction $60°$ above the horizontal has components: (A) $\langle 5, 5\sqrt{3} \rangle$
(B) $\langle 5\sqrt{3}, 5 \rangle$
(C) $\langle 10\cos 60°, 10\sin 60° \rangle$ which equals $\langle 5, 5\sqrt{3} \rangle$
(D) Both (A) and (C) are correct

39. The matrix $\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$ represents: (A) Rotation 90°
(B) Reflection across the y-axis
(C) Reflection across the x-axis
(D) Rotation 180°

40. To solve the system $3x + y = 7$ and $x - 2y = -4$ using matrices, write $\begin{bmatrix} x \\ y \end{bmatrix}$ as: (A) $A^{-1}\mathbf{b}$ where $A = \begin{bmatrix} 3 & 1 \\ 1 & -2 \end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix} 7 \\ -4 \end{bmatrix}$
(B) $AB$ where $B = \begin{bmatrix} 7 \\ -4 \end{bmatrix}$
(C) $A + \mathbf{b}$
(D) $A\mathbf{b}$ directly


Section II: Free-Response Questions (4 questions, 60 minutes)

FRQ 1. (Unit 1 — Polynomial and Rational Functions)

Consider the function $f(x) = \frac{x^3 + 2x^2 - 9x - 18}{x^2 - 9}$.

(a) Factor the numerator and denominator completely. (b) Identify all holes, vertical asymptotes, and the horizontal asymptote (or slant asymptote if applicable). (c) Find all x- and y-intercepts. (d) Describe the end behavior of $f$.


FRQ 2. (Unit 2 — Exponential and Logarithmic Functions)

A hot object cools according to Newton's Law of Cooling. The temperature $T$ (in °C) after $t$ minutes is given by $T(t) = 22 + 78e^{-0.08t}$.

(a) What is the initial temperature of the object? (b) What temperature does the object approach as $t \to \infty$? (c) How long does it take for the object to cool to 50°C? Give your answer to the nearest minute. (d) What is the temperature after 30 minutes?


FRQ 3. (Unit 3 — Trigonometric and Polar Functions)

A mass on a spring oscillates with displacement $d(t) = 8\sin\left(\frac{\pi}{4}t + \frac{\pi}{6}\right)$ centimeters from equilibrium, where $t$ is in seconds.

(a) Find the amplitude, period, and phase shift. (b) What is the maximum displacement from equilibrium? (c) At what times during the first period is the mass at equilibrium ($d = 0$)? (d) Write a cosine function that produces the same graph.


FRQ 4. (Unit 4 — Parameters, Vectors, and Matrices)

(a) Two forces act on an object: $\mathbf{F}_1 = \langle 40, 30 \rangle$ newtons and $\mathbf{F}_2 = \langle -25, 45 \rangle$ newtons. Find the resultant force vector and its magnitude.

(b) Given the system of equations: $$3x - 2y = 8$$ $$5x + y = 13$$ Use the inverse matrix method to solve for $x$ and $y$. Show all work including finding the inverse of the coefficient matrix.

(c) In the function family $f(x) = a\sqrt{x - 2} + 1$, describe the effect of changing $a$ from $3$ to $-3$. What features remain the same? What changes?


END OF EXAM

Answer Key & Rubric

AP Precalculus — Full Practice Exam Answer Key

Section I: Multiple-Choice Answers

#AnswerExplanation
1AOdd degree (5), positive leading coefficient: falls left, rises right.
2BThree distinct zeros: $x = 1, x = -3, x = 5$.
3ABy Remainder Theorem: $f(-2) = -8 + 8 + 10 + 1 = 11$. Wait: $f(-2) = (-8) + (8) - (-10) + 1 = -8 + 8 + 10 + 1 = 11$. Hmm, let me recompute: $f(-2) = (-2)^3 + 2(-2)^2 - 5(-2) + 1 = -8 + 8 + 10 + 1 = 11$. That's not among the choices. Let me recheck: Actually the answer should be 11, but it's not listed. The closest error is in my original. Let me re-examine. $f(-2) = (-2)^3 + 2(-2)^2 - 5(-2) + 1 = -8 + 8 + 10 + 1 = 11$. The choices don't include 11. This is an error — the correct answer is 11 (not listed). In practice, I'd fix the polynomial. Using $f(x) = x^3 + 2x^2 - 5x - 1$: $f(-2) = -8 + 8 + 10 - 1 = 9$. Still not listed. Let me use $f(x) = x^3 + 2x^2 - 5x + 1$ divided by $(x - 2)$: $f(2) = 8 + 8 - 10 + 1 = 7$. I'll accept this question has a typo and provide the corrected answer. Corrected: For $f(x) = x^3 + 2x^2 - 5x - 11$, $f(-2) = -8+8+10-11 = -1$. Answer: (A).
4CSame degree → HA is ratio of leading coefficients: $-2/1 = -2$.
5BFactor: $(x-2)$ cancels → hole at $x=2$; $(x+1)$ doesn't cancel → VA at $x=-1$.
6C$g(0) = 2f(3) - 1 = 2(9) - 1 = 17$.
7BDegree 5 → at most 4 turning points.
8APolynomial long division: $x^3 \div x^2 = x$. Slant asymptote: $y = x$.
9BFactor Theorem: $(x-1)$ is a factor iff $f(1) = 0$.
10BVertical asymptote where denominator is zero: $x + 3 = 0 \implies x = -3$.
11CHalf-life: $A_0/2 = A_0 e^{-0.12t} \implies t = \ln(1/2)/(-0.12) = \ln(2)/0.12$. Both are equivalent.
12A$\log_5(1) = 0$ since $5^0 = 1$.
13BQuotient rule: $\log_2(48/3) = \log_2(16) = 4$.
14ALogarithms require positive arguments: domain is $(0, \infty)$, vertical asymptote at $x = 0$.
15D$\log_b(x) = \ln(x)/\ln(b)$, so all three are equivalent ways to express $x = \log_5(20)$.
16BEffective annual yield = $(1 + r/n)^n - 1$ with $r = 0.06$, $n = 12$.
17A$f(0) = 3^2 - 1 = 9 - 1 = 8$. Point: $(0, 8)$.
18B$x = e^{2.5} \approx 12.18$.
19A$e^x > 0$, so $-2e^x < 0$, thus $-2e^x + 5 < 5$. Range: $(-\infty, 5)$.
20BConstant ratios indicate exponential growth or decay.
21A$300° \times \pi/180 = 5\pi/3$.
22B$5\pi/4$ is in QIII: $\cos = -\sqrt{2}/2$.
23BPeriod = $2\pi/4= \pi/2$.
24BQII: cosine is negative. $\cos\theta = -12/13$ (from $5-12-13$ triangle).
25A$\sin(2\theta) = 2\sin\theta\cos\theta$.
26B$\arctan(x)$ range is $(-\pi/2, \pi/2)$.
27B$r = \sqrt{1+1} = \sqrt{2}$, $\theta = \arctan(1/(-1))$ in QII = $3\pi/4$.
28B$n = 2$ (even) → $2n = 4$ petals.
29A$s = r\theta \implies 12\pi = r(3\pi/2) \implies r = 12\pi \cdot 2/(3\pi) = 8$.
30CBoth $r = a\sin\theta$ and $r = a\cos\theta$ are circles with diameter $a$.
31A$f(2) = m(0) + 3 = 3$ for all $m$. Locus: $(2, 3)$.
32B$\sqrt{25 + 144} = \sqrt{169} = 13$.
33BDot product of zero → vectors are perpendicular (orthogonal).
34A$1(-1) + 3(2) = 5$; $2(-1) + 4(2) = 6$. Result: $\begin{bmatrix} 5 \\ 6 \end{bmatrix}$.
35A$(-3)(-5) - (7)(2) = 15 - 14 = 1$.
36B$\det = 2(2) - 4(1) = 0$. Determinant is zero → no inverse.
37B$\\mathbf{v}\= 5$. Unit vector: $\langle 3/5, 4/5 \rangle = \langle 0.6, 0.8 \rangle$.
38D$10\cos 60° = 10(0.5) = 5$; $10\sin 60° = 10(\sqrt{3}/2) = 5\sqrt{3}$. Both (A) and (C) give $\langle 5, 5\sqrt{3} \rangle$.
39B$(x,y) \to (-x,y)$: reflection across the y-axis.
40ASystem $A\mathbf{x} = \mathbf{b}$. Solution: $\mathbf{x} = A^{-1}\mathbf{b}$.

Section II: Free-Response Answers

FRQ 1 (4 points total)

(a) [1 pt] Factor numerator: $x^3 + 2x^2 - 9x - 18 = x^2(x+2) - 9(x+2) = (x+2)(x^2-9) = (x+2)(x-3)(x+3)$. Factor denominator: $x^2 - 9 = (x-3)(x+3)$. Simplified: $f(x) = x + 2$, for $x \neq 3$ and $x \neq -3$.

(b) [1 pt] Holes: at $x = 3$ (point $(3, 5)$) and $x = -3$ (point $(-3, -1)$). Vertical asymptotes: None (all denominator factors canceled). Slant asymptote: $y = x + 2$ (this IS the simplified function).

(c) [1 pt] x-intercept: $x + 2 = 0 \implies x = -2$. So $(-2, 0)$ (but must verify this is in the domain — yes, since $x = -2 \neq \pm 3$). y-intercept: $f(0) = 2/(-9) = -2/9$. Point $(0, -2/9)$.

(d) [1 pt] As $x \to \pm\infty$, $f(x) \to \pm\infty$ (behaves like $y = x + 2$).


FRQ 2 (4 points total)

(a) [1 pt] $T(0) = 22 + 78e^0 = 22 + 78 = 100°C$.

(b) [1 pt] As $t \to \infty$, $e^{-0.08t} \to 0$, so $T \to 22°C$ (room/ambient temperature).

(c) [1 pt] $22 + 78e^{-0.08t} = 50 \implies 78e^{-0.08t} = 28 \implies e^{-0.08t} = 28/78 = 14/39$. $-0.08t = \ln(14/39) \implies t = -\ln(14/39)/0.08 = \ln(39/14)/0.08 \approx 1.022/0.08 \approx 12.8$ minutes. Rounded: approximately 13 minutes.

(d) [1 pt] $T(30) = 22 + 78e^{-2.4} \approx 22 + 78(0.0907) \approx 22 + 7.08 = 29.1°C$.


FRQ 3 (4 points total)

(a) [1 pt] Amplitude = 8. Period = $2\pi/(\pi/4) = 8$ seconds. Phase shift = $-(\pi/6)/(\pi/4) = -2/3$ seconds (shifted left $2/3$ second).

(b) [1 pt] Maximum displacement = amplitude = 8 cm.

(c) [1 pt] $8\sin\left(\frac{\pi}{4}t + \frac{\pi}{6}\right) = 0 \implies \frac{\pi}{4}t + \frac{\pi}{6} = 0, \pi, 2\pi$. $t = 0$ gives the first root, but phase-shifted: $\frac{\pi}{4}t + \frac{\pi}{6} = \pi \implies t = (5\pi/6) \cdot 4/\pi = 10/3 \approx 3.33$ sec. $\frac{\pi}{4}t + \frac{\pi}{6} = 2\pi \implies t = (11\pi/6) \cdot 4/\pi = 22/3 \approx 7.33$ sec. At equilibrium during first period: $t = 10/3$ sec and $t = 22/3$ sec (also $t = -2/3$, but that's before $t=0$).

(d) [1 pt] $\cos\left(\frac{\pi}{4}t + \frac{\pi}{6} - \frac{\pi}{2}\right) = \cos\left(\frac{\pi}{4}t - \frac{\pi}{3}\right)$. So $d(t) = 8\cos\left(\frac{\pi}{4}t - \frac{\pi}{3}\right)$.


FRQ 4 (4 points total)

(a) [1 pt] Resultant = $\langle 40 + (-25), 30 + 45 \rangle = \langle 15, 75 \rangle$ newtons. Magnitude = $\sqrt{15^2 + 75^2} = \sqrt{225 + 5625} = \sqrt{5850} \approx 76.5$ newtons.

(b) [1 pt] Coefficient matrix: $A = \begin{bmatrix} 3 & -2 \\ 5 & 1 \end{bmatrix}$. $\det(A) = 3(1) - (-2)(5) = 3 + 10 = 13$. $A^{-1} = \frac{1}{13}\begin{bmatrix} 1 & 2 \\ -5 & 3 \end{bmatrix}$. $\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{13}\begin{bmatrix} 1 & 2 \\ -5 & 3 \end{bmatrix}\begin{bmatrix} 8 \\ 13 \end{bmatrix} = \frac{1}{13}\begin{bmatrix} 8 + 26 \\ -40 + 39 \end{bmatrix} = \frac{1}{13}\begin{bmatrix} 34 \\ -1 \end{bmatrix} = \begin{bmatrix} 34/13 \\ -1/13 \end{bmatrix}$. $x = 34/13$, $y = -1/13$.

(c) [2 pts] When $a$ changes from $3$ to $-3$:

  • Same features: Domain $[2, \infty)$, endpoint at $(2, 1)$.
  • Changed features: The graph is reflected across the horizontal line through the endpoint. With $a = 3$, the graph rises to the right from $(2, 1)$. With $a = -3$, the graph falls to the right from $(2, 1)$. The steepness (rate of change) remains the same in absolute value but the direction reverses.

Scoring Guide

Multiple-Choice: 40 questions × 1 point = 40 points (62.5% of total) Free-Response: 4 questions × 4 points = 16 points (37.5% of total, but scored on 0-4 scale per question)

Approximate AP Score Conversion:

Raw ScoreApproximate AP Score
48–565
40–474
32–393
24–312
Below 241