AP Precalculus Exam (College Board Advanced Placement) — Questions and Answers

Question 1: What is the solution to e^(2x) = 7?

ln 7 − 2
ln(7/2)
ln 14
(ln 7)/2 (Correct answer)

Correct answer: (ln 7)/2

Take the natural log of both sides: 2x = ln 7, so x = (ln 7)/2.

Question 2: For an exponential function f(x) = a·b^x with b > 0, f(1) = 10, f(3) = 40 and f(5) = 160. What is f(2)?

30
20 (Correct answer)
25
15

Correct answer: 20

Outputs are multiplied by 4 over each 2-unit step, so b² = 4 and b = 2. Then a = 10/2 = 5, and f(2) = 5·2² = 20.

Question 3: Which polar function has a graph that is symmetric about the y-axis (the line θ = π/2) but not about the x-axis?

r = 2 - 2 cos θ
r = 5 cos θ
r = 3 + 2 cos θ
r = 4 + 3 sin θ (Correct answer)

Correct answer: r = 4 + 3 sin θ

Because sin(π - θ) = sin θ, a function of the form r = a + b sin θ gives the same r for θ and π - θ, so its graph is symmetric about the y-axis.

Question 4: On an interval, a function's average rates of change over consecutive equal-length subintervals are all positive and increasing. What does this indicate about the graph of the function on that interval?

Decreasing and concave down
Decreasing and concave up
Increasing and concave up (Correct answer)
Increasing and concave down

Correct answer: Increasing and concave up

Positive rates of change mean the function is increasing, and rates that grow larger mean the graph is concave up.

Question 5: Which describes the vertical asymptotes of y = tan(2x), where k is any integer?

x = kπ/2
x = π/2 + 2kπ
x = π/4 + kπ/2 (Correct answer)
x = π/2 + kπ

Correct answer: x = π/4 + kπ/2

Tangent is undefined when its input equals π/2 + kπ. Setting 2x = π/2 + kπ gives x = π/4 + kπ/2.

Question 6: Let g(x) = −3·2^x + 5. Which statement about g is true?

g has a horizontal asymptote at y = 5 and decreases without bound as x → ∞. (Correct answer)
g has a horizontal asymptote at y = −3 and increases without bound as x → ∞.
g has a horizontal asymptote at y = 5 and increases without bound as x → ∞.
g has a horizontal asymptote at y = 2 and decreases without bound as x → ∞.

Correct answer: g has a horizontal asymptote at y = 5 and decreases without bound as x → ∞.

The vertical shift of +5 places the horizontal asymptote at y = 5, as x → −∞. The negative coefficient reflects the graph, so g(x) → −∞ as x → ∞.

Question 7: What is the end behavior of f(x) = −2x⁵ + 3x² − 1?

f(x) → −∞ as x → ∞ and f(x) → ∞ as x → −∞ (Correct answer)
f(x) → −∞ as x → ∞ and f(x) → −∞ as x → −∞
f(x) → ∞ as x → ∞ and f(x) → −∞ as x → −∞
f(x) → ∞ as x → ∞ and f(x) → ∞ as x → −∞

Correct answer: f(x) → −∞ as x → ∞ and f(x) → ∞ as x → −∞

The leading term −2x⁵ has odd degree and a negative coefficient, so the graph falls to the right and rises to the left.

Question 8: On a semi-log plot, log₁₀(y) is graphed against x, and the points form a straight line passing through (0, 1) and (2, 5). Which exponential model fits the data?

y = 10·100ˣ (Correct answer)
y = 10·2ˣ
y = 2·10ˣ
y = 100·10ˣ

Correct answer: y = 10·100ˣ

The line has slope (5 − 1)/(2 − 0) = 2 and intercept 1, so log y = 2x + 1. Then y = 10^(2x+1) = 10·(10²)ˣ = 10·100ˣ.

Question 9: The polar point (6, 2π/3) is converted to rectangular coordinates. Which is the result?

(-3, 3√3) (Correct answer)
(3, 3√3)
(-3√3, 3)
(3√3, -3)

Correct answer: (-3, 3√3)

x = 6cos(2π/3) = 6(-1/2) = -3 and y = 6sin(2π/3) = 6(√3/2) = 3√3.

Question 10: For the polar function r = 2 + 4 sin θ, which values of θ in [0, 2π) give r = 0, so the graph passes through the pole?

5π/6 and 7π/6
7π/6 and 11π/6 (Correct answer)
π/6 and 5π/6
2π/3 and 4π/3

Correct answer: 7π/6 and 11π/6

Setting 2 + 4 sin θ = 0 gives sin θ = -1/2, which occurs at θ = 7π/6 and 11π/6.

Question 11: A rational function has zeros at x = 1 and x = −3, a vertical asymptote at x = 2, and a horizontal asymptote at y = 4. Which function fits all of these?

f(x) = (x − 1)(x + 3)/(4(x − 2)²)
f(x) = 4(x − 1)(x + 3)/(x − 2)
f(x) = 4(x − 1)(x + 3)/(x − 2)² (Correct answer)
f(x) = 4(x − 2)/((x − 1)(x + 3))

Correct answer: f(x) = 4(x − 1)(x + 3)/(x − 2)²

The numerator's zeros are 1 and −3, and the denominator's zero is 2. Both have degree 2, and the ratio of leading coefficients is 4/1 = 4, which gives the horizontal asymptote y = 4.

Question 12: Which expression is equivalent to log(x³√y / z) for positive x, y, z?

3 log x · ½ log y ÷ log z
(3 log x + ½ log y) ÷ log z
3 log x + ½ log y + log z
3 log x + ½ log y − log z (Correct answer)

Correct answer: 3 log x + ½ log y − log z

The quotient rule turns division into subtraction, the product rule turns multiplication into addition, and the power rule brings down exponents (√y = y^(1/2)).

Question 13: Which line is the horizontal asymptote of f(x) = (3x² − 5x + 1)/(x² + 4)?

y = 0
y = 3 (Correct answer)
y = 1/4
y = −5

Correct answer: y = 3

The numerator and denominator both have degree 2, so the horizontal asymptote is the ratio of leading coefficients, 3/1 = 3.

Question 14: What is the period of f(x) = 4 sin(πx/3) + 2?

2/3
6 (Correct answer)
3
6π

Correct answer: 6

For f(x) = a sin(bx) + d, the period is 2π/|b|. Here b = π/3, so the period is 2π ÷ (π/3) = 6.

Question 15: A town has 2400 residents, and its population decreases by 6% each year. Which function models the population P after t years?

P(t) = 2400(0.06)^t
P(t) = 2400(0.94)^t (Correct answer)
P(t) = 2400 − 0.06t
P(t) = 2400(1.06)^t

Correct answer: P(t) = 2400(0.94)^t

A 6% decrease each year multiplies the population by 1 − 0.06 = 0.94 every year. That gives P(t) = 2400(0.94)^t.

Question 16: For f(x) = (x + 1)/(x − 4), what happens to f(x) as x approaches 4 from the left (x → 4⁻)?

f(x) increases without bound (→ +∞)
f(x) decreases without bound (→ −∞) (Correct answer)
f(x) approaches 5
f(x) approaches 0

Correct answer: f(x) decreases without bound (→ −∞)

Near x = 4 from the left, the numerator is close to 5 (positive) and the denominator is a small negative number. The quotient is a large negative number, so f(x) → −∞.

Question 17: What is the remainder when f(x) = x³ − 2x² + 5x − 1 is divided by x − 2?

7
5
9 (Correct answer)
13

Correct answer: 9

By the Remainder Theorem, the remainder equals f(2) = 8 − 8 + 10 − 1 = 9.

Question 18: What is the greatest possible number of relative extrema (turning points) of a polynomial function of degree 6?

5 (Correct answer)
6
3
7

Correct answer: 5

A polynomial of degree n has at most n − 1 turning points, so a degree 6 polynomial has at most 5.

Question 19: A cell-phone signal strength reading s(t) oscillates sinusoidally with a maximum of 9 and a minimum of 3. It reaches a maximum at t = 0 and completes one cycle every 8 seconds. Which function models s(t)?

s(t) = 3 cos(πt/8) + 6
s(t) = 3 cos(πt/4) + 6 (Correct answer)
s(t) = 3 sin(πt/4) + 6
s(t) = 6 cos(πt/4) + 3

Correct answer: s(t) = 3 cos(πt/4) + 6

The midline is (9+3)/2 = 6 and the amplitude is (9−3)/2 = 3. Starting at a maximum calls for cosine, and b = 2π/8 = π/4.

Question 20: A Ferris wheel has a diameter of 40 m, its center is 25 m above the ground, and it makes one full revolution every 10 minutes. A rider is at the lowest point at t = 0. Which function gives the rider's height h(t) in meters?

h(t) = −20 sin(πt/5) + 25
h(t) = −20 cos(2πt/5) + 25
h(t) = −20 cos(πt/5) + 25 (Correct answer)
h(t) = 20 cos(πt/5) + 25

Correct answer: h(t) = −20 cos(πt/5) + 25

The radius is 20 m and the center is at 25 m, so amplitude is 20 and midline is 25. Starting at the minimum requires −cos, and b = 2π/10 = π/5.

Question 21: For g(x) = −2 cos(x) + 5, which pair gives the midline and the amplitude?

Midline y = −2, amplitude 5
Midline y = 3, amplitude 2
Midline y = 5, amplitude 2 (Correct answer)
Midline y = 5, amplitude −2

Correct answer: Midline y = 5, amplitude 2

The midline is y = d = 5. The amplitude is |a| = |−2| = 2, so the graph oscillates between 3 and 7.

Question 22: A student fits an exponential model to a data set and plots the residuals. Which observation about the residual plot best supports the model being appropriate?

The residuals form a clear U-shaped curve.
The residuals increase steadily as the input increases.
The residuals are scattered randomly above and below zero with no pattern. (Correct answer)
All residuals are positive.

Correct answer: The residuals are scattered randomly above and below zero with no pattern.

A residual is the actual value minus the predicted value. Random scatter around zero with no pattern shows the model captures the trend in the data.

Question 23: Where is the hole in the graph of f(x) = (x² + x − 6)/(x² − 4)?

(−2, 5/4)
(2, 0)
(2, 5)
(2, 5/4) (Correct answer)

Correct answer: (2, 5/4)

Factoring gives f(x) = (x + 3)(x − 2)/((x − 2)(x + 2)). The factor (x − 2) cancels, so the hole is at x = 2. The simplified function (x + 3)/(x + 2) gives y = 5/4 there.

Question 24: What is the average rate of change of f(x) = x² − 4x + 1 over the interval [1, 5]?

−2
6
2 (Correct answer)
4

Correct answer: 2

f(5) = 6 and f(1) = −2, so the average rate of change is (6 − (−2))/(5 − 1) = 2.

Question 25: As x → ∞ and as x → −∞, the graph of f(x) = (x² + 1)/(x − 1) approaches which line?

y = 1
y = x
y = x − 1
y = x + 1 (Correct answer)

Correct answer: y = x + 1

Polynomial division gives x² + 1 = (x − 1)(x + 1) + 2, so f(x) = x + 1 + 2/(x − 1). The remainder term 2/(x − 1) approaches 0, which leaves the slant line y = x + 1.

Question 26: What is the exact value of log₄ 32?

3
2
8
5/2 (Correct answer)

Correct answer: 5/2

Write both as powers of 2: 32 = 2⁵ and 4 = 2². Then log₄ 32 = 5/2 by the change of base formula, or because 4^(5/2) = 2⁵ = 32.

Question 27: What is the vertical asymptote of g(x) = log₂(x + 4) − 5?

x = −4 (Correct answer)
x = −5
x = 5
x = 4

Correct answer: x = −4

The graph of log₂ x has its asymptote at x = 0. Replacing x with x + 4 shifts it left 4 units, giving x = −4. The −5 only moves the graph vertically.

Question 28: For r = 1 + 2 sin θ, consider the interval 7π/6 < θ < 3π/2. On this interval, r is negative and decreasing as θ increases. What is happening to the distance of the point (r, θ) from the pole?

It is decreasing, because r is decreasing.
It is increasing, because |r| is increasing. (Correct answer)
It is decreasing, because r is negative.
It is constant, because the angle is in Quadrant III.

Correct answer: It is increasing, because |r| is increasing.

Distance from the pole is |r|. When r is negative and decreasing, it becomes more negative, so |r| increases.

Question 29: Which set contains all solutions of 2 sin x = √3 on the interval [0, 2π)?

x = π/3 and x = 4π/3
x = π/6 and x = 5π/6
x = 2π/3 and x = 5π/3
x = π/3 and x = 2π/3 (Correct answer)

Correct answer: x = π/3 and x = 2π/3

Dividing by 2 gives sin x = √3/2. Sine equals √3/2 at the reference angle π/3 in Quadrant I and at π − π/3 = 2π/3 in Quadrant II.

Question 30: Let f(x) = 2x + 3 and g(x) = x². What is the value of (f ∘ g)(2)?

11 (Correct answer)
7
13
49

Correct answer: 11

(f ∘ g)(2) = f(g(2)) = f(4) = 2(4) + 3 = 11. The inner function g is applied first.

Question 31: Consider f(x) = (x − 2)/((x − 2)(x + 3)). Which statement correctly describes the graph of f?

A vertical asymptote at x = −3 and a hole at x = 2 (Correct answer)
A vertical asymptote at x = −3 only, with no hole
A vertical asymptote at x = 2 and a hole at x = −3
Vertical asymptotes at x = 2 and x = −3

Correct answer: A vertical asymptote at x = −3 and a hole at x = 2

The factor (x − 2) cancels, which leaves a hole at x = 2. The factor (x + 3) remains in the denominator, which gives a vertical asymptote at x = −3.

AP Precalculus Exam (College Board Advanced Placement)

Tests college-level precalculus skills: polynomial and rational functions, exponential and logarithmic functions, and trigonometric and polar functions. It covers modeling, analyzing graphs, and function behavior. Scores are reported on a 1-5 scale.

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60 min

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