Free-Response Strategies: the Four Question Types

📝 Free-Response Strategies: the Four Question Types

Part 1 of 7 — The Free-Response Section and How Points Are Earned

Topic Overview

The 2027 Free-Response Section

Section II is 4 questions in 70 minutes and counts for 37.5% of the exam score. The four questions always play the same four roles, in the same order:

Each question is scored out of 6 points, in three lettered parts (A), (B), (C), most of which split into (i) and (ii). A point is tied to one specific thing: a value, a justification, an interpretation. You never lose a point for an error in a different part, and you can earn a later part with a wrong earlier value if your method is right.

Key Fact: Four questions, four roles, 6 points each. Knowing the role of a question tells you what it will ask before you read it.

What Earns the Point

What a Point Is Not

• A point is not awarded for a correct number written next to an unrelated or wrong reason when the part says justify or explain.
• A point is not awarded for a correct unlabeled answer buried in scratch work: label the part and box or underline the answer.
• Crossed-out work is not read. If you change your mind, cross out and rewrite.

Format and conventions check. 🎯

Scoring conventions. 🔽

Key Takeaways — Part 1

• Four questions, four fixed roles: Function Concepts, Non-Periodic Modeling (calculator, Part A, 35 min); Periodic Modeling, Symbolic Manipulations (no calculator, Part B, 35 min)
• Each question is 6 points; points are earned part by part, so attempt every part
• Part A decimals: three decimal places, model outputs are approximately; Part B answers are exact
• Rates of change carry units and an interval; justifications come from the data or the model, not from how a graph looks

📝 Free-Response Strategies: the Four Question Types

Part 2 of 7 — Question 1: Function Concepts

The Shape of Question 1

Question 1 (Part A, graphing calculator) presents two functions in different forms: usually a table of selected values of one function and an algebraic rule for the other. It then asks you to combine, compare and describe them. Nothing in it needs regression; the calculator is for arithmetic and for checking.

Key Fact: Question 1 is where the reader checks whether you can read a table, use a rule, and say why. Every justification point is earned with the specific numbers that prove the claim.

Worked Question 1

The function $g$ is a polynomial function. Selected values of $g$ are given in the table. The function $h$ is given by $h(x) = \dfrac{2x + 6}{x - 4}$.

(A)(i) Find the value of $h(g(0))$.

Inside first, from the table: $g(0) = 5$. Then from the rule: $h(5) = \dfrac{2(5) + 6}{5 - 4} = \dfrac{16}{1} = 16$.

Answer: $h(g(0)) = 16$.

(A)(ii) Find the value of $h^{-1}(4)$.

$h^{-1}(4)$ is the input $x$ with $h(x) = 4$:

$\frac{2x + 6}{x - 4} = 4 \;\Rightarrow\; 2x + 6 = 4x - 16 \;\Rightarrow\; 22 = 2x \;\Rightarrow\; x = 11$

Answer: $h^{-1}(4) = 11$. (Check: $h(11) = \frac{28}{7} = 4$.)

(B)(i) Find the average rate of change of $g$ over the interval $0 \le x \le 6$.

$\frac{g(6) - g(0)}{6 - 0} = \frac{11 - 5}{6} = 1$

Answer: The average rate of change is $1$. (Show the quotient: the point is "answer with the difference quotient", not a bare number.)

(B)(ii) Explain why $g$ does not have an inverse function on its domain.

Answer: "From the table, $g(0) = 5$ and $g(4) = 5$. Two different inputs produce the same output, so $g$ is not one-to-one and has no inverse function."

Naming the two inputs and the shared output is what earns the point. "It fails the horizontal line test" without the numbers does not.

(C)(i) Determine the end behavior of $h$ with limit notation.

The degrees of numerator and denominator are equal, so the quotient of leading coefficients gives the horizontal asymptote:

$\lim_{x \to \infty} h(x) = 2 \quad \text{and} \quad \lim_{x \to -\infty} h(x) = 2$

Answer: $y = 2$ is a horizontal asymptote.

(C)(ii) Determine the vertical asymptote of $h$ and justify.

At $x = 4$ the denominator is $0$ and the numerator is $2(4) + 6 = 14 \ne 0$, so $x = 4$ is a vertical asymptote: $\lim_{x \to 4^+} h(x) = \infty$ (for $x$ slightly more than $4$, $\frac{14}{\text{small positive}}$) and $\lim_{x \to 4^-} h(x) = -\infty$.

Scoring guide

Key Fact: A vertical asymptote is justified by "denominator zero, numerator nonzero" or by a one-sided limit; a horizontal asymptote by the limit as $x \to \pm\infty$.

Question 1 practice. 🎯 Use the table of $g$ and $h(x) = \dfrac{2x + 6}{x - 4}$ from the worked question.

Features of $h(x) = \dfrac{2x + 6}{x - 4}$. ✍️

1) The vertical asymptote is $x =$ ?

2) The horizontal asymptote is $y =$ ?

3) The value of $h^{-1}(6)$ (the input $x$ with $h(x) = 6$):

Key Takeaways — Part 2

• Question 1 pairs a table with a rule: evaluate compositions inside-out, find inverse values by solving $h(x) = c$
• An average rate of change is a quotient with the interval shown, not a bare number
• "No inverse" is justified by two named inputs with the same output
• Asymptotes and end behavior are stated with limit notation and justified from the rule

📝 Free-Response Strategies: the Four Question Types

Part 3 of 7 — Question 2: Modeling a Non-Periodic Context

The Shape of Question 2

Question 2 (Part A, graphing calculator) gives data from a real context, usually a table, and asks you to build a linear, quadratic, cubic, exponential or logarithmic model with regression, then use it. The calculator does the regression; you do the choosing, the stating, the predicting and the interpreting.

Key Fact: The justification point in (A) is always about the numbers in the table: differences, second differences, or ratios. The words "it looks exponential" earn nothing.

Worked Question 2

The table gives the number of downloads $D$, in thousands, of a new app during week $t$ after its release.

(A)(i) Explain why an exponential model is appropriate.

Ratios of consecutive values: $\dfrac{5.2}{4.0} = 1.300$, $\dfrac{6.8}{5.2} \approx 1.308$, $\dfrac{8.8}{6.8} \approx 1.294$, $\dfrac{11.4}{8.8} \approx 1.295$, $\dfrac{14.8}{11.4} \approx 1.298$.

Answer: "Over equal one-week intervals, the output changes by a roughly constant ratio of about $1.30$, so the downloads grow proportionally and an exponential model is appropriate."

(By contrast, the first differences $1.2, 1.6, 2.0, 2.6, 3.4$ are not constant, so the model is not linear.)

(A)(ii) Use exponential regression to find a model $D(t) = a \cdot b^t$.

Calculator (ExpReg on $t$ and $D$):

$D(t) = 3.087\,(1.299)^t$

Store the full-precision model in $Y_1$. State it to three decimal places; compute with the stored version.

(B)(i) Use the model to predict the downloads in week $9$.

$Y_1(9) \approx 32.493$. Answer: "The model predicts approximately $32.493$ thousand downloads in week $9$."

(The rounded coefficients give $3.087(1.299)^9 \approx 32.510$: a different third decimal. This is why you compute with the stored model.)

(B)(ii) Find and interpret the average rate of change of $D$ over $2 \le t \le 6$.

$\frac{D(6) - D(2)}{6 - 2} = \frac{14.8 - 5.2}{4} = \frac{9.6}{4} = 2.4$

Answer: "Between week $2$ and week $6$, the number of downloads per week increased at an average rate of $2.4$ thousand downloads per week."

(C)(i) According to the model, during which week does $D$ first reach $50$ thousand?

Solve $Y_1 = 50$ on the calculator (intersection with $Y_2 = 50$, or $\ln$ algebra): $t \approx 10.648$.

Answer: "$D(t) = 50$ when $t \approx 10.648$, so weekly downloads first reach $50$ thousand during week $11$."

(C)(ii) Give a limitation of the model.

Answer: "The model assumes downloads keep growing by about $30\%$ every week. The number of people who could download the app is finite, so the growth must slow, and the model will overpredict for large $t$ (it gives about $71$ thousand in week $12$)."

Scoring guide

Key Fact: In (B)(ii), the rate is computed from the data when the question says "the table", and from the model when it says "the model". Read which.

Question 2 practice. 🎯 Use the download data and the model $D(t) = 3.087(1.299)^t$.

Reading the model. 🔽 The regression model is $D(t) = 3.087(1.299)^t$ and $D(3) \approx 6.765$.

Key Takeaways — Part 3

• Justify the model type with differences, second differences or ratios from the table
• State the regression to three decimals; predict with the stored model and the word approximately
• A rate of change is reported with units of output per input and its interval
• A limitation names the assumed pattern and a contextual reason it fails; "the model is only an estimate" earns nothing

📝 Free-Response Strategies: the Four Question Types

Part 4 of 7 — Question 3: Modeling a Periodic Context

The Shape of Question 3

Question 3 opens Part B: no calculator. It describes something that repeats (a rotating wheel or blade, a tide, a bouncing spring, a seasonal temperature) and gives a graph of the quantity, usually marking a maximum, a minimum and the period. Everything must be exact: $\pi$ stays $\pi$.

Key Fact: The period is the time for one complete cycle (maximum to the next maximum), not the time from a maximum to the next minimum, which is half a period.

Worked Question 3

The graph shows the height $h$, in meters above the ground, of the tip of a wind-turbine blade $t$ seconds after the tip passes its highest point. The tip is $120$ meters high at $t = 0$, $40$ meters high at $t = 3$, and back to $120$ meters at $t = 6$; the motion repeats every $6$ seconds.

(A)(i) State the midline, amplitude and period of $h$.

$\text{midline } h = \frac{120 + 40}{2} = 80, \qquad \text{amplitude} = \frac{120 - 40}{2} = 40, \qquad \text{period} = 6 \text{ seconds}$

(A)(ii) Write a sinusoidal function that models $h(t)$.

The graph is at its maximum at $t = 0$, so a cosine with no horizontal shift fits. $b = \dfrac{2\pi}{6} = \dfrac{\pi}{3}$.

$h(t) = 40\cos\left(\frac{\pi}{3}t\right) + 80$

Equivalent correct forms also earn the point: $h(t) = 40\sin\left(\frac{\pi}{3}(t + 1.5)\right) + 80$ or $h(t) = -40\cos\left(\frac{\pi}{3}(t - 3)\right) + 80$.

(B)(i) Find $h(1)$.

$h(1) = 40\cos\left(\frac{\pi}{3}\right) + 80 = 40 \cdot \frac{1}{2} + 80 = 100 \text{ meters}$

(B)(ii) Find all times $t$ in $0 \le t \le 6$ when the tip is $100$ meters high.

$40\cos\left(\frac{\pi}{3}t\right) + 80 = 100 \;\Rightarrow\; \cos\left(\frac{\pi}{3}t\right) = \frac{1}{2}$

On one cycle $0 \le \frac{\pi}{3}t < 2\pi$: $\dfrac{\pi}{3}t = \dfrac{\pi}{3}$ or $\dfrac{\pi}{3}t = \dfrac{5\pi}{3}$, so $t = 1$ or $t = 5$.

Answer: $t = 1$ second and $t = 5$ seconds. (Giving only $t = 1$ loses the point: the question says all times.)

(C) On the interval $3 < t < 4.5$, is $h$ increasing or decreasing, and is the rate of change of $h$ increasing or decreasing? Give a reason.

Answer: "On $3 < t < 4.5$ the tip rises from its minimum of $40$ meters toward the midline of $80$ meters, so $h$ is increasing. The graph is concave up there (the curve bends upward as it leaves the minimum), so the rate of change of $h$ is increasing: the tip is rising faster and faster."

Scoring guide

Key Fact: Between a minimum and the midline a sinusoid is increasing and concave up (rate increasing). Between the midline and a maximum it is increasing and concave down (rate decreasing).

Question 3 practice. 🎯 Use the graph of the blade-tip height above.

Exact values from the model $h(t) = 40\cos\left(\frac{\pi}{3}t\right) + 80$. ✍️

1) $h(2)$, in meters:

2) $h(4.5)$, in meters:

3) The first time $t > 0$, in seconds, at which the tip is $40$ meters high:

Key Takeaways — Part 4

• Read the graph: midline $\frac{\max + \min}{2}$, amplitude $\frac{\max - \min}{2}$, period $=$ one full cycle, $b = \frac{2\pi}{\text{period}}$
• A maximum at $t = 0$ suggests cosine; any equivalent form (shifted sine, negative cosine) earns the point
• "Find all times" means every solution in the interval; exact values only in Part B
• Describe change with two statements: increasing/decreasing and rate of change increasing/decreasing, each with a reason from the graph

📝 Free-Response Strategies: the Four Question Types

Part 5 of 7 — Question 4: Symbolic Manipulations

The Shape of Question 4

Question 4 closes Part B, no calculator. There is no context and no data: three short algebraic tasks, each worth 2 points, drawn from the whole course. Every answer is exact, and every solution of an equation must be checked against the domain of the original equation.

Key Fact: In a logarithmic equation, a solution that makes any original log argument $\le 0$ is extraneous. Reporting it loses the point even when the correct solution is also listed.

Worked Question 4

(A)(i) Solve $5 \cdot 3^{2x} = 45$.

$3^{2x} = 9 = 3^2 \;\Rightarrow\; 2x = 2 \;\Rightarrow\; x = 1$

(Equivalent: $2x = \log_3 9$, so $x = \frac{1}{2}\log_3 9 = 1$.)

(A)(ii) Solve $\log_2 x + \log_2(x - 2) = 3$.

$\log_2\big(x(x - 2)\big) = 3 \;\Rightarrow\; x^2 - 2x = 8 \;\Rightarrow\; (x - 4)(x + 2) = 0$

$x = 4$ or $x = -2$. The original equation needs $x > 0$ and $x - 2 > 0$, so $x = -2$ is extraneous.

Answer: $x = 4$ only. (Write the rejection: "$x = -2$ is rejected because $\log_2(-2)$ is undefined.")

(B)(i) Rewrite $2\ln x - \ln(x + 1)$ as a single logarithm.

$2\ln x - \ln(x + 1) = \ln x^2 - \ln(x + 1) = \ln\left(\frac{x^2}{x + 1}\right), \quad x > 0$

(B)(ii) Rewrite $k(x) = 4 \cdot 8^{x/3}$ in the form $a \cdot b^x$.

$8^{x/3} = \left(8^{1/3}\right)^x = 2^x, \qquad \text{so } k(x) = 4 \cdot 2^x$

(C)(i) Solve $\cos(2\theta) - \cos\theta = 0$ on $0 \le \theta < 2\pi$.

Replace $\cos(2\theta)$ with $2\cos^2\theta - 1$ so the equation has one function:

$2\cos^2\theta - \cos\theta - 1 = 0 \;\Rightarrow\; (2\cos\theta + 1)(\cos\theta - 1) = 0$

$\cos\theta = -\dfrac{1}{2}$ gives $\theta = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$; $\cos\theta = 1$ gives $\theta = 0$.

Answer: $\theta = 0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$.

(C)(ii) Given $\sin\theta = \dfrac{3}{5}$ and $\dfrac{\pi}{2} < \theta < \pi$, find $\sin(2\theta)$.

In Quadrant II, $\cos\theta < 0$: $\cos\theta = -\sqrt{1 - \frac{9}{25}} = -\dfrac{4}{5}$.

$\sin(2\theta) = 2\sin\theta\cos\theta = 2 \cdot \frac{3}{5} \cdot \left(-\frac{4}{5}\right) = -\frac{24}{25}$

Scoring guide

Key Fact: Choose the identity that leaves one trigonometric function: $\cos(2\theta) = 2\cos^2\theta - 1$ next to $\cos\theta$, $\cos(2\theta) = 1 - 2\sin^2\theta$ next to $\sin\theta$.

Question 4 practice. 🎯 No calculator.

Exact solving. ✍️

1) The solution of $2^{x + 3} = 32$:

2) The solution of $\log_3(x + 1) = 2$:

3) The number of solutions of $2\sin^2\theta - \sin\theta = 0$ on $0 \le \theta < 2\pi$:

Key Takeaways — Part 5

• Question 4 is three 2-point algebra tasks, no calculator, exact answers
• Exponential equations: match bases or take a logarithm; logarithmic equations: combine, convert, then check the domain and reject extraneous solutions in writing
• Rewriting means reaching the form asked for: single log, $a \cdot b^x$, one trigonometric function
• Trigonometric equations: use the identity that leaves one function, factor, and list every solution in the interval

📝 Free-Response Strategies: the Four Question Types

Part 6 of 7 — Common Point Losses and the 35 + 35 Minute Budget

Where Points Go

Most lost points on the free-response section are not lost to hard mathematics. They are lost to conventions: a short decimal, a missing word, a half period, an extra solution. Each row below is a point that students who knew the mathematics still lost.

Key Fact: Before leaving any part, ask three questions: Did I name the quantity? Did I give units and the interval? Did I answer the whole question (all solutions, both statements)?

A Student Response, Scored

Here is a student's response to the blade-tip question of Part 4 (maximum $120$ m at $t = 0$, minimum $40$ m at $t = 3$, repeating every $6$ seconds), with the reader's decisions.

Score: 3 of 6. The student understood sinusoids. Three points went to a half period, a missing second solution, and a concavity statement made without looking at the curve.

The 35 + 35 Minute Budget

Pacing rules that hold up under pressure:

1. Enter the Q2 data first, run the regression, store it. Every later part depends on it.
2. Every part is worth attempting. Parts rarely depend on each other, and a wrong value from (A) does not stop (B) from being scored on its own merits.
3. Spend at most 5 minutes stuck. Write what you know, label it, move on, come back.
4. Last two minutes of each section: check decimals, units, "approximately", and that every equation's solutions are all listed.

Point-loss check. 🎯

Exam habits. 🔽

Key Takeaways — Part 6

• The common losses are conventions: three decimals, approximately, units and interval, all solutions, domain check, full period, both halves of a rate-of-change description
• Points are whole and independent; a mistake in one part does not spread
• Part A: Q1 ≈ 15 min, Q2 ≈ 20 min (data entry first); Part B: Q3 ≈ 17 min, Q4 ≈ 18 min
• Finish each section with a two-minute convention check

📝 Free-Response Strategies: the Four Question Types

Part 7 of 7 — Mixed Mini-FRQ Set and Mastery Check

Two Mini-FRQs, Scored

Work each one on paper before reading the solution. Time yourself: about 8 minutes each.

Mini-FRQ A (Function Concepts style, calculator allowed)

The function $k$ is an exponential function. Selected values of $k$ are given in the table.

(A) Find the value of $k(5)$.

(B) Find the average rate of change of $k$ over the interval $1 \le x \le 3$.

(C)(i) Explain why $k$ has an inverse function. (ii) Find $k^{-1}(48)$.

Solution

(A) Consecutive ratios are $2$, so $k(x) = 3 \cdot 2^x$ and $k(5) = 3 \cdot 32 = 96$. (1 point)

(B) $\dfrac{k(3) - k(1)}{3 - 1} = \dfrac{24 - 6}{2} = 9$. (1 point, quotient shown)

(C)(i) "$k(x) = 3 \cdot 2^x$ is strictly increasing (each output is double the previous), so every output occurs for exactly one input: $k$ is one-to-one and has an inverse." (1 point)

(C)(ii) $3 \cdot 2^x = 48 \Rightarrow 2^x = 16 \Rightarrow x = 4$, so $k^{-1}(48) = 4$. (1 point)

Mini-FRQ B (Symbolic Manipulations style, no calculator)

(A) Solve $2\log_5 x = \log_5(x + 6)$.

(B) Rewrite $3 \cdot 9^x$ as a single power of $3$.

(C) Solve $\cos\theta = \sin\theta$ on $0 \le \theta < 2\pi$.

Solution

(A) $\log_5 x^2 = \log_5(x + 6) \Rightarrow x^2 = x + 6 \Rightarrow (x - 3)(x + 2) = 0$. The original equation needs $x > 0$, so $x = -2$ is rejected. $x = 3$. (2 points: solution and rejection)

(B) $9^x = (3^2)^x = 3^{2x}$, so $3 \cdot 9^x = 3^{2x + 1}$. (2 points)

(C) Where $\cos\theta = \sin\theta$, the point on the unit circle has equal coordinates: $\theta = \dfrac{\pi}{4}$ or $\theta = \dfrac{5\pi}{4}$. (Equivalently, dividing by $\cos\theta \ne 0$ gives $\tan\theta = 1$.) (2 points)

Key Fact: Each solution above contains the same ingredients every time: the computation, the check (domain, interval), and the answer in the form asked for.

Mastery check. 🎯 Use the table of $k$ from Mini-FRQ A where needed.

Final checks. ✍️

1) With $k(x) = 3 \cdot 2^x$, the value of $k^{-1}(48)$:

2) The number of solutions of $\cos\theta = \sin\theta$ on $0 \le \theta < 2\pi$:

3) The number of real solutions of $\log_2 x + \log_2(x - 2) = 3$ (after rejecting any extraneous value):

Key Takeaways — Free-Response Strategies

• 6 points per question, earned part by part; attempt every part
• Part A: three decimals, approximately, stored model; Part B: exact
• Every rate of change: units + interval; every justification: from the data or the model
• Budget 35 + 35 minutes: Q1 ≈ 15, Q2 ≈ 20, Q3 ≈ 17, Q4 ≈ 18, with a two-minute convention check at the end of each section