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AP Precalculus -1.2 Rates of Change- FRQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.2 Rates of Change- FRQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.2 Rates of Change- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – FRQ Exam Style Questions- All Topics

Question 

A student won $500$ in an art contest. At first, the student kept the money in a desk. After $10$ months, the student deposited the money in a savings account that earned interest. Six months after depositing the money ($t = 6$), the amount in the account is $508.67$. Twelve months after depositing the money ($t = 12$), the amount in the account is $517.50$.
The amount of money the student has can be modeled by the piecewise function $M$ given by: $$M(t) = \begin{cases} 500 & \text{for } -10 \le t < 0 \\ ab^{(t/12)} & \text{for } t \ge 0 \end{cases}$$ where $M(t)$ is the amount, in dollars, at time $t$ months since the $500$ was deposited into the savings account. A negative value for $t$ represents the number of months before the student deposited the $500$ into the savings account.

Part A

(i) Use the given data to write two equations that can be used to find the values for constants $a$ and $b$ in the expression for $M(t)$.
(ii) Find the values for $a$ and $b$ as decimal approximations.

Part B

(i) Use the given data to find the average rate of change of the amount of money the student has, in dollars per month, from $t = -2$ to $t = 12$ months. Express your answer as a decimal approximation. Show the computations that lead to your answer.
(ii) Use $M(12)$ and the average rate of change found in (i) to estimate the amount of money, in dollars, the student has when $t = 20$ months. Show the work that leads to your answer.
(iii) Let $A(t)$ be the estimate of the amount of money, in dollars, the student has at time $t$ months using the average rate of change found in (i). If $A(t)$ is used to estimate values for $M(t)$ for $t > 12$, the error in the estimates will increase as $t$ increases. Explain why this is true.

Part C

The student plans to close the account when the amount of money in the account reaches $565$. Explain how this information can be used to determine the domain limitations for the model $M$.
▶️ Answer/Explanation
Detailed solution

Part A

(i)
Using the data $M(6) = 508.67$ and $M(12) = 517.50$:
$ab^{(6/12)} = 508.67$ (or $ab^{0.5} = 508.67$)
$ab^{(12/12)} = 517.50$ (or $ab = 517.50$)

(ii)
Divide the second equation by the first: $\frac{ab}{ab^{0.5}} = \frac{517.50}{508.67}$
$b^{0.5} \approx 1.017359…$
$b \approx (1.017359…)^2 \approx 1.035019…$
Using $ab = 517.50 \implies a = \frac{517.50}{1.035019…} \approx 500$
Final values: $a \approx 500.00$ and $b \approx 1.035$

Part B

(i)
$t = -2$ falls in the interval $-10 \le t < 0$, so $M(-2) = 500$.
$t = 12$ is given as $M(12) = 517.50$.
Average Rate of Change $= \frac{M(12) – M(-2)}{12 – (-2)}$
$= \frac{517.50 – 500}{12 + 2}$
$= \frac{17.50}{14} = 1.25$ dollars per month.

(ii)
The linear estimate $A(t)$ uses the point $(12, 517.50)$ and slope $1.25$.
$A(20) = M(12) + 1.25(20 – 12)$
$A(20) = 517.50 + 1.25(8)$
$A(20) = 517.50 + 10 = 527.50$ dollars.

(iii)
The model $M(t)$ for $t \ge 0$ is an exponential function ($b > 1$), which is concave up.
The estimate $A(t)$ is a linear function (a secant line).
Since $M(t)$ is increasing at an increasing rate (exponential growth), the linear model will fall further behind the actual values as $t$ increases.

Part C

The model is only valid as long as the account is open.
Setting $M(t) = 565$ allows us to solve for the maximum value of $t$.
$500(1.035)^{(t/12)} = 565$
This value of $t$ serves as the upper bound (maximum) for the domain of the model $M$.

Question 

The figure shows a robotic arm rotating in a circular counterclockwise direction that completes one rotation every $2$ seconds. Point $S$ is on the tip of the arm, and point $X$ does not move. As the arm rotates at a constant speed, the height of $S$ above $X$ periodically increases and decreases. At time $t = 0$ seconds, $S$ is at its lowest position, $6$ inches directly below $X$. At its highest position, $S$ is $20$ inches directly above $X$.
The sinusoidal function $h$ models the height of $S$ above $X$, in inches, as a function of time $t$, in seconds. A positive value of $h(t)$ indicates $S$ is above $X$; a negative value of $h(t)$ indicates $S$ is below $X$.
Part A
The graph of $h$ and its dashed midline for two full cycles is shown. Five points, $F, G, J, K,$ and $P$, are labeled on the graph. Determine possible coordinates $(t, h(t))$ for the five points.
Part B
The function $h$ can be written in the form $h(t) = a \cos(b(t + c)) + d$. Find values of constants $a, b, c,$ and $d$.
Part C
Refer to the graph of $h$ in part (A). The $t$-coordinate of $K$ is $t_1$, and the $t$-coordinate of $P$ is $t_2$.
(i) On the interval $(t_1, t_2)$, which of the following is true about $h$?
      a. $h$ is positive and increasing.
      b. $h$ is positive and decreasing.
      c. $h$ is negative and increasing.
      d. $h$ is negative and decreasing.
(ii) Describe how the rate of change of $h$ is changing on the interval $(t_1, t_2)$.
▶️ Answer/Explanation
Detailed solution

Part A: Coordinates of Points

At $t = 0$, $S$ is at its minimum height $-6$ (below $X$).
The maximum height is $20$.
The midline $d = \frac{20 + (-6)}{2} = 7$.
The period is $2$ seconds.
The graph starts at a minimum at $t = 0$, reaches midline at $t = 0.5$, maximum at $t = 1$, midline at $t = 1.5$, and minimum at $t = 2$.
Based on the visual positions in the provided graph:
$F$ (first maximum): $(1, 20)$
$G$ (midline, decreasing): $(1.5, 7)$
$J$ (minimum): $(2, -6)$
$K$ (midline, increasing): $(2.5, 7)$
$P$ (second maximum): $(3, 20)$

Part B: Finding Constants

$a$ (Amplitude) $= \frac{20 – (-6)}{2} = 13$. Since we use $\cos$ and start at a minimum, $a = -13$ (or use a phase shift).
$d$ (Vertical shift/Midline) $= 7$.
$b$ (Frequency factor) $= \frac{2\pi}{\text{period}} = \frac{2\pi}{2} = \pi$.
$c$ (Phase shift): For $h(t) = a \cos(b(t+c)) + d$, if $a = -13$, then at $t=0$, $-13\cos(b(0+c))+7 = -6 \implies \cos(bc)=1 \implies c = 0$.
Final values: $a = -13, b = \pi, c = 0, d = 7$.

Part C: Interval Analysis

(i) At $K$, $h(t)=7$ and is increasing. At $P$, $h(t)=20$ (maximum).
On $(t_1, t_2)$, the height is between $7$ and $20$, so it is positive.
The graph is moving from the midline up to the peak, so it is increasing.
Correct Option: a

(ii) On the interval $(t_1, t_2)$, the graph is concave down as it approaches the maximum.
Therefore, the rate of change of $h$ (the slope) is decreasing.
It starts at its maximum positive value at $K$ and decreases toward zero at $P$.

Question 

A student won $500$ in an art contest. At first, the student kept the money in a desk. After $10$ months, the student deposited the money in a savings account that earned interest. Six months after depositing the money ($t = 6$), the amount in the account is $508.67$. Twelve months after depositing the money ($t = 12$), the amount in the account is $517.50$.
The amount of money the student has can be modeled by the piecewise function $M$ given by: $$M(t) = \begin{cases} 500 & \text{for } -10 \le t < 0 \\ ab^{(t/12)} & \text{for } t \ge 0 \end{cases}$$ where $M(t)$ is the amount, in dollars, at time $t$ months since the $500$ was deposited into the savings account. A negative value for $t$ represents the number of months before the student deposited the $500$ into the savings account.

Part A

(i) Use the given data to write two equations that can be used to find the values for constants $a$ and $b$ in the expression for $M(t)$.
(ii) Find the values for $a$ and $b$ as decimal approximations.

Part B

(i) Use the given data to find the average rate of change of the amount of money the student has, in dollars per month, from $t = -2$ to $t = 12$ months. Express your answer as a decimal approximation. Show the computations that lead to your answer.
(ii) Use $M(12)$ and the average rate of change found in (i) to estimate the amount of money, in dollars, the student has when $t = 20$ months. Show the work that leads to your answer.
(iii) Let $A(t)$ be the estimate of the amount of money, in dollars, the student has at time $t$ months using the average rate of change found in (i). If $A(t)$ is used to estimate values for $M(t)$ for $t > 12$, the error in the estimates will increase as $t$ increases. Explain why this is true.

Part C

The student plans to close the account when the amount of money in the account reaches $565$. Explain how this information can be used to determine the domain limitations for the model $M$.
▶️ Answer/Explanation
Detailed solution

Part A

(i)
Using the data $M(6) = 508.67$ and $M(12) = 517.50$:
$ab^{(6/12)} = 508.67$ (or $ab^{0.5} = 508.67$)
$ab^{(12/12)} = 517.50$ (or $ab = 517.50$)

(ii)
Divide the second equation by the first: $\frac{ab}{ab^{0.5}} = \frac{517.50}{508.67}$
$b^{0.5} \approx 1.017359…$
$b \approx (1.017359…)^2 \approx 1.035019…$
Using $ab = 517.50 \implies a = \frac{517.50}{1.035019…} \approx 500$
Final values: $a \approx 500.00$ and $b \approx 1.035$

Part B

(i)
$t = -2$ falls in the interval $-10 \le t < 0$, so $M(-2) = 500$.
$t = 12$ is given as $M(12) = 517.50$.
Average Rate of Change $= \frac{M(12) – M(-2)}{12 – (-2)}$
$= \frac{517.50 – 500}{12 + 2}$
$= \frac{17.50}{14} = 1.25$ dollars per month.

(ii)
The linear estimate $A(t)$ uses the point $(12, 517.50)$ and slope $1.25$.
$A(20) = M(12) + 1.25(20 – 12)$
$A(20) = 517.50 + 1.25(8)$
$A(20) = 517.50 + 10 = 527.50$ dollars.

(iii)
The model $M(t)$ for $t \ge 0$ is an exponential function ($b > 1$), which is concave up.
The estimate $A(t)$ is a linear function (a secant line).
Since $M(t)$ is increasing at an increasing rate (exponential growth), the linear model will fall further behind the actual values as $t$ increases.

Part C

The model is only valid as long as the account is open.
Setting $M(t) = 565$ allows us to solve for the maximum value of $t$.
$500(1.035)^{(t/12)} = 565$
This value of $t$ serves as the upper bound (maximum) for the domain of the model $M$.

Question 

After Mr. Passwater missed a day of school, a rumor began to spread that he had won the Powerball lottery and moved to Japan. Initially, seven students knew about the rumor (they were the ones that started it!). After two hours (\(t=2\)), a total of 15 students had heard the rumor. After six hours (\(t=6\)), 67 students had heard the rumor.
The number of students that have heard the rumor can be modeled by the piecewise function \(R\) given by: $$R(t) = \begin{cases} 7(a)^{t/2} & \text{for } 0 \le t < 6 \\ -213.29 + b \ln t & \text{for } t \ge 6 \end{cases}$$ where \(R(t)\) is the number of students that have heard the rumor at time \(t\) hours since the rumor first began.

(A) (i) Use the given data to write two equations that can be used to find the values for constants \(a\) and \(b\) in the expression for \(R(t)\).
        (ii) Find the values for \(a\) and \(b\) as decimal approximations.

(B) (i) Use the given data to find the average rate of change in the number of students that have heard the rumor, in students per hour, from \(t=2\) to \(t=6\) hours. Express your answer as a decimal approximation. Show the computations that lead to your answer.
        (ii) Interpret the meaning of your answer from (i) in the context of the problem.
        (iii) Consider the values that result from using the average rate of change found in (i) to estimate the number of students that have heard the rumor for times \(t=p\) hours, where \(0 < p < 6\). Are these estimates less than or greater than the number of students predicted by the model \(R\) for times \(t=p\) hours? Explain your reasoning using characteristics of the average rate of change and characteristics of the model \(R\).

(C) The model \(R\) is valid for \(0 \le t \le 12\) hours. Explain how the range values of the function \(R\) should be limited by the context of the problem.
▶️ Answer/Explanation
Detailed solution

Part (A)

(i) Writing the equations:
We are given the following data points:
• At \(t=2\), \(R(2) = 15\).
• At \(t=6\), \(R(6) = 67\).

For \(t=2\), since \(0 \le 2 < 6\), we use the first part of the piecewise function: \(R(t) = 7(a)^{t/2}\).
$$15 = 7(a)^{2/2} \quad \Rightarrow \quad 15 = 7a^1$$
Equation 1: \(15 = 7a\)

For \(t=6\), since \(t \ge 6\), we use the second part of the piecewise function: \(R(t) = -213.29 + b \ln t\).
Equation 2: \(67 = -213.29 + b \ln(6)\)

(ii) Finding the values for \(a\) and \(b\):
From Equation 1:
$$a = \frac{15}{7} \approx 2.1428$$
From Equation 2:
$$67 + 213.29 = b \ln(6)$$
$$280.29 = b \ln(6)$$
$$b = \frac{280.29}{\ln(6)} \approx \frac{280.29}{1.79176} \approx 156.4328$$
Answer: \(a \approx 2.143\), \(b \approx 156.433\)

Part (B)

(i) Average Rate of Change:
The formula for the average rate of change from \(t=2\) to \(t=6\) is:
$$\text{Avg Rate} = \frac{R(6) – R(2)}{6 – 2}$$
Substituting the given values (\(R(6)=67\) and \(R(2)=15\)):
$$\text{Avg Rate} = \frac{67 – 15}{4} = \frac{52}{4} = 13$$
Answer: 13 students per hour.

(ii) Interpretation:
On average, the number of students who have heard the rumor increases by 13 students per hour between the 2nd hour and the 6th hour.

(iii) Estimates vs. Model Prediction:
Answer: The estimates are greater than the number of students predicted by the model.
Reasoning:
• On the interval \(0 < t < 6\), the function \(R(t) = 7(a)^{t/2}\) is an exponential growth function with a base greater than 1.
• Exponential growth functions are concave up (the rate of change is increasing).
• The average rate of change corresponds to the slope of the secant line connecting the points at \(t=2\) and \(t=6\).
• For a concave up curve, the secant line lies above the curve on the interval between the two points. Therefore, linear estimates based on the average rate (secant line) will be greater than the actual function values.

Part (C)

Explanation:
The range values (outputs) of \(R(t)\) represent the number of students. In the context of the problem, this range must be limited in two ways:
1. Population Cap: The number of students who heard the rumor cannot exceed the total student population of the school.
2. Discrete Values: You cannot have a fraction of a student, so strictly speaking, the context implies the range should consist of whole numbers (non-negative integers).

Question 

A market analyst working for a small appliance manufacturer finds that if the firm produces and sells \(x\) blenders annually, the total profit (in dollars) is \(P(x) = -0.0013x^3 + 0.3507x^2 – 0.4591x – 421.888\). (4 marks each part)
a. Use a graphing device to help graph the polynomial function \(P\).
b. Find the average rate of change of \(P\) between two relative (local) extrema when \(0 < x < 200\).
c. Find the equation of the secant line of the graph of \(P\) between the two points in part b.
d. The inflection point of a function is the point on the graph where the graph changes concavity. Find out the inflection point of the graph of \(P\).
e. How will the rate of change vary before and after the inflection point?
▶️ Answer/Explanation
Detailed solution

a. Graphing the function
Using a graphing utility for \(P(x) = -0.0013x^3 + 0.3507x^2 – 0.4591x – 421.888\) on the interval \([0, 200]\) reveals a cubic curve shape.
The graph starts with a slight dip to a local minimum near \(x=0\), then rises steeply to a local maximum near \(x=180\), before falling again.

b. Average rate of change between extrema
First, find the derivative: \(P'(x) = -0.0039x^2 + 0.7014x – 0.4591\).
Set \(P'(x) = 0\) and use the quadratic formula to find the extrema: \(x \approx 0.66\) (local min) and \(x \approx 179.19\) (local max).
Calculate the profit at these points: \(P(0.66) \approx -422.04\) and \(P(179.19) \approx 3276.57\).
The average rate of change is: \(\frac{3276.57 – (-422.04)}{179.19 – 0.66} = \frac{3698.61}{178.53} \approx 20.72\).

c. Equation of the secant line
The slope \(m\) was found in part (b) to be approximately \(20.72\).
Using the point-slope form \(y – y_1 = m(x – x_1)\) with the minimum point \((0.66, -422.04)\):
\(y – (-422.04) = 20.72(x – 0.66)\)
\(y = 20.72x – 13.68 – 422.04\)
The equation is approximately \(y = 20.72x – 435.72\).

d. Inflection point
Find the second derivative: \(P”(x) = -0.0078x + 0.7014\).
Set \(P”(x) = 0\) to find the change in concavity: \(0 = -0.0078x + 0.7014 \Rightarrow x = \frac{0.7014}{0.0078} \approx 89.92\).
Find the corresponding y-value: \(P(89.92) \approx 1427.27\).
The inflection point is approximately \((89.92, 1427.27)\).

e. Variation of rate of change
The rate of change is represented by the derivative, \(P'(x)\).
Before the inflection point (\(x < 89.92\)), the graph is concave up (\(P”(x) > 0\)), so the rate of change is increasing.
After the inflection point (\(x > 89.92\)), the graph is concave down (\(P”(x) < 0\)), so the rate of change is decreasing.

Question 

The Chinese bamboo tree exhibits a unique growth pattern. Once a seed has been planted, the Chinese bamboo tree does not break through the ground for several years. However, once it breaks through the ground, the Chinese bamboo tree grows exponentially. In one particular experiment, a group of biologists recorded the height of a Chinese bamboo tree once it broke through the ground. After one week (\(t = 1\)), the Chinese bamboo tree measured 3 feet, and after five weeks (\(t = 5\)), the same tree measured 89 feet.
The height of the Chinese bamboo tree can be modeled by the function \(H\) given by \(H(t) = ab^x\), where \(H(t)\) is the height of the tree, in feet, \(t\) weeks after it first breaks ground.
(A) (i) Use the given data to write two equations that can be used to find the values for constants \(a\) and \(b\) in the expression for \(H(t)\).
(ii) Find the values for \(a\) and \(b\).
(B) (i) Use the given data to find the average rate of change of the height of the Chinese bamboo tree, in feet per week, from \(t = 1\) to \(t = 5\) weeks. Express your answer as a decimal approximation. Show the computations that lead to your answer.
(ii) Interpret the meaning of your answer from (i) in the context of the problem.
(iii) Consider the average rates of change of \(H\) from \(t = 5\) to \(t = p\) weeks, where \(p > 5\). Are these average rates of change less than or greater than the average rate of change from \(t = 1\) to \(t = 5\) weeks found in (i)? Explain your reasoning.
(C) For which \(t\)-value, \(t = 4\) weeks or \(t = 11\) weeks, should the biologists have more confidence in when using the model \(H\)? Give a reason for your answer in the context of the problem.
▶️ Answer/Explanation
Detailed solution

(A)(i) Equations
Substituting the points \((1, 3)\) and \((5, 89)\) into \(H(t) = ab^t\):
1. \(3 = ab^1\) (or \(3 = ab\))
2. \(89 = ab^5\)

(A)(ii) Values for a and b
Dividing equation 2 by equation 1: \(\frac{ab^5}{ab} = \frac{89}{3} \implies b^4 = 29.67\).
Solving for \(b\): \(b = (29.67)^{0.25} \approx 2.33\).
Solving for \(a\): \(a = \frac{3}{2.33} \approx 1.29\).

(B)(i) Average Rate of Change
\(\text{Rate} = \frac{H(5) – H(1)}{5 – 1} = \frac{89 – 3}{4} = \frac{86}{4} = 21.5\)
Answer: 21.5 feet per week.

(B)(ii) Interpretation
The answer indicates that between the first and fifth weeks, the bamboo tree grew at an average speed of 21.5 feet per week.

(B)(iii) Comparison
Greater. The function represents exponential growth (\(b > 1\)), which is concave up. This means the rate of growth increases over time, so the rate after week 5 will be steeper than the rate before week 5.

(C) Confidence
\(t = 4\) weeks.
The biologists should be more confident in \(t=4\) because it is an interpolation (within the observed data range). \(t=11\) is an extrapolation; biological growth cannot remain exponential indefinitely, so the model is likely inaccurate that far out.

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