AP Precalculus -1.3 Rates of Change in Linear and Quadratic Functions- FRQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.3 Rates of Change in Linear and Quadratic Functions- FRQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.3 Rates of Change in Linear and Quadratic Functions- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

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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

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.

AP Precalculus -1.3 Rates of Change in Linear and Quadratic Functions- FRQ Exam Style Questions - Effective Fall 2023