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AP Precalculus

AP Precalculus free-response questions: how every point is scored

Updated · For the May 2027 AP exam

The AP Precalculus exam ends with four free-response questions, each worth 6 points, together 37.5% of your score. Questions 1 and 2 (function concepts and a real-world model) require a graphing calculator; questions 3 and 4 (a periodic model and symbolic manipulation) allow none. The College Board’s scoring data for 2024 and 2025 shows the same weak spots both years: explanations that don’t cite specific values, turning a period and a starting point into the constants of a sinusoid, and the last part of question 4.

The four questions

The four questions are always the same four types, in the same order, and each has parts A, B and C. You get two 35-minute parts: questions 1 and 2 with a graphing calculator, then questions 3 and 4 without one.

The four free-response questions, May 2027
QuestionUnitsCalculatorWhat it asks you to do
1. Function Concepts1–2RequiredWork with one function given as a graph or table and another given by a formula: composition, inverses, solving on the calculator, limits, choosing a function type
2. Modeling a Non-Periodic Context1–2RequiredBuild a model from real data, work with its average rate of change, and explain where the model stops making sense
3. Modeling a Periodic Context3Not allowedTurn a story about something that repeats into five points on a sinusoid, then into aa, bb, cc and dd, then describe the function on an interval
4. Symbolic Manipulations2–3Not allowedSolve exponential, logarithmic and trigonometric equations by hand and rewrite expressions in a required form, with exact answers
Questions 1 and 2 make up Section II, Part A; questions 3 and 4 make up Part B. Each part is 35 minutes and 18.75% of the exam score. Sources: AP Precalculus Course and Exam Description (fall 2026) and the AP Central exam page.

Every free-response point counts the same. One point is 124\frac{1}{24} of 37.5%, about 1.6% of the exam score, and one multiple-choice question is worth about 1.5%. So a small point here, like the sentence that finishes an explanation or the third decimal place on an answer, counts as much as a multiple-choice question.

The questions appear on screen in the Bluebook app, but you write your answers by hand in a paper booklet, and only the booklet is scored. Within a part you can move between its two questions until time runs out. On the 2026 exam the on-screen clock turned red at five minutes, and the proctor gave no time warnings, so keep an eye on it yourself.

Thirty-five minutes for two questions is 17½ minutes each. Students had 15 through 2026, so the 2027 exam gives you a little more time to write explanations.

Where students lose points

Each fall the College Board publishes a Chief Reader report: the faculty member who leads the scoring describes how students did on each question and gives the average score on every point. The reports for 2024 and 2025, the exam’s first two years, show the same pattern both times, and you can plan your studying around it.

Share of students who earned each point, 2024 and 2025
PointWhat it was for20242025
Question 1Function Concepts, mean out of 62.953.51
1 A(i)Value of a composition, read from a graph or table0.630.64
1 A(ii)Inputs for a given output, or an inverse value0.610.72
1 B(i)Solve on the calculator: every solution, to three decimals0.510.56
1 B(ii)End behavior in limit notation0.480.56
1 C(i)The answer: has an inverse or not, or which function type0.590.73
1 C(ii)The reason, citing specific values0.150.30
Question 2Modeling a Non-Periodic Context, mean out of 62.032.16
2 A(i)Equations from the data0.580.53
2 A(ii)Values of the constants0.360.48
2 B(i)Average rate of change, computation shown0.580.63
2 B(ii)An estimate from that rate0.390.41
2 B(iii)Why those estimates are all too low0.040.06
2 CWhere and why the model stops working0.080.05
Question 3Modeling a Periodic Context, mean out of 62.963.13
3 A, first pointHeights of the five labeled points0.710.77
3 A, second pointTimes of the five labeled points0.390.34
3 B, first pointaa and dd (amplitude and midline)0.600.67
3 B, second pointbb and cc (period and phase shift)0.200.18
3 C(i)Positive or negative, increasing or decreasing0.680.67
3 C(ii)How the rate of change is changing0.380.51
Question 4Symbolic Manipulations, mean out of 61.281.93
4 A(i)Solve an exponential or logarithmic equation0.370.53
4 A(ii)Solve an inverse-trig or trig equation0.330.43
4 B(i)Rewrite as a single logarithm0.160.45
4 B(ii)Rewrite a trig expression in one function0.240.28
4 C, first pointOne solution (2024) or the quadratic form (2025)0.150.14
4 C, second pointEvery solution0.040.10
Each point is worth 1, so its average is the share of students awarded it: 0.04 means about 4 in 100. Some recorded point awards include partial credit for other work in that part, so they do not always measure success on the named task alone. Question means are from the College Board’s scoring statistics; point averages are rounded and may not add up exactly. The 2026 report and statistics had not been published as of October 24, 2026. Sources: 2024 and 2025 Chief Reader reports and scoring statistics.

Most students earn the points for reading and computing. Reading a value off a graph, writing equations from data, computing an average rate of change, finding the heights on a sinusoid: more than half of students earned each of these, both years.

The points for explaining are rarely earned. Question 2’s two explanation points went to between 4% and 8% of students, and the reason in question 1 to 15% in 2024 and 30% in 2025. The 2024 report noted that many explanations stated a correct general rule without evidence from the function itself, and the 2025 report stressed that specific values were required. The section on writing reasons shows what the guidelines accept.

Question 4 is the hardest question on the exam. It had the lowest mean both years, and its last point went to 4% and then 10% of students. In question 3, the point for bb and cc, the constants that come from the period and the starting position, went to 20% of students in 2024 and 18% in 2025.

Added up, the average student earned 9.2 of the 24 points in 2024 and 10.7 in 2025.

Rules that apply to every question

These rules come from the general notes in the scoring guidelines and from the question 4 directions. Students lose points to them every year even when the math is right.

Give decimals to three places

In questions 1 and 2, a decimal answer has to be correct to three places after the decimal point. Rounding and truncating are both accepted, so 5.4828 can be written 5.483 or 5.482, and trailing zeros can go: 2.25 is fine for 2.250.

A correct answer with too few digits is a “decimal presentation error.” The first one in a question costs that point; later decimal-presentation errors in the same question do not automatically lose another point if the response is otherwise complete and correct. An inaccurate calculation from early rounding is a different error. The 2026 guidelines, and the 2025 guidelines for question 2, also count an exact answer where a decimal was asked for, such as 36ln⁡92\frac{36}{\ln 92} for 7.961, as a decimal presentation error.

Show the work when the question asks for it

When a part says “show the work that leads to your answer” or “show the computations,” the answer alone doesn’t earn the point. In 2025, question 1 asked for h(1)=g(f(1))h(1) = g(f(1)), and students who wrote only 0.333 lost the point: the readers needed to see f(1)=1.75f(1) = 1.75 or g(1.75)g(1.75) on the way. The question 4 directions ask for work in every part, and most of its points require it. In parts without that instruction, a correct answer stands on its own.

A wrong answer next to a right one cancels it

List a correct answer beside an incorrect one and you usually lose the point. The guidelines say it in several places: a list of answers earns the point only with “no incorrect values” in it, and an extra limit statement “must be correct.” In 2025 some students answered f−1(3.5)f^{-1}(3.5) with both “0” and “no solution.” Commit to one answer.

Say what is changing

Readers are strict about the subject of a sentence. In 2024, “increasing at a decreasing rate” with no subject was read as describing the rate of change of hh, which was actually decreasing, so it didn’t earn the point. With the subject, “hh is increasing at a decreasing rate,” or simply “the rate of change of hh is decreasing,” it did. The 2025 and 2026 released versions also asked for concavity, and their points needed both. Also, don’t describe how the rate of change itself is changing. “The rate of change is decreasing at an increasing rate” loses the point even when it’s true, because the guidelines say that conclusion takes calculus.

Which calculator, and when

Calculator rules, May 2027
PartQuestionsCalculator
Section I, Part A29 multiple choice, 65 minutesNot permitted
Section I, Part B13 multiple choice, 40 minutesGraphing calculator required
Section II, Part AFree response 1 and 2, 35 minutesGraphing calculator required
Section II, Part BFree response 3 and 4, 35 minutesNot permitted
Sources: the AP Precalculus exam page and Graphing Calculators for AP Precalculus, AP Central.

In the calculator parts you can use the Desmos graphing calculator built into Bluebook or a handheld from the College Board’s approved list, which includes the TI-84 and TI-Nspire families. The website and app versions of Desmos are fine all year but not on exam day. Put a handheld in radian mode.

Calculator work still has to be written down, because the reader sees your booklet, not your screen. Write the equation you solved and the solution, or the regression you ran and the values it gave. And 62.5% of the exam score comes from parts with no calculator at all, so practice the algebra by hand even when a calculator could do it.

Question 1: Function Concepts

Question 1 gives you two functions in two forms: one as a graph or a table, the other as a formula. The formula was exponential in 2024, a cubic in 2025 and a logarithm in 2026. The parts move from reading values, to calculator work, to a reason in words.

Question 1, point by point
PartWhat earns the pointWhat doesn’t
A(i) compositionThe decimal value of g(f(a))g(f(a)), with the inner value f(a)f(a) shown when the part asks for workThe final number with no sign of f(a)f(a); too few decimal places
A(ii) inputs and outputsEvery input that gives the output, or the inverse value read from the graph or tableOne input when there are three; a right answer plus “no solution”
B(i) solve on the calculatorEvery solution, each to three decimalsA solution missed because it was outside the graphing window
B(ii) limitA complete limit statement: “lim,” where xx is heading, the function, and the valueThe right idea in words only; notation with a piece missing, such as the ++ in x→1+x \to 1^{+}
C(i) the answer“ff does not have an inverse function,” “exponential”: one clear answer“ff is not an inverse,” which says something else
C(ii) the reasonThe property, tied to specific values from this graph or table“Not one-to-one,” “fails the horizontal line test,” or a regression, with no values
From the 2024–2026 scoring guidelines and the 2024 and 2025 Chief Reader reports.

Part C(ii) is the point students miss most on this question. In 2024 the question asked whether ff has an inverse, and most students knew it doesn’t. What many of them wrote was a general rule, such as “ff isn’t one-to-one,” and the guidelines say a rule isn’t enough. The reason has to point at this graph: f(−3)=f(0)=f(3)=1f(-3) = f(0) = f(3) = 1, so the output 1 comes from three different inputs, and an inverse would have to send 1 back to three places at once, which no function can do. Only 15% of students wrote something like that.

Worked example

We wrote this question to match the real question 1 in layout and difficulty, with our own six-point practice rubric based on the published task model. Try it on paper before you read the response.

Selected values of ff
xx01234
f(x)f(x)95.531.51
Practice question 1: Function Concepts (graphing calculator)

The function ff is defined and decreasing on its domain, 0≤x≤40 \le x \le 4, and the table gives values of f(x)f(x) at selected values of xx. The function gg is given by g(x)=2x2+1x2−4g(x) = \frac{2x^2 + 1}{x^2 - 4}. A (i) The function hh is defined by h(x)=g(f(x))h(x) = g(f(x)). Find the value of h(3)h(3) as a decimal approximation, or indicate that it is not defined. Show the work that leads to your answer. (ii) Find the value of f−1(3)f^{-1}(3), or indicate that it is not defined. B (i) Find all values of xx, as decimal approximations, for which g(x)=3g(x) = 3, or indicate that there are no such values. (ii) Determine the behavior of gg as xx increases and gets arbitrarily close to 2. Express your answer using the mathematical notation of a limit. C (i) Based on the table, which function type best models ff: linear, quadratic, exponential, or logarithmic? (ii) Give a reason for your answer based on how the output values of ff change over equal-length input intervals. Refer to values in the table.

A (i) f(3)=1.5f(3) = 1.5, so h(3)=g(1.5)=2(1.5)2+1(1.5)2−4=5.5−1.75=−3.143h(3) = g(1.5) = \frac{2(1.5)^2 + 1}{(1.5)^2 - 4} = \frac{5.5}{-1.75} = -3.143

A (ii) f(2)=3f(2) = 3, so f−1(3)=2f^{-1}(3) = 2.

B (i) Graphing y=g(x)y = g(x) and y=3y = 3 and finding where they intersect: x=−3.606x = -3.606 and x=3.606x = 3.606.

B (ii) lim⁡x→2−g(x)=−∞\lim_{x \to 2^{-}} g(x) = -\infty

C (i) Quadratic.

C (ii) The inputs go up by 1 each time. The outputs change by −3.5-3.5, −2.5-2.5, −1.5-1.5 and −0.5-0.5, so the average rates of change over these equal intervals go up by the same amount, 1, each step. These constant second differences make a quadratic the best of the listed models for these data. The outputs aren’t proportional either (5.59≈0.61\frac{5.5}{9} \approx 0.61 but 35.5≈0.55\frac{3}{5.5} \approx 0.55), so it isn’t exponential.

How it scores

A(i): value of h(3)h(3)1/1The right decimal, and f(3)=1.5f(3) = 1.5 is on the page, which is the work a reader looks for. Truncating to −3.142 would also earn it.
A(ii): value of f−1(3)f^{-1}(3)1/1Read from the table: an inverse swaps inputs and outputs, so look for 3 among the outputs. No work required.
B(i): solutions of g(x)=3g(x) = 31/1Both solutions, each to three decimals. gg is symmetric about the yy-axis, so stopping at 3.606 would have lost the point.
B(ii): limit1/1x→2−x \to 2^{-}, from the left, because xx increases toward 2. Just below 2 the numerator is close to 9 and the denominator is a tiny negative number, so gg falls without bound.
C(i): function type1/1One answer, with no alternatives listed beside it.
C(ii): reason1/1It names the property and shows it with the table’s values for more than one pair. The 2025 guidelines required exactly that for the exponential version of this part, and they ruled out reasons based on a regression.

Why it works

  • In A(i), if f(3)f(3) had been 2, then h(3)=g(2)h(3) = g(2), which is undefined because the denominator is 0. That’s why the question offers “indicate that it is not defined.” When you see that phrase, check the domain of the outer function before you compute.

Lessons on the ideas question 1 keeps testing: composition, inverses, end behavior and asymptotes, and how outputs change for each function type.

Question 2: Modeling a Non-Periodic Context

Question 2 is a story with numbers: daily sales of a video game in 2024, plays of a song in 2025, a car losing value in 2026. You’re given a model with two or three unknown constants, and you build it from the data, use it, and say where it stops working. It’s the second-lowest-scoring question, and the explanation points in its last two parts are almost never earned. For 2027 the College Board changed its layout; the 2024–2026 version comes first because the released questions use it, and the new layout follows.

The parts, 2024 to 2026

Question 2 as it appeared 2024–2026
PartTaskWhat earns the point
A(i)Equations for the constantsSubstitute each measured input and output into the given model
A(ii)Values of the constantsThe values to three decimals. Work is optional, so the calculator or a regression is fine
B(i)Average rate of changeThe difference quotient with the data substituted, then its decimal; follow the particular question’s instructions about units
B(ii)An estimate from that rateA point on the secant line: start at an endpoint and add rate × elapsed time
B(iii)Why the estimates are too low (or too high)The concavity of the model on the interval between the data points; in 2024 and 2025 the secant line there too
CA limitation of the modelThe model’s behavior set against what the real quantity can do
From the 2024, 2025 and 2026 scoring guidelines. In 2027 the question changes; see below.

The 2025 Chief Reader named a mistake in B(ii) worth knowing: students used a line with the right slope through the wrong point. The song data had three points, and they started from the middle one, (2,30)(2, 30), which isn’t on the secant line between t=0t = 0 and t=4t = 4. The estimate has to start from one of the two endpoints of the interval you used for the rate.

The secant-line explanation in B(iii)

Part B(iii) asked the same thing all three years: explain why every estimate made with the average rate of change is below (or above) the model between the two data points. It’s gone from the 2027 sample, but the idea is still part of the course.

The average rate of change between two points is the slope of the secant line through them, so every estimate from B(ii) is a point on that line. Whether the line sits below or above the graph depends only on concavity. A concave-down graph bulges above any secant line between two of its points, so every estimate on the secant line is below the graph. A concave-up graph sags below its secant lines, so the estimates are too high. That was 2026’s twist: the car’s value was concave up, and the estimates were too high.

The 2024 report’s diagnosis was that students mixed up the function with its rate of change. Keep them apart: “DD is increasing” is about the function, “concave down” is about its rate of change, and only the concavity decides which side of the graph the secant line is on.

Part C: where the model stops making sense

Part C asks about the model’s limits in the story. The point goes to answers that put two facts side by side: what the model does, and what the real quantity can or can’t do.

  • 2024, video game sales. Daily sales fell after day 91, but the model GG keeps increasing. The two matched on day 91, so the gap between them grows every day after. Earned by 8% of students.
  • 2025, song plays. A total number of plays can never go down. The quadratic model peaks at t=11t = 11 and decreases after that, so it only works up to t=11t = 11. Earned by 5%.
  • 2026, a car’s value. The owner will donate the car when its value reaches 2 thousand dollars, and at that moment it loses all its value. Solving V(k)=2V(k) = 2 gives k≈22.444k \approx 22.444, so the model applies on 0≤t≤22.4440 \le t \le 22.444.

Each of these answers puts two facts together: what the model does, and what the real quantity can or can’t do. The 2025 report noted that few responses did both: DD decreases after t=11t = 11, while the total number of plays never decreases.

What changes in 2027

The College Board restructured question 2 for May 2027. In its updated sample question, part B(iii) is gone. Part B is the average rate of change and what it means in context. Part C is worth two points: write and solve an equation for when the model reaches a threshold, then explain the domain restriction that the answer gives. The secant-line reasoning is still part of the course, so keep it; the example below is in the new layout.

Worked example in the 2027 layout

Our own question and six-point practice rubric, built around the College Board’s updated sample tasks. Use a graphing calculator.

Practice question 2: Modeling a Non-Periodic Context (graphing calculator)

A park ranger tracks a patch of algae spreading across a pond. At t=2t = 2 weeks after the first measurement, the algae covers 3.1 square meters. At t=5t = 5 weeks it covers 9.6 square meters. The area covered can be modeled by A(t)=abtA(t) = ab^t, where A(t)A(t) is in square meters and tt is in weeks since the first measurement. The surface of the pond is 150 square meters. A (i) Use the given data to write two equations that can be used to find the values of the constants aa and bb. (ii) Find the values of aa and bb as decimal approximations. B (i) Use the given data to find the average rate of change of the area covered, in square meters per week, from t=2t = 2 to t=5t = 5. Express your answer as a decimal approximation. Show the computations that lead to your answer. (ii) Interpret the meaning of your answer from B(i) in the context of the problem. C The model is appropriate until the algae covers the whole pond. (i) Write and solve an equation whose solution gives the time when the model predicts the algae covers the whole pond. Express your answer as a decimal approximation. (ii) Explain how your answer to C(i) can be used to determine an appropriate domain restriction for AA. Refer to the context of the problem.

A (i) A(2)=3.1A(2) = 3.1 and A(5)=9.6A(5) = 9.6, so ab2=3.1ab^2 = 3.1 and ab5=9.6ab^5 = 9.6.

A (ii) Dividing the second equation by the first, b3=9.63.1b^3 = \frac{9.6}{3.1}, so b=1.458b = 1.458 (stored as 1.457594…) and a=3.1b2=1.459a = \frac{3.1}{b^2} = 1.459 (stored as 1.459112…).

B (i) A(5)−A(2)5−2=9.6−3.13=2.167\frac{A(5) - A(2)}{5 - 2} = \frac{9.6 - 3.1}{3} = 2.167 square meters per week

B (ii) From week 2 to week 5, the area covered by the algae grew by an average of 2.167 square meters per week.

C (i) abt=150ab^t = 150 with the stored values of aa and bb. The graphs of y=abty = ab^t and y=150y = 150 intersect at t=12.296t = 12.296 weeks.

C (ii) The model predicts the algae covers the whole 150-square-meter pond at t=12.296t = 12.296 weeks. After that the model keeps growing, but the algae can’t cover more than the pond, so the model only makes sense for 0≤t≤12.2960 \le t \le 12.296.

How it scores

A(i): two equations1/1Both data points go into the model.
A(ii): values of aa and bb1/1Both to three decimals. An exponential regression on the two points gives the same values and needs no other work.
B(i): average rate of change1/1The quotient shows the given values, and the answer has three decimals. Leaving it as 6.53\frac{6.5}{3} would be a decimal presentation error.
B(ii): interpretation1/1It says what changed (the area covered), over which interval, at what average rate, with units. The College Board’s sample answer has the same pieces.
C(i): equation and value1/1The equation comes before the answer, and the answer uses unrounded constants. This is the rounding trap from the rules above: rounded constants give 12.287.
C(ii): domain1/1It says why the model stops (the algae can’t cover more than the whole pond) and states the domain that follows, 0≤t≤12.2960 \le t \le 12.296.

Why it works

  • In B(ii), “the algae grows 2.167 square meters every week” would be wrong: an exponential model doesn’t grow by the same amount each week. “On average” over a named interval is what makes the sentence true.

Lessons on building a model from data, average rates of change, and the assumptions behind a model.

Question 3: Modeling a Periodic Context

Question 3 describes something that goes around or back and forth: a point on a rolling tire in 2024, a vibrating guitar string in 2025, a waterwheel in 2026. There’s no calculator. You turn the story into a sinusoid in three steps, and the first two steps carry 4 of the 6 points.

The graph has been the same picture on all three exams so far. It shows two full cycles and a dashed midline, with no axes and no scale, and five labeled points in the same places. F is a maximum, G is on the midline going down, J is a minimum, K is on the midline going up, and P is the next maximum. Each point is a quarter of a period after the one before it.

Step 1: the five points (part A)

  1. Read the story for the highest value, the lowest value and the period. The midline is halfway between the highest and lowest values.
  2. The heights go maximum, midline, minimum, midline, maximum. For the 2025 guitar string that’s 2, 0, −2, 0, 2 millimeters. This is the first point, and about three students in four earned it.
  3. Find one time when the point is at its maximum. That’s F. Add a quarter period for each point after it. In 2025 the string starts at its highest position at t=0t = 0 and vibrates 200 times a second, so the period is 1200\frac{1}{200} second and the five times are 00, 1800\frac{1}{800}, 2800\frac{2}{800}, 3800\frac{3}{800} and 4800\frac{4}{800}.

The times are where students lose points: 39% earned it in 2024 and 34% in 2025. The 2025 report reminds teachers that the period is the reciprocal of the frequency. “200 times in 1 second” is a frequency; one cycle takes 1200\frac{1}{200} of a second.

Step 2: aa, bb, cc and dd (part B)

Every year so far the form has been h(t)=asin⁡(b(t+c))+dh(t) = a\sin(b(t + c)) + d.

  • dd is the midline, and aa is the amplitude, half the distance from the maximum to the minimum. Take aa positive.
  • b=2πperiodb = \frac{2\pi}{\text{period}}. A period of 2 seconds gives b=πb = \pi; a period of 1200\frac{1}{200} second gives b=400πb = 400\pi.
  • cc comes from the first maximum. With a>0a > 0, the sine curve peaks when b(t+c)=π2b(t + c) = \frac{\pi}{2}, which happens when t+ct + c is a quarter of a period. So if F is at time tFt_F, then c=period4−tFc = \frac{\text{period}}{4} - t_F.

Check it on the 2024 tire. The period is 2 seconds, so b=πb = \pi, and the first peak on the graph is at t=32t = \frac{3}{2}. Then c=24−32=−1c = \frac{2}{4} - \frac{3}{2} = -1 and h(t)=9sin⁡(π(t−1))+9h(t) = 9\sin(\pi(t - 1)) + 9. At t=32t = \frac{3}{2} that gives 9sin⁡π2+9=189\sin\frac{\pi}{2} + 9 = 18, the top of the tire.

A negative aa is allowed too, with a different cc. And if your part A times are shifted but still a quarter period apart, the guidelines let you earn the part B points with constants that match your own points. Consistency is rewarded, so finish part B even if you doubt part A.

Step 3: behavior and concavity (part C)

Part C picks two neighboring labeled points and asks about the function between them. C(i) is multiple choice: positive or negative, increasing or decreasing. Positive and negative mean above or below zero, not above or below the midline. On the 2024 tire the height was positive except at the instants W touched the ground.

C(ii) asks how the rate of change is changing, and sinusoids have a shortcut. Above the midline the graph is concave down; below it, the graph is concave up. Concave down means the rate of change is decreasing, and concave up means it’s increasing. In 2026 the interval ran from J, the minimum, to K, the midline going up. That stretch is below the midline, so the answer was “concave up, so the rate of change of hh is increasing,” even though hh itself was negative there.

Worked example

Our own question and six-point practice rubric, in the same layout and without a calculator. Its part A graph is the standard one described above: two cycles with a dashed midline, F at a maximum, G on the midline going down, J at a minimum, K on the midline going up and P at the next maximum.

Practice question 3: Modeling a Periodic Context (no calculator)

Point TT is on the tip of one blade of a wind turbine. The blade is 40 meters long, and the center of the hub is 80 meters above the ground. The blades turn at a constant rate of 15 full turns per minute. At time t=0t = 0 seconds, TT is at its lowest point. The sinusoidal function hh gives the height of TT above the ground, in meters, as a function of tt in seconds. A The graph of hh and its dashed midline for two full cycles, the standard pattern described above, has five labeled points F, G, J, K and P. Determine possible coordinates (t,h(t))(t, h(t)) for the five points. B The function hh can be written in the form h(t)=asin⁡(b(t+c))+dh(t) = a\sin(b(t + c)) + d. Find values of the constants aa, bb, cc and dd. C The tt-coordinate of G is t1t_1, and the tt-coordinate of J is t2t_2. (i) On the interval (t1,t2)(t_1, t_2), is hh positive or negative, and is it increasing or decreasing? (ii) On (t1,t2)(t_1, t_2), describe the concavity of the graph of hh and determine whether the rate of change of hh is increasing or decreasing.

A Fifteen turns per minute is one turn every 4 seconds. The highest point is 80+40=12080 + 40 = 120 meters, the lowest is 80−40=4080 - 40 = 40, and the midline is 80. TT is lowest at t=0t = 0, so it’s highest half a turn later, at t=2t = 2. F (2,120)(2, 120), G (3,80)(3, 80), J (4,40)(4, 40), K (5,80)(5, 80), P (6,120)(6, 120).

B a=40a = 40, d=80d = 80, b=2π4=π2b = \frac{2\pi}{4} = \frac{\pi}{2}. The first peak is at t=2t = 2, so c=44−2=−1c = \frac{4}{4} - 2 = -1. That gives h(t)=40sin⁡(π2(t−1))+80h(t) = 40\sin\left(\frac{\pi}{2}(t - 1)\right) + 80. Check: h(2)=40sin⁡π2+80=120h(2) = 40\sin\frac{\pi}{2} + 80 = 120.

C (i) hh is positive and decreasing.

C (ii) The graph of hh is concave up on (t1,t2)(t_1, t_2), so the rate of change of hh is increasing.

How it scores

A: heights1/1120, 80, 40, 80, 120: maximum, midline, minimum, midline, maximum.
A: times1/1A quarter period (1 second) apart, starting from a maximum at a time that fits the story. (−2,120)(-2, 120), (−1,80)(-1, 80), (0,40)(0, 40), (1,80)(1, 80), (2,120)(2, 120) would also be right.
B: aa and dd1/1Amplitude 40, the blade length; midline 80, the height of the hub.
B: bb and cc1/1bb from the 4-second period, cc from the first peak, and the check at t=2t = 2 confirms both. a=−40a = -40 with c=1c = 1 would also be right.
C(i): behavior1/1The height is between 80 and 40 meters here, so it’s positive, and it’s falling from the midline to the lowest point.
C(ii): concavity and rate of change1/1Concave up and an increasing rate of change, both stated, with hh as the subject. Between G and J the graph is below its midline, so it’s concave up. hh is still decreasing, but more and more slowly, which means its rate of change is increasing toward 0.

Why it works

  • The trap in part C is answering “decreasing” twice. On this interval the function is decreasing and its rate of change is increasing. The 2024 report singles out exactly this confusion between how a function changes and how its rate of change changes.

Lessons on concavity and rates of change, and on building sinusoidal models from a story.

Question 4: Symbolic Manipulations

Question 4 is algebra with exponentials, logarithms and trig functions, with no calculator and no real-world context. It has the lowest average of the four questions, 1.28 out of 6 in 2024 and 1.93 in 2025. The 2024 Chief Reader wrote that students across the exam showed more success with concepts than with algebraic manipulation, and pointed to this question’s scores as the evidence.

The layout hasn’t changed: part A solves two equations, part B rewrites two expressions in a required form, and part C is one harder equation, worth two points.

In 2025, x=32x = 3^2 instead of 9, and x=cos⁡−1(32)x = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) instead of π6\frac{\pi}{6}, didn’t earn the points, because neither was evaluated. Rationalizing denominators isn’t required, though: 12\frac{1}{\sqrt{2}} is fine.

The moves that keep coming back

Question 4 moves, with examples from the real exams
MoveExampleThe trap
Undo an exponential with a log2024: ex+3=10e^{x+3} = 10, so x+3=ln⁡10x + 3 = \ln 10 and x=ln⁡10−3x = \ln 10 - 3Stopping at x+3=ln⁡10x + 3 = \ln 10
Undo a log with an exponential2026: log⁡2(5x)=3\log_2(5x) = 3, so 5x=235x = 2^3 and x=85x = \frac{8}{5}Leaving x=235x = \frac{2^3}{5} unevaluated
Undo an inverse trig function2024: arcsin⁡(x2)=π4\arcsin\left(\frac{x}{2}\right) = \frac{\pi}{4}, so x2=sin⁡π4=22\frac{x}{2} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2} and x=2x = \sqrt{2}Not knowing the inverse trig functions well, a gap the 2024 report names
Combine logs into one2024: log⁡(8x5)+log⁡(2x2)−9log⁡x=log⁡(16x2)\log(8x^5) + \log(2x^2) - 9\log x = \log\left(\frac{16}{x^2}\right)Stopping at log⁡(16x7x9)\log\left(\frac{16x^7}{x^9}\right), or writing log⁡(16x−2)\log(16x^{-2}) with a negative exponent
Use a Pythagorean identity2025: csc⁡2x−1=cot⁡2x\csc^2 x - 1 = \cot^2 xKnowing sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 but not the forms that come from it
Use a double-angle identity2026: sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x\cos xReaching 2sin⁡xcos⁡x⋅1cos⁡x2\sin x\cos x \cdot \frac{1}{\cos x} and not canceling
Spot a quadratic2025: e2x−ex−12=(ex−4)(ex+3)e^{2x} - e^x - 12 = (e^x - 4)(e^x + 3)Not seeing that e2x=(ex)2e^{2x} = (e^x)^2; keeping ex=−3e^x = -3, which has no solution
Find every solution2024: tan⁡(2x)=1\tan(2x) = 1 gives x=π8+nπ2x = \frac{\pi}{8} + \frac{n\pi}{2}Writing π8+nπ\frac{\pi}{8} + n\pi, which misses half the solutions
Which identities are tested: the Pythagorean identities, and the sum, difference and double-angle identities for sine and cosine. Half-angle identities and the tangent sum and difference identities are not tested (CED, fall 2026).

Part C needs every solution

Part C’s last point went to 4% of students in 2024 and 10% in 2025. The reports and the 2026 guidelines point to these habits.

  • Undo the outside function first. In 2024, cos⁡−1(tan⁡(2x))=0\cos^{-1}(\tan(2x)) = 0 means tan⁡(2x)=cos⁡0=1\tan(2x) = \cos 0 = 1; the report named this step as a gap.
  • Look for a quadratic. In 2025, e2x−ex−12=0e^{2x} - e^x - 12 = 0 is a quadratic in exe^x, which few responses recognized.
  • Solve for the inside angle first. For tan⁡(2x)=1\tan(2x) = 1, solve for 2x2x: 2x=π4+nπ2x = \frac{\pi}{4} + n\pi. Only then divide by 2, and divide the nπn\pi as well.
  • Look for every value of the trig function. tan⁡2(3x)=1\tan^2(3x) = 1 means tan⁡(3x)=1\tan(3x) = 1 or tan⁡(3x)=−1\tan(3x) = -1. In 2026 one of the three answers came from the −1-1.
  • Check the domain and the interval. ex=−3e^x = -3 has no solution. When the question gives an interval, such as [0,π2]\left[0, \frac{\pi}{2}\right] in 2026, the guidelines want the actual values in it, not only a general formula.

Worked example

Our own question and six-point practice rubric. Give exact real answers, finish the requested forms, and show your work without a calculator.

Practice question 4: Symbolic Manipulations (no calculator)

A The functions gg and hh are given by g(x)=52x−1g(x) = 5^{2x - 1} and h(x)=arccos⁡(x3)h(x) = \arccos\left(\frac{x}{3}\right). (i) Solve g(x)=125g(x) = 125 for values of xx in the domain of gg. (ii) Solve h(x)=2π3h(x) = \frac{2\pi}{3} for values of xx in the domain of hh. B The functions jj and kk are given by j(x)=3ln⁡x−ln⁡(4x)+2ln⁡(2x)j(x) = 3\ln x - \ln(4x) + 2\ln(2x) and k(x)=1−cos⁡(2x)2sin⁡xtan⁡xk(x) = \frac{1 - \cos(2x)}{2\sin x\tan x}. (i) Rewrite j(x)j(x) as a single natural logarithm without negative exponents in any part of the expression. (ii) Rewrite k(x)k(x) as an expression in which cos⁡x\cos x appears exactly once and no other trigonometric functions are involved. C The function mm is given by m(x)=2sin⁡2x−sin⁡x−1m(x) = 2\sin^2 x - \sin x - 1. Find all input values in the domain of mm that yield an output value of 0.

A (i) 52x−1=125=535^{2x - 1} = 125 = 5^3, so 2x−1=32x - 1 = 3 and x=2x = 2.

A (ii) arccos⁡(x3)=2π3\arccos\left(\frac{x}{3}\right) = \frac{2\pi}{3}, so x3=cos⁡2π3=−12\frac{x}{3} = \cos\frac{2\pi}{3} = -\frac{1}{2} and x=−32x = -\frac{3}{2}.

B (i) j(x)=ln⁡(x3)−ln⁡(4x)+ln⁡(4x2)=ln⁡(x3⋅4x24x)=ln⁡(x4)j(x) = \ln(x^3) - \ln(4x) + \ln(4x^2) = \ln\left(\frac{x^3 \cdot 4x^2}{4x}\right) = \ln(x^4), for x>0x > 0

B (ii) 1−cos⁡(2x)=1−(1−2sin⁡2x)=2sin⁡2x1 - \cos(2x) = 1 - (1 - 2\sin^2 x) = 2\sin^2 x, so k(x)=2sin⁡2x2sin⁡x⋅sin⁡xcos⁡x=2sin⁡2xcos⁡x2sin⁡2x=cos⁡xk(x) = \frac{2\sin^2 x}{2\sin x \cdot \frac{\sin x}{\cos x}} = \frac{2\sin^2 x\cos x}{2\sin^2 x} = \cos x, on the original domain sin⁡x≠0\sin x \ne 0 and cos⁡x≠0\cos x \ne 0

C 2sin⁡2x−sin⁡x−1=(2sin⁡x+1)(sin⁡x−1)=02\sin^2 x - \sin x - 1 = (2\sin x + 1)(\sin x - 1) = 0, so sin⁡x=−12\sin x = -\frac{1}{2} or sin⁡x=1\sin x = 1. Then x=7π6+2nπx = \frac{7\pi}{6} + 2n\pi, x=11π6+2nπx = \frac{11\pi}{6} + 2n\pi or x=π2+2nπx = \frac{\pi}{2} + 2n\pi, where nn is any integer.

How it scores

A(i): solution1/1Both sides written as powers of 5, then the exponents set equal. x=log⁡5125+12x = \frac{\log_5 125 + 1}{2} is correct too, but it would have to be evaluated to 2.
A(ii): solution1/1Cosine undoes arccosine, and cos⁡2π3\cos\frac{2\pi}{3} is evaluated exactly. A solution exists because 2π3\frac{2\pi}{3} is inside arccosine’s range, [0,π][0, \pi].
B(i): single logarithm1/1One natural log, no negative exponents. 4ln⁡x4\ln x is equal for x>0x > 0 but isn’t the form asked for.
B(ii): one cosine1/1Uses the double-angle identity cos⁡(2x)=1−2sin⁡2x\cos(2x) = 1 - 2\sin^2 x, which is tested because it comes from the sum identity.
C: first point1/1Seeing a quadratic in sin⁡x\sin x and factoring it, the same move as 2025’s quadratic in exe^x.
C: every solution1/1All three families of solutions. sin⁡x=−12\sin x = -\frac{1}{2} has two solutions in each cycle, one in the third quadrant and one in the fourth, and dropping either loses this point.

Why it works

  • Before you write the last line of part C, count. sin⁡x=1\sin x = 1 has one solution per cycle and sin⁡x=−12\sin x = -\frac{1}{2} has two, so the answer needs three families.

Lessons on the algebra question 4 tests. For the identities and exact values you need from memory, see the AP Precalculus formula sheet.

Writing reasons that earn the point

The points for explaining in words are among the least earned on the exam: question 1 C(ii) (15% and 30% in 2024 and 2025), question 2’s last parts (4% to 8%) and question 3 C(ii) (38% and 51%). The scoring guidelines reward the same pattern each time, so it can be learned.

  1. Name the property in the course’s words: concave down, increasing on its whole domain, proportional over equal-length input intervals, the secant line.
  2. Point at this function. Give the specific values or the interval: f(−3)=f(0)=f(3)=1f(-3) = f(0) = f(3) = 1, or “on 0<t<40 < t < 4,” or 714=3.57=0.5\frac{7}{14} = \frac{3.5}{7} = 0.5.
  3. Finish the thought. End with the claim you were asked to support, joined by “so” or “because.”
Reasons that do and don’t earn the point
QuestionDoesn’t earn itEarns it
2024 Q1 C(ii): does ff have an inverse?“ff is not one-to-one.” “ff fails the horizontal line test.”“f(−3)=f(0)=f(3)=1f(-3) = f(0) = f(3) = 1, so the output 1 comes from three inputs, and ff has no inverse function.”
2025 Q1 C(ii): why exponential?“An exponential regression fits best.”“The inputs go up by 1 and each output is half the one before: 714=3.57=1.753.5=0.5\frac{7}{14} = \frac{3.5}{7} = \frac{1.75}{3.5} = 0.5.”
2026 Q1 C(ii): is ff invertible?“ff passes the horizontal line test.”“ff is increasing on its domain, so no two points on its graph have the same yy-coordinate.”
2025 Q2 B(iii): why are the estimates too low?“DD is increasing, so the line is below it.”“AtA_t is on the secant line through (0,D(0))(0, D(0)) and (4,D(4))(4, D(4)), and the graph of DD is concave down on 0<t<40 < t < 4, so that line is below the graph.”
2025 Q2 C: where does the model stop?“The model can’t go on forever.”“The total number of plays can never decrease, but DD reaches its maximum at t=11t = 11 and decreases after that, so the domain of DD ends at t=11t = 11.”
2024 Q3 C(ii): how is the rate of change changing?“Increasing at a decreasing rate,” with no subject“The graph of hh is concave down on (t1,t2)(t_1, t_2), so the rate of change of hh is decreasing.”
Our wording, following the decision rules in the 2024–2026 scoring guidelines. The left column shows the kinds of answers those rules say are not enough; it doesn’t quote real students.

The reports add two pieces of advice. Name things instead of writing “it”: the 2024 report suggests teachers show students vague responses, full of “it,” next to precise ones. And reuse the course’s vocabulary (input and output values, rate of change, average rate of change, concave up and down, secant line) because the guidelines are written in it.

What the instruction words ask for

Task words in AP Precalculus free-response questions
Word in the questionWhat it asks for
Determine, find, identifyAn answer by any correct method; show work when the part asks for it
SolveAll the solutions of an equation
RewriteAn equivalent expression in the form the question names
EstimateAn approximate value from a model or a rate, with the work that produced it
Explain, give a reason, justifyReasons drawn from the function or the story that support your answer
InterpretWhat a number means in the story, with units
DescribeThe behavior in words, with the function named as the subject
Express, indicateThe result in the form asked for, such as limit notation, a decimal or units
Our plain-language versions of the task verbs listed in the CED (fall 2026), p. 132.

What changes in May 2027

The course content is the same. The College Board changed the multiple-choice counts, timing and question 2, and made a minor wording change to sample question 1 C(i) without changing its scoring.

  • Each free-response part is 35 minutes instead of 30, so you have about 17½ minutes a question instead of 15.
  • Question 2 is restructured. In the updated sample, the secant-line explanation, B(iii), is gone, B(ii) asks what the average rate of change means in context, and part C is worth two points. The details are in the question 2 section.
  • Multiple choice changed too. Part A is 29 questions in 65 minutes (it was 28 in 80), and Part B is 13 questions in 40 minutes (it was 12).

The released questions from 2024 to 2026 are still the closest practice to the real exam for questions 1, 3 and 4, and for most of question 2. For question 2’s new parts, use the College Board’s updated sample in the Course and Exam Description and the practice question above. For the rest of the exam, see the 2027 exam format.

Practice with the released questions

All twelve released questions are free on AP Central with their scoring guidelines, and our solutions go through every point. Do them on paper with a 35-minute timer for each pair, then score yourself with the official guidelines, strictly. The guidelines show which wording earns a point; be strictest with yourself on the explanation points.

Released AP Precalculus free-response questions
YearQuestion 1Question 2Question 3Question 4
2026 solutionsIncreasing ff from a graph; logarithmic ggA car’s value, exponential modelA waterwheele2xe^{2x}, log⁡2(5x)\log_2(5x), 7(3x+1)⋅7x7^{(3x+1)} \cdot 7^x, tan⁡2(3x)=1\tan^2(3x) = 1
2025 solutionsHalving ff from a table; cubic ggSong plays, quadratic modelA vibrating guitar stringlog⁡3x\log_3 x, cos⁡2x\cos^2 x, a quadratic in exe^x
2024 solutionsff from a graph; exponential ggVideo game sales, logarithmic modelA rolling tireex+3e^{x+3}, arcsine, logs, cos⁡−1(tan⁡2x)=0\cos^{-1}(\tan 2x) = 0
Official PDFs on AP Central. Questions: 2026, 2025, 2024. Scoring guidelines: 2026, 2025, 2024.

Try the 2026 questions first, mark them with the official guidelines, and record each question’s score below. The 2025 and 2024 solution pages have the same score tracker for their questions.

Record your 2026 free-response scores

Turn your points into an estimated AP score with the score calculator. For the formulas, identities and exact values to know by heart, use the formula sheet and the AP Precalculus cheat sheets.

Sources

  1. AP Precalculus Course and Exam Description, fall 2026: exam format, free-response task models, task verbs, sample questions and scoring (PDF)
  2. AP Precalculus Clarifications and Corrections, fall 2026: changes for the May 2027 exam (PDF)
  3. AP Precalculus Exam: format, timing and weights (AP Central)
  4. Graphing Calculators for AP Precalculus (AP Central)
  5. 2026 AP Precalculus Free-Response Questions (PDF)
  6. 2026 AP Precalculus Scoring Guidelines (PDF)
  7. 2025 AP Precalculus Free-Response Questions (PDF)
  8. 2025 AP Precalculus Scoring Guidelines (PDF)
  9. 2025 AP Precalculus Chief Reader Report (PDF)
  10. 2025 AP Precalculus Scoring Statistics (PDF)
  11. 2024 AP Precalculus Free-Response Questions (PDF)
  12. 2024 AP Precalculus Scoring Guidelines (PDF)
  13. 2024 AP Precalculus Chief Reader Report (PDF)
  14. 2024 AP Precalculus Scoring Statistics (PDF)
  15. AP Precalculus past exam questions, 2024–2026 (AP Central)