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

Question 1: 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(πt/5) + 25
h(t) = −20 cos(2πt/5) + 25
h(t) = −20 cos(πt/5) + 25 (Correct answer)

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

4
−2
2 (Correct answer)
6

Correct answer: 2

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

Question 3: As x → ∞ and as x → −∞, the graph of f(x) = (x² + 1)/(x − 1) approaches which line?

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

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

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

6 (Correct answer)
3
2/3
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 6: 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 0
f(x) approaches 5

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

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

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

3
5 (Correct answer)
7
6

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 9: The graphs of r = 2 and r = 4 cos θ intersect at two points. Which values of θ in [0, 2π) give these intersection points?

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

Correct answer: π/3 and 5π/3

Setting 4 cos θ = 2 gives cos θ = 1/2, so θ = π/3 and θ = 5π/3 in [0, 2π).

Question 10: Which expression is equivalent to (x² + 5x + 7)/(x + 2) for x ≠ −2 and shows its slant asymptote?

x + 2 + 3/(x + 2)
x + 3 + 1/(x + 2) (Correct answer)
x + 5 + 1/(x + 2)
x + 3 + 7/(x + 2)

Correct answer: x + 3 + 1/(x + 2)

Dividing gives quotient x + 3 and remainder 1, because (x + 3)(x + 2) = x² + 5x + 6. So the expression equals x + 3 + 1/(x + 2).

Question 11: What is the value of arcsin(sin(5π/6))?

5π/6
7π/6
−π/6
π/6 (Correct answer)

Correct answer: π/6

sin(5π/6) = 1/2, and arcsin returns the angle in [−π/2, π/2] whose sine is 1/2, which is π/6.

Question 12: What is the end behavior of f(x) = (x³ − 1)/(x² + 2)?

As x → ∞, f(x) → ∞, and as x → −∞, f(x) → −∞ (Correct answer)
As x → ±∞, f(x) → 0
As x → ±∞, f(x) → 1
As x → ∞ and as x → −∞, f(x) → ∞

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

The numerator has a higher degree than the denominator. For large |x|, f(x) behaves like x³/x² = x, which goes to ∞ as x → ∞ and to −∞ as x → −∞.

Question 13: 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 − 2)/((x − 1)(x + 3))
f(x) = 4(x − 1)(x + 3)/(x − 2)² (Correct answer)

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

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

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.

Question 15: Which function is equivalent to f(t) = 5·2^(3t) written in the form a·b^t?

15·2^t
125·2^t
5·6^t
5·8^t (Correct answer)

Correct answer: 5·8^t

By the power rule for exponents, 2^(3t) = (2^3)^t = 8^t. So f(t) = 5·8^t.

Question 16: For g(x) = −2 cos(x) + 5, which pair gives the midline and the amplitude?

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

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 17: What is the exact value of log₄ 32?

2
8
3
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 18: 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 19: 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 = 10·2ˣ
y = 100·10ˣ
y = 10·100ˣ (Correct answer)
y = 2·10ˣ

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 20: 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?

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

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 21: 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 = 5 cos θ
r = 4 + 3 sin θ (Correct answer)
r = 3 + 2 cos θ
r = 2 - 2 cos θ

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 22: What is the domain of f(x) = ln(5 − 2x)?

x ≤ 5/2
x < 5/2 (Correct answer)
x < −5/2
x > 5/2

Correct answer: x < 5/2

The argument of a logarithm must be strictly positive, so 5 − 2x > 0, which gives x < 5/2.

Question 23: What is the period, as a function of θ, of the polar function r = 5 + 2 sin(3θ)?

6π
2π/3 (Correct answer)
3π
π/3

Correct answer: 2π/3

For a sinusoid with argument bθ, the period is 2π/|b|. Here b = 3, so the period is 2π/3.

Question 24: The rectangular point (-2, 2) is written in polar form (r, θ) with r > 0 and 0 ≤ θ < 2π. Which is correct?

(2, 3π/4)
(2√2, 3π/4) (Correct answer)
(2√2, π/4)
(2√2, 5π/4)

Correct answer: (2√2, 3π/4)

r = √((-2)² + 2²) = 2√2. The point lies in Quadrant II and tan θ = -1, so θ = 3π/4.

Question 25: What are the zeros of f(x) = (x² − 4)/(x² − x − 6)?

x = −2 only
x = 2 only (Correct answer)
x = −2 and x = 2
x = 3 only

Correct answer: x = 2 only

Factoring gives f(x) = (x − 2)(x + 2)/((x − 3)(x + 2)). The factor (x + 2) cancels, so x = −2 is a hole, not a zero. The only zero is x = 2.

Question 26: What is the range of the quadratic function g(x) = −(x + 2)² + 5?

(−∞, 5] (Correct answer)
[5, ∞)
(−∞, ∞)
(−∞, −2]

Correct answer: (−∞, 5]

The graph of y = x² is shifted left 2, reflected over the x-axis, and shifted up 5, so it opens downward with vertex (−2, 5) and maximum value 5.

Question 27: The table gives values of a function: f(0) = 4, f(1) = 12, f(2) = 36, f(3) = 108. Which model best fits the data?

y = 4 + 8x
y = 4x² + 8x
y = 4·3^x (Correct answer)
y = 3·4^x

Correct answer: y = 4·3^x

Each output is 3 times the previous one over equal input steps, so the data are exponential with initial value 4 and base 3. That gives y = 4·3^x.

Question 28: 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(0.94)^t (Correct answer)
P(t) = 2400 − 0.06t
P(t) = 2400(1.06)^t
P(t) = 2400(0.06)^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 29: 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 negative.
It is increasing, because |r| is increasing. (Correct answer)
It is constant, because the angle is in Quadrant III.
It is decreasing, because r is decreasing.

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 30: Which expression is equivalent to (2x² + 3x − 5)/(x − 1) for all x ≠ 1?

2x + 5 (Correct answer)
2x − 5
2x² + 3x − 4
2x + 3

Correct answer: 2x + 5

The numerator factors as (2x + 5)(x − 1), so the factor (x − 1) cancels and leaves 2x + 5 for x ≠ 1.

Question 31: Which function is the inverse of h(x) = eˣ + 1?

h⁻¹(x) = ln(x) − 1
h⁻¹(x) = ln(x) + 1
h⁻¹(x) = 1/(eˣ + 1)
h⁻¹(x) = ln(x − 1) (Correct answer)

Correct answer: h⁻¹(x) = ln(x − 1)

Set y = eˣ + 1, subtract 1 to get y − 1 = eˣ, then take ln: x = ln(y − 1). Swapping variables gives h⁻¹(x) = ln(x − 1).

Question 32: 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 are scattered randomly above and below zero with no pattern. (Correct answer)
The residuals increase steadily as the input increases.
The residuals form a clear U-shaped curve.
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 33: Let f(x) = 3·2ˣ. Which expression gives the inverse function f⁻¹(x)?

log₂(3x)
log₂(x) − 3
log₂(x/3) (Correct answer)
3·log₂(x)

Correct answer: log₂(x/3)

Set y = 3·2ˣ, divide by 3 to get y/3 = 2ˣ, then take log₂ of both sides: x = log₂(y/3). Swapping variables gives f⁻¹(x) = log₂(x/3).

Question 34: On the interval (π/2, π), the graph of y = sin x is which of the following?

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

Correct answer: Decreasing and concave down

Sine falls from 1 at π/2 to 0 at π, so it is decreasing. The graph bends downward there, so it is concave down.

Question 35: What is the end behavior of f(x) = −2x⁵ + 3x² − 1?

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 → −∞
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 36: Which set contains all solutions of 2 sin x = √3 on the interval [0, 2π)?

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

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

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

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 38: 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)?

20 (Correct answer)
15
30
25

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 39: A quadratic function has average rates of change of 3 on [0, 1] and 7 on [1, 2]. If this pattern continues, what is its average rate of change on [2, 3]?

9
10
11 (Correct answer)
8

Correct answer: 11

For a quadratic, average rates of change over consecutive equal-length intervals change by a constant amount; here it increases by 4, so the next value is 11.

Question 40: How many petals does the graph of r = 5 sin(4θ) have?

5
4
10
8 (Correct answer)

Correct answer: 8

For r = a sin(nθ) with n even, the graph is a rose with 2n petals, so n = 4 gives 8 petals.

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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