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Unit 1 · Polynomial & rational
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| f(x) | 5 | 9 | 15 | 23 |
Answer: B
The first differences are 4, 6, and 8. Their differences are 2 and 2. Constant nonzero second differences for equally spaced inputs support a quadratic model.
Unit 1 · Polynomial & rational
Answer: D
The leading term is -2x5. An odd degree with a negative leading coefficient gives a left end that rises and a right end that falls.
Unit 1 · Polynomial & rational
Answer: A
Cancel x - 2 while retaining the restrictions x ≠ 2, 4. The reduced expression is (x + 3)/(x - 4). At x = 2 its value is -5/2, so the hole is (2, -5/2). The uncanceled denominator gives x = 4 as a vertical asymptote.
Unit 1 · Polynomial & rational
Answer: C
The numerator and denominator have the same degree. Dividing their leading coefficients gives the horizontal asymptote y = 3/2.
Unit 1 · Polynomial & rational
Answer: A
Away from x = 1, the squared factor is positive, so the sign is the sign of (x + 2)(x - 4). This product is positive outside its two zeros. Strict positivity excludes x = -2 and x = 4; x = 1 is not in either resulting interval.
Unit 1 · Polynomial & rational
Answer: D
To use f(2) = -3, set 3(x - 1) = 2. This gives x = 5/3. Then g(5/3) = -2(-3) + 5 = 11.
Unit 1 · Polynomial & rational
Answer: B
Compute f(1) = 6 and f(4) = 9. The average rate is [f(4) - f(1)]/(4 - 1) = (9 - 6)/3 = 1.
Unit 1 · Polynomial & rational
Answer: C
Real coefficients require the conjugate zero 1 - 3i. Thus p(x) = (x - 2)[(x - 1)2 + 9] = (x - 2)(x2 - 2x + 10), which expands to choice C.
Unit 1 · Polynomial & rational
Answer: A
On (-3, -1), the graph rises as x increases, but it becomes less steep as it approaches its local maximum. Its secant slopes decrease, so it is concave down on that interval.
Unit 1 · Polynomial & rational
Answer: B
Polynomial division gives r(x) = x + 2 + 3/(x + 1). Therefore r(x) - (x + 2) = 3/(x + 1), which approaches 0.
Unit 2 · Exponential & logarithmic
Answer: C
The common ratio r satisfies r3 = 96/12 = 8, so r = 2. Therefore an = 12 · 2n-2 = 3 · 2n.
Unit 2 · Exponential & logarithmic
Answer: D
Divide the given outputs: b3 = 486/18 = 27, so b = 3. Since a · 32 = 18, a = 2.
Unit 2 · Exponential & logarithmic
Answer: B
Because 2-x = (1/2)x, the exponential term decreases and is always positive. Adding 5 gives outputs greater than 5 and a horizontal asymptote y = 5.
Unit 2 · Exponential & logarithmic
Answer: A
The domain requires x > 3. Combine logarithms to obtain (x - 1)(x - 3) = 8. Then x2 - 4x - 5 = 0, so x = -1 or 5. Only x = 5 is in the original domain.
Unit 2 · Exponential & logarithmic
Answer: C
From y = 4 · 3x-2 + 1, obtain (y - 1)/4 = 3x-2. Taking log base 3 and adding 2 gives x = log3((y - 1)/4) + 2. The inverse domain is x > 1.
Unit 2 · Exponential & logarithmic
| x | -1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|
| f(x) | 2 | 3 | -1 | 0 | 1 |
| g(x) | 1 | -1 | 2 | 3 | 0 |
Answer: D
Read the inner function first: g(1) = 2. Then read f(2) = 0. Thus f(g(1)) = 0.
Unit 2 · Exponential & logarithmic
Answer: A
The transformed line has slope 1/2 and intercept 1: log10(y) = 1 + x/2. Exponentiating base 10 gives y = 101+x/2 = 10 · 10x/2.
Unit 2 · Exponential & logarithmic
Answer: B
The outputs are f(0) = 8, f(1) = 12, and f(2) = 18. For intervals of length 1, the average rates are 12 - 8 = 4 and 18 - 12 = 6. Positive, increasing rates are consistent with this increasing, concave-up exponential function.
Unit 2 · Exponential & logarithmic
Answer: C
The logarithm argument must be positive: x + 2 > 0. A logarithm with base between 0 and 1 is decreasing. A horizontal translation preserves that decreasing behavior.
Unit 3 · Trigonometric & polar
Answer: D
Use arc length s = rθ. Therefore θ = s/r = 5π/6 radians.
Unit 3 · Trigonometric & polar
Answer: A
The identity sin2θ + cos2θ = 1 gives |cos θ| = 4/5. In quadrant III, cos θ = -4/5. Thus tan θ = (-3/5)/(-4/5) = 3/4.
Unit 3 · Trigonometric & polar
Answer: B
The maximum is 4 and the minimum is -2, so the amplitude is 3 and the midline is y = 1. Consecutive maxima occur at x = 1 and x = 5, giving period 4 and angular coefficient 2π/4 = π/2. The graph crosses the midline upward at x = 0, matching the sine model.
Unit 3 · Trigonometric & polar
Answer: C
Factor as (2 sin x + 1)(sin x - 1) = 0. Thus sin x = -1/2 or sin x = 1. On the specified interval these give x = 7π/6, 11π/6, and π/2.
Unit 3 · Trigonometric & polar
Answer: D
First, sin(5π/6) = 1/2. The principal range of arcsin is [-π/2, π/2], so arcsin(1/2) = π/6.
Unit 3 · Trigonometric & polar
Answer: A
Use 1 - cos(2x) = 2 sin2x and sin(2x) = 2 sin x cos x. Their quotient simplifies to sin x/cos x = tan x, on the domain of the original expression.
Unit 3 · Trigonometric & polar
Answer: B
Tangent is undefined when its argument is π/2 + kπ. Set 2x - π/3 = π/2 + kπ and solve: x = 5π/12 + kπ/2.
Unit 3 · Trigonometric & polar
Answer: C
Use x = r cos θ and y = r sin θ. Then x = (-4)(-1/2) = 2 and y = (-4)(√3/2) = -2√3. A negative r places the point opposite the ray at angle 2π/3.
Unit 3 · Trigonometric & polar
Answer: D
Cosine decreases from 0 to -1, so r decreases from 0 to -2. Distance from the origin is |r|, which increases from 0 to 2. Signed radius and distance have different behavior here.
Unit 1 · Polynomial & rational
Answer: C
Graph y = x3 - 4x - 1 and locate its zero in (1, 3), or use a numerical equation solver. The solution is x = 2.114907541..., which rounds to 2.115.
Unit 1 · Polynomial & rational
Answer: B
A graphing calculator shows a local maximum at x = 2.113248654... and a local minimum at x = 7.886751346.... The change from increasing to decreasing occurs at the local maximum, x ≈ 2.113. No differentiation is needed.
Unit 1 · Polynomial & rational
Answer: D
Average cost is (4200 + 18n)/n = 4200/n + 18. Since n > 0, the inequality 4200/n + 18 < 30 gives n > 350. Strict inequality makes 351 the least possible integer.
Unit 1 · Polynomial & rational
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| y | 2.1 | 5.0 | 10.2 | 17.1 | 26.0 |
Answer: A
Quadratic regression gives y ≈ 0.978571429x2 + 2.075714286x + 2.057142857. Substituting x = 5 using unrounded coefficients gives y = 36.900.
Unit 2 · Exponential & logarithmic
| x | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| y | 120 | 158 | 212 | 278 |
Answer: B
Regressing ln y on x and exponentiating gives y ≈ 119.8710567(1.151101106)x. At x = 8, the prediction is 369.5065178..., so 369.5 is closest. Fitting the logarithms is explicitly specified to avoid differences between regression methods.
Unit 2 · Exponential & logarithmic
Answer: C
Solve e0.16t = 500/180. Taking natural logarithms gives t = ln(500/180)/0.16 = 6.385320297... hours.
Unit 2 · Exponential & logarithmic
Answer: D
Set 21 + 74(0.86)t = 30, so (0.86)t = 9/74. Then t = ln(9/74)/ln(0.86) = 13.968970622... minutes. The function decreases, so this is its only crossing.
Unit 2 · Exponential & logarithmic
Answer: A
Rearrange to ln x = (7 - 2.7)/1.8. Exponentiating gives x = e(7-2.7)/1.8 = 10.901374573....
Unit 3 · Trigonometric & polar
Answer: C
Find intersections of y = 3 sin(0.7x) and y = 0.4x in radian mode. Besides x = 0, the first positive intersection is x = 3.741795188.... The zero solution is excluded by the word positive.
Unit 3 · Trigonometric & polar
Answer: B
The inequality becomes cos((π/6)(t - 2)) > 2/3. The interval [0, 12] is one full period. The angular width is 2 arccos(2/3), so the time length is (6/π) · 2 arccos(2/3) = 3.212645674... units.
Unit 3 · Trigonometric & polar
Answer: D
The average rate is [r(1.1) - r(0.4)]/(1.1 - 0.4) = 2[sin(2.2) - sin(0.8)]/0.7 = 0.260400894... radial units per radian.
Unit 3 · Trigonometric & polar
Answer: A
Solve 18 - 15 cos(πt/20) = 25 to obtain cos(πt/20) = -7/15. During the first rising half-cycle, t = (20/π) arccos(-7/15) = 13.090904365... seconds. The later solution is not the first time.
6 practice points · Suggested time: 15 minutes
The table gives selected values of a one-to-one function f. The function g is defined by g(x) = 1.8x - 2 for all real x.
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| f(x) | 1 | 4 | 2 | 5 | 8 |
Since f(3) = 5, f-1(5) = 3.
1 point: correct inverse value.
The table gives f(2) = 2. Therefore g(f(2)) = g(2) = 1.82 - 2 = 1.240.
1 point: correct composite value.
The average rate is [f(4) - f(1)]/(4 - 1) = (8 - 4)/3 = 4/3.
1 point: correct average rate.
The secant through (2, 2) and (3, 5) has slope 3. Its value at 2.5 is 2 + 3(2.5 - 2) = 3.5, the requested estimate.
1 point: correct secant-line estimate.
Since f(4) = 8, solve 1.8x - 2 = 8. Thus x = ln 10 / ln 1.8 = 3.917 (approximately).
1 point: correct solution.
Solving y = 1.8x - 2 for x gives g-1(x) = ln(x + 2)/ln 1.8, with domain x > -2.
1 point: correct inverse formula and domain.
Review focus: A table lookup must follow the order of composition. Inverse inputs are outputs of the original function.
Use the rubric to self-assess. These are practice points, not an official AP score.
6 practice points · Suggested time: 15 minutes
A dissolved substance has concentration C(t) milligrams per liter t hours after observation begins. On 0 ≤ t ≤ 24, an exponential model C(t) = abt fits the data below exactly. Assume 0 < b < 1.
| t (hours) | 0 | 2 | 4 |
|---|---|---|---|
| C(t) (mg/L) | 80 | 60.8 | 46.208 |
C(0) = a = 80. Also, 60.8 = 80b2, so b2 = 0.76 and b = √0.76 ≈ 0.871780. Thus C(t) = 80(0.76)t/2.
2 points: 1 for a = 80 with a supporting equation; 1 for b = √0.76 with a supporting equation.
C(6) = 80(0.76)3 = 35.11808, so the prediction is 35.118 mg/L.
1 point: correct predicted concentration.
Solve 80(0.76)t/2 = 12. Then t = 2 ln(12/80)/ln(0.76) = 13.825548680..., or 13.826 hours.
1 point: correct time.
Over two hours, the multiplier is b2 = 0.76. The percent decrease is (1 - 0.76) × 100% = 24%.
1 point: correct percent decrease.
The slope is (60.8 - 80)/2 = -9.6, so L(t) = 80 - 9.6t. For t > 80/9.6 = 25/3 hours, L(t) is negative. Negative concentration is impossible, so this model is unsuitable for all of [0, 24].
1 point: correct linear model and a valid contextual explanation.
Review focus: The two-hour multiplier is 0.76; the one-hour multiplier is √0.76. Retain the exact form when calculating later predictions.
Use the rubric to self-assess. These are practice points, not an official AP score.
6 practice points · Suggested time: 15 minutes
A wave-testing tank is programmed so that its water depth d(t), in centimeters, varies sinusoidally with time t in minutes. A maximum depth of 11 cm occurs at t = 1, the next minimum of 3 cm occurs at t = 5, and the next maximum occurs at t = 9.
Amplitude = (11 - 3)/2 = 4 cm; midline value = (11 + 3)/2 = 7 cm; period = 9 - 1 = 8 minutes.
1 point: all three values correct.
A cosine model that starts a cycle at the maximum t = 1 is d(t) = 7 + 4 cos((π/4)(t - 1)). Its period is 2π/(π/4) = 8. Equivalent sinusoidal formulas are accepted.
1 point: a correct model.
d(3) = 7 + 4 cos(π/2) = 7 cm.
1 point: correct depth.
Set cos((π/4)(t - 1)) = 1/2. Over one cycle, the arguments are π/3 and 5π/3. Thus t = 1 + 4/3 or 1 + 20/3, giving t = 7/3 and 23/3 minutes.
1 point: both exact times and no extra times.
The depth rises from the minimum at t = 5 to the next maximum at t = 9. It is increasing on (5, 9). Endpoint notation is not a scoring distinction here.
1 point: correct interval.
[d(5) - d(1)]/(5 - 1) = (3 - 11)/4 = -2 cm/min. From minute 1 to minute 5, the depth decreases by an average of 2 cm per minute.
1 point: correct rate and contextual interpretation.
Review focus: The elapsed time from a maximum to the next minimum is half a period. Exact times should retain fractions and π when appropriate.
Use the rubric to self-assess. These are practice points, not an official AP score.
6 practice points · Suggested time: 15 minutes
Show algebraic steps and state any domain restrictions needed in your work. Give exact answers.
The product rule gives f(x) = log2(x2 - 1). Both original arguments must be positive, so the original domain is x > 1.
1 point: correct combined logarithm and original domain.
log2(x2 - 1) = 3 gives x2 - 1 = 8, so x = ±3. Only x = 3 satisfies x > 1.
1 point: the valid solution with no extraneous solution.
Factor r(x) = [(x - 2)(x - 3)]/[(x - 2)(x + 2)]. For x ≠ ±2, this equals (x - 3)/(x + 2). The canceled factor gives a hole at x = 2; its y-value is (2 - 3)/(2 + 2) = -1/4. The hole is (2, -1/4).
1 point: correct hole supported by factoring or cancellation.
The remaining denominator vanishes at x = -2, while its numerator does not. The vertical asymptote is x = -2.
1 point: correct vertical asymptote.
Factor as (2 cos θ - 1)(cos θ - 1) = 0. Thus cos θ = 1/2 or 1. On 0 ≤ θ < 2π, the complete solution set is {0, π/3, 5π/3}. The endpoint 2π is excluded.
2 points: 1 for the correct factorization or equivalent reduction; 1 for all three solutions with no extras.
Review focus: Combining logarithms can produce an expression with a larger apparent domain. Keep the restrictions from the original expression.
Use the rubric to self-assess. These are practice points, not an official AP score.
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These original questions cover polynomial and rational functions, exponential and logarithmic functions, and trigonometric and polar functions. The downloadable paper uses the May 2026 format: 28 no-calculator MCQs in 80 minutes, 12 calculator MCQs in 40 minutes, and two 30-minute FRQ parts with two questions each. This is the 2026 format, not the revised 2027 format.
The online questions match the PDF. A first-try answer counts as unaided only if its solution had not been opened before checking. Scores are for practice; they do not predict an AP score of 1–5. Free-response notes are not automatically graded. Progress is stored on this device and is not shared with teachers.
Official format references: College Board exam information and format changes effective from 2027.