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

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

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

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The table gives the average rates of change for the functions $ f, g, h, $ and $ k $ for certain intervals of $ x $. Which of the functions is best modeled by a piecewise-linear function with two linear segments with different slopes?

(A) $ f $
(B) $ g $
(C) $ h $
(D) $ k $

▶️ Answer/Explanation

Detailed solution

For a piecewise-linear function with two linear segments, the average rate of change should be roughly constant in each segment but different between segments.

$ f $: rates ~constant ⇒ one linear function, not two segments.
$ g $: rates increase steadily ⇒ curved, not piecewise-linear.
$ h $: rates ~2.1 for intervals 1–3 (0<x<3), then ~4.2 for intervals 4–7 (3<x<7) ⇒ two different constant slopes ⇒ piecewise-linear with two segments.
$ k $: rates increase then decrease ⇒ not constant in two blocks.

✅ Answer: (C)

Question

The function $ f $ is defined for all real values of $ x $. For a constant $ a $, the average rate of change of $ f $ from $ x = a $ to $ x = a + 1 $ is given by the expression $ 2a + 1 $. Which of the following statements is true?

(A) The average rate of change of $ f $ over consecutive equal-length input-value intervals is positive, so the graph of $ f $ could be a line with a positive slope.
(B) The average rate of change of $ f $ over consecutive equal-length input-value intervals is positive, so the graph of $ f $ could be a parabola that opens up.
(C) The average rate of change of $ f $ over consecutive equal-length input-value intervals is increasing at a constant rate, so the graph of $ f $ could be a line with a positive slope.
(D) The average rate of change of $ f $ over consecutive equal-length input-value intervals is increasing at a constant rate, so the graph of $ f $ could be a parabola that opens up.

▶️ Answer/Explanation

Detailed solution

Average rate of change from $ a $ to $ a+1 $ is $ 2a + 1 $, which is linear in $ a $.
If $ f $ were linear, the average rate of change would be constant (not depend on $ a $), so (A) and (C) are false.
If $ f $ is quadratic, then the average rate of change over a 1-unit interval is indeed linear in $ a $ (this can be derived). Here $ 2a + 1 $ increases as $ a $ increases, at a constant rate (slope 2 with respect to $ a $), meaning $ f $ is concave up (opens upward).
Thus the correct reasoning: the average rate of change is increasing at a constant rate ⇒ $ f $ could be a parabola opening up.
✅ Answer: (D)

Question

The table gives values of the function $ f $ for selected values of $ x $. If the function $ f $ is linear, what is the value of $ f(13) $?

$ x $	-3	3	6
$ f(x) $	-10	-2	2

(A) 4
(B) $\frac{29}{4}$
(C) $\frac{28}{3}$
(D) $\frac{34}{3}$

▶️ Answer/Explanation

Detailed solution

Using points (3, -2) and (6, 2): slope $ m = \frac{2 – (-2)}{6 – 3} = \frac{4}{3} $.
Equation: $ f(x) – (-2) = \frac{4}{3}(x – 3) $ ⇒ $ f(x) = \frac{4}{3}x – 4 – 2 = \frac{4}{3}x – 6 $.
Check with (-3, -10): $ \frac{4}{3}(-3) – 6 = -4 – 6 = -10 $ ✓.
Now $ f(13) = \frac{4}{3}(13) – 6 = \frac{52}{3} – \frac{18}{3} = \frac{34}{3} $.
✅ Answer: (D)

Question

The average rate of change of the quadratic function $ p $ is $-4$ on the interval $ 0 \leq x \leq 2 $ and $-1$ on the interval $ 2 \leq x \leq 4 $. What is the average rate of change of $ p $ on the interval $ 6 \leq x \leq 8 $?

(A) 2
(B) 3
(C) 5
(D) The average rate of change on the interval $ 6 \leq x \leq 8 $ cannot be determined from the information given.

▶️ Answer/Explanation

Detailed solution

For a quadratic function $ p(x) = ax^2 + bx + c $, the average rate of change over equal-length intervals changes linearly.
Given:
Average rate on $ [0,2] $ is $ -4 $.
Average rate on $ [2,4] $ is $ -1 $.
The change from one interval to the next is constant: $ -1 – (-4) = 3 $.
Thus, for each subsequent interval of length 2, the average rate increases by 3.
Sequence of average rates:
$ [0,2]: -4 $
$ [2,4]: -1 $
$ [4,6]: -1 + 3 = 2 $
$ [6,8]: 2 + 3 = 5 $
✅ Answer: (C) 5

Question

An object is moving in a straight line from a starting point. The distance, in meters, from the starting point at selected times, in seconds, is given in the table. If the pattern is consistent, which of the following statements about the rate of change of the rates of change of distance over time is true?

(A) The rate of change of the rates of change is 0 meters per second, and the object is neither speeding up nor slowing down.
(B) The rate of change of the rates of change is 0 meters per second per second, and the object is neither speeding up nor slowing down.
(C) The rate of change of the rates of change is 4 meters per second, and the object is neither speeding up nor slowing down.
(D) The rate of change of the rates of change is 4 meters per second per second, and the object is speeding up.

▶️ Answer/Explanation

Detailed solution

Calculate the average rate of change (speed) over each time interval:
From $ t=1 $ to $ t=3 $: $ \frac{9-1}{3-1} = \frac{8}{2} = 4 $ m/s
From $ t=3 $ to $ t=6 $: $ \frac{21-9}{6-3} = \frac{12}{3} = 4 $ m/s
From $ t=6 $ to $ t=11 $: $ \frac{41-21}{11-6} = \frac{20}{5} = 4 $ m/s
The speed is constant at 4 m/s, so the acceleration (rate of change of speed) is 0 m/s².
Therefore, the rate of change of the rates of change is $ 0 $ m/s², and the object moves with constant speed (neither speeding up nor slowing down).
✅ Answer: (B)

Question

$x$

$-2$

$-1$

$0$

$1$

$f(x)$

$5$

$2$

$1$

$2$

The table gives values of a function $f$ for selected values of $x$. Which of the following conclusions with reason is consistent with the data in the table?

(A) $f$ could be a linear function because the rates of change over consecutive equal-length intervals in the table can be described by $y = 2x$.
(B) $f$ could be a linear function because the rates of change over consecutive equal-length intervals in the table can be described by $y = 2x + 1$.
(C) $f$ could be a quadratic function because the rates of change over consecutive equal-length intervals in the table can be described by $y = 2x$.
(D) $f$ could be a quadratic function because the rates of change over consecutive equal-length intervals in the table can be described by $y = 2x + 1$.

▶️ Answer/Explanation

Detailed solution

$
\begin{array}{c|cccc}
x & -2 & -1 & 0 & 1 \\
\hline
f(x) & 5 & 2 & 1 & 2
\end{array}
$

Step 1: Rates of change over equal intervals

All $x$-intervals have length $1$.

$[-2,-1] : \quad 2 – 5 = -3 $
$[-1,0]: \quad 1 – 2 = -1 $
$[0,1]: \quad 2 – 1 = 1$

So the rates of change are
$
-3,\,-1,\,1.
$

Step 2: Identify the pattern

The rates of change are not constant, so $f$ is not linear.

Now check whether the rates follow a linear rule.
Label each interval by its left endpoint $x=-2,-1,0$:

$
2x+1 =
\begin{cases}
2(-2)+1=-3, \\
2(-1)+1=-1, \\
2(0)+1=1.
\end{cases}
$

This matches the observed rates of change.

Step 3: Conclusion

Since the first differences follow a linear pattern, $f$ could be a quadratic function.

$
\boxed{\text{Correct answer: (D)}}
$

Question

The function $ f $ is not explicitly given. The function $ g $ is given by $ g(x) = f(x + 1) – f(x) $. The function $ h $ is given by $ h(x) = g(x + 1) – g(x) $. If $ h(x) = -6 $ for all values of $ x $, which of the following statements must be true?

(A) Because $ h $ is negative and constant, the graphs of $ g $ and $ f $ always have negative slope.
(B) Because $ h $ is negative and constant, the graphs of $ g $ and $ f $, as well as $ h $, are concave down.
(C) Because $ h $ is negative and constant, $ g $ is decreasing, and the graph of $ f $ always has negative slope.
(D) Because $ h $ is negative and constant, $ g $ is decreasing, and the graph $ f $ is concave down.

▶️ Answer/Explanation

Detailed solution

We have:
$ g(x) = f(x+1) – f(x) $ → average rate of change of $ f $ over interval $[x, x+1]$.
$ h(x) = g(x+1) – g(x) $ → change in $ g $ from $ x $ to $ x+1 $.
Given $ h(x) = -6 $ for all $ x $.
Thus $ g(x+1) – g(x) = -6 $ ⇒ $ g $ is decreasing linearly with slope $-6$.
But $ g $ itself is the rate of change of $ f $ (over 1-unit intervals).
Since $ g $ is decreasing, the rate of change of $ f $ is decreasing ⇒ the graph of $ f $ is concave down.
Also, $ g $ decreasing does **not** mean $ g $ is negative, only that it’s getting smaller. So $ f $ could still have positive slope if $ g > 0 $.
Thus: $ h $ constant negative ⇒ $ g $ decreasing ⇒ $ f $ concave down.
✅ Answer: (D)

Question

The graph of the piecewise-linear function $f$ is shown in the figure. Let $g$ be the inverse function of $f$. What is the maximum value of $g$?

(A) $\frac{1}{7}$

(B) $\frac{1}{5}$

(D) $7$

▶️ Answer/Explanation

Detailed solution

The function $g$ is defined as the inverse of $f$, meaning $g = f^{-1}$.
By the property of inverse functions, the range of $f$ becomes the domain of $g$.
Similarly, the domain of $f$ becomes the range of $g$.
The maximum value of $g$ is the maximum value in its range, which is the maximum $x$-value of $f$.
Looking at the provided graph, the function $f$ is defined on the interval $x \in [0, 5]$.
The largest $x$-coordinate reached by the graph of $f$ is $5$.
Therefore, the maximum value of the inverse function $g$ is $5$.
The correct option is (C).

Question

The average rate of change of the quadratic function $p$ is $-4$ on the interval $0 \leq x \leq 2$ and $-1$ on the interval $2 \leq x \leq 4$. What is the average rate of change of $p$ on the interval $6 \leq x \leq 8$?

(A) $2$

(B) $3$

(D) The average rate of change on the interval $6 \leq x \leq 8$ cannot be determined from the information given.

▶️ Answer/Explanation

Detailed solution

For a quadratic function, the average rate of change over equal-length adjacent intervals forms an arithmetic progression.
The length of the given intervals is constant at $\Delta x = 2$.
The first interval $[0, 2]$ has an average rate of change of $-4$.
The second interval $[2, 4]$ has an average rate of change of $-1$.
The common difference between these rates is $(-1) – (-4) = 3$.
The third interval $[4, 6]$ would have a rate of $(-1) + 3 = 2$.
The fourth interval $[6, 8]$ would have a rate of $2 + 3 = 5$.
Therefore, the average rate of change on $6 \leq x \leq 8$ is $5$.

Question

The table gives values of a function $f$ for selected values of $x$. Which of the following conclusions with reason is consistent with the data in the table?

(A) $f$ could be a linear function because the rates of change over consecutive equal-length intervals in the table can be described by $y = 2x$.

(B) $f$ could be a linear function because the rates of change over consecutive equal-length intervals in the table can be described by $y = 2x + 1$.

(C) $f$ could be a quadratic function because the rates of change over consecutive equal-length intervals in the table can be described by $y = 2x$.

(D) $f$ could be a quadratic function because the rates of change over consecutive equal-length intervals in the table can be described by $y = 2x + 1$.

▶️ Answer/Explanation

Detailed solution

The correct option is (D).
First, calculate the average rates of change for each interval of width $\Delta x = 1$:
For $[-2, -1]$, the rate is $\frac{2 – 5}{-1 – (-2)} = -3$.
For $[-1, 0]$, the rate is $\frac{1 – 2}{0 – (-1)} = -1$.
For $[0, 1]$, the rate is $\frac{2 – 1}{1 – 0} = 1$.
Since the rates of change are not constant, the function cannot be linear.
The rates of change $\{-3, -1, 1\}$ follow a linear pattern described by $y = 2x + 1$ (where $x$ is the left endpoint).
Because the first differences are linear, the function $f$ could be quadratic.

Question

The table describes rates of change of a function $f$ for selected intervals of $x$. The function $f$ is defined for $0 \le x \le 4$. On which of the following intervals is the graph of $f$ concave down?

(A) $0 < x < 1$
(B) $1 < x < 2$
(C) $2 < x < 3$
(D) $3 < x < 4$

▶️ Answer/Explanation

Detailed solution

Concavity is determined by the behavior of the rate of change (the first derivative, $f’$).
A function is concave down on an interval if its rate of change is decreasing.

In the interval $0 < x < 1$, the rate of change is increasing, so the graph is concave up.
In the interval $1 < x < 2$, the rate of change is constant, so the graph is linear.
In the interval $2 < x < 3$, the rate of change is decreasing, which fits the definition of concave down.
In the interval $3 < x < 4$, the rate of change is constant, so the graph is linear.
Therefore, the graph of $f$ is concave down only on the interval $2 < x < 3$.
The correct option is (C).

Question (Calc allowed)

A ball is thrown through an open window to the ground below. The height of the ball, in meters, at time $t$ seconds after it is thrown can be modeled by the function $h$, given by $h(t) = -4.9t^2 + 4.4t + 15.24$. Which of the following describes the height of the ball above the ground?

(A) The ball begins at its maximum height of $15.240$ meters. The height of the ball decreases until it reaches the ground $1.820$ seconds after it leaves the window.

(B) The ball begins at its maximum height of $15.240$ meters. The height of the ball decreases until it reaches the ground $2.269$ seconds after it leaves the window.

(C) After leaving the window, the height of the ball increases to its maximum height of $16.228$ meters. Then the height of the ball decreases until it reaches the ground $1.820$ seconds after reaching its maximum height.

(D) After leaving the window, the height of the ball increases to its maximum height of $16.228$ meters. Then the height of the ball decreases until it reaches the ground $2.269$ seconds after reaching its maximum height.

▶️ Answer/Explanation

Detailed solution

The initial height at $t = 0$ is $h(0) = 15.24$ meters, but the positive linear term $+4.4t$ indicates the ball is thrown upward first.
The time to reach maximum height is $t = \frac{-b}{2a} = \frac{-4.4}{2(-4.9)} \approx 0.449$ seconds.
The maximum height is $h(0.449) = -4.9(0.449)^2 + 4.4(0.449) + 15.24 \approx 16.228$ meters.
To find when it hits the ground, set $h(t) = 0$ and solve $-4.9t^2 + 4.4t + 15.24 = 0$ using the quadratic formula.
The positive root is $t \approx 2.269$ seconds, which is the total time from the window to the ground.
The time from the maximum height to the ground is $2.269 – 0.449 = 1.820$ seconds.
Therefore, the height increases to $16.228$ m, then decreases for $1.820$ seconds until it hits the ground.
Correct Option: (C)

Question

The table gives characteristics of the rates of change of the function $f$ on different intervals. Which of the following is true about $f$ on the interval $3 < x < 4$?

(A) $f$ is increasing, and the graph of $f$ is concave down.
(B) $f$ is increasing, and the graph of $f$ is concave up.
(C) $f$ is decreasing, and the graph of $f$ is concave down.
(D) $f$ is decreasing, and the graph of $f$ is concave up.

▶️ Answer/Explanation

Detailed solution

The “rate of change of $f(x)$” corresponds to the first derivative, $f'(x)$.
On the interval $3 < x < 4$, the rate of change is positive, meaning $f'(x) > 0$.
A positive first derivative indicates that the function $f$ is increasing.
The table also specifies that the rate of change is decreasing.
This implies the second derivative, $f”(x)$, is negative because $f'(x)$ is falling.
A negative second derivative ($f”(x) < 0$) means the graph of $f$ is concave down.
Combining these facts, $f$ is increasing and concave down, which is option (A).

Question

Water is poured into an empty vase at a constant rate. A graph (not shown) models the depth of the water in the vase over time. The graph can be described as follows: the graph is always increasing; the first portion of the graph is clearly concave up; and the next portion of the graph has a fairly steady and steep increase. Which of the following vases is appropriate for the context described by the graph?

▶️ Answer/Explanation

Detailed solution

The rate of change of depth $\frac{dh}{dt}$ is inversely proportional to the cross-sectional area of the vase.
A concave up graph indicates that $\frac{d^2h}{dt^2} > 0$, meaning the vase must be narrowing as height increases.
Vase (D) starts with a wide base that curves inward, causing the water level to rise faster and faster (concave up).
A steady and steep increase implies a constant, high rate of change, which occurs in a narrow, cylindrical neck.
Vase (D) features a long, uniform, and very narrow neck, which produces a linear and steep depth-time relationship.
Vase (B) lacks the uniform vertical neck required for a “steady” increase, as its width continues to change at the top.
Vase (A) and (C) do not match the concave up start because they begin by widening at the bottom.
Therefore, Vase (D) is the correct choice as it perfectly aligns with both geometric descriptions in the graph.

Question

The figure shows a sphere that can expand. As it expands, the volume inside the sphere and the radius of the sphere both increase. In particular, the radius increases at a decreasing rate with respect to the volume.

Which of the following graphs could depict this situation, where volume, in cubic centimeters, is the independent variable and radius, in centimeters, is the dependent variable?

(A)

(B)

(C)

(D)

▶️ Answer/Explanation

Detailed solution

The volume of a sphere is given by $V = \frac{4}{3}\pi r^{3}$, so the radius is $r = \sqrt[3]{\frac{3V}{4\pi}}$.
Since the radius $r$ increases as volume $V$ increases, the graph must have a positive slope.
The problem states the radius increases at a decreasing rate with respect to volume.
This means the slope $\frac{dr}{dV}$ is positive but getting smaller, indicating a concave down shape.
Graph (A) shows a decreasing radius, which contradicts the expansion of the sphere.
Graph (B) shows an increasing radius with a decreasing rate (concave down), matching the description.
Graph (C) shows an increasing rate (concave up), and Graph (D) shows a decreasing radius.
Therefore, the correct graph depicting this relationship is (B).

Question

At time $t = 0$, water begins pouring into an empty container at a constant rate. The water pours into the container until it is full. The situation is modeled by the given graph, where time, in seconds, is the independent variable and the depth of water in the container, in centimeters, is the dependent variable. For which of the following containers would the graph be appropriate?

▶️ Answer/Explanation

Detailed solution

The graph shows an S-shaped (sigmoidal) curve.
Initially, the slope is increasing, meaning the depth rises faster as the container narrows.
This indicates the bottom of the container is wide and gets narrower.
In the middle, the slope is steepest where the container is narrowest.
Then, the slope decreases, meaning the depth rises slower as the container widens again.
This matches a container that is wide at the bottom, narrow in the middle, and wide at the top.
Therefore, the correct container is (B).

Question

The table gives values for a function $ g $ at selected values of $ x $. Which of the following statements is true?

(A) $ g $ is best modeled by a linear function, because the average rate of change over any length input-value interval is constant.

(B) $ g $ is best modeled by a quadratic function, because the average rates of change over consecutive equal-length input-value intervals are constant.

(C) $ g $ is best modeled by a linear function, because the successive 2nd differences of the output values over equal-interval input values are constant.

(D) $ g $ is best modeled by a quadratic function, because the successive 2nd differences of the output values over equal-interval input values are constant.

▶️ Answer/Explanation

Correct Answer: (D)

Detailed Solution

1. Compute first differences: $ -2 – (-3) = 1 $, $ 1 – (-2) = 3 $, $ 6 – 1 = 5 $, $ 13 – 6 = 7 $.
2. First differences are $ 1, 3, 5, 7 $ (not constant).
3. Compute second differences: $ 3 – 1 = 2 $, $ 5 – 3 = 2 $, $ 7 – 5 = 2 $.
4. Second differences are constant and equal to $ 2 $.
5. A constant second difference indicates a quadratic function.
6. Therefore, the correct statement is that $ g $ is best modeled by a quadratic function.
7. Hence, option (D) is correct.

Question

The table gives values of the function $ f $ for selected values of $ x $. If the function $ f $ is linear, what is the value of $ f(13) $?

(A) $ 4 $

(B) $ \frac{29}{4} $

(D) $ \frac{34}{3} $

▶️ Answer/Explanation

Detailed solution

First, calculate the slope ($ m $) of the linear function using the points $ (-3, -10) $ and $ (3, -2) $:
$ m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{-2 – (-10)}{3 – (-3)} = \frac{8}{6} = \frac{4}{3} $.
Next, use the point-slope form or slope-intercept form $ y = mx + b $ to find the equation. Using point $ (3, -2) $:
$ -2 = \frac{4}{3}(3) + b \implies -2 = 4 + b \implies b = -6 $.
The function is defined as $ f(x) = \frac{4}{3}x – 6 $.
Substitute $ x = 13 $ into the equation to find the required value:
$ f(13) = \frac{4}{3}(13) – 6 = \frac{52}{3} – \frac{18}{3} = \frac{34}{3} $.
Therefore, the correct value is $ \frac{34}{3} $, which matches option (D).

Question

Two drones are flying over a given area, and their heights above the ground are changing. The table gives the change in height, in feet, for the drones over successive $6$-second intervals. Which of the following is true about the average rates of change for drone A and drone B over the time interval from $t = 0$ seconds to $t = 30$ seconds?

(A) The average rates of change are equal.

(B) The average rate of change for drone A is greater than for drone B.

(D) The average rates of change cannot be determined because changes in heights are given, not heights of the drones.

▶️ Answer/Explanation

Detailed solution

The average rate of change is defined as the total change in height divided by the total time elapsed.
First, calculate the total change in height for Drone A by summing the changes in each interval: $17 + (-4) + 11 + (-5) + (-3) = 16 \text{ feet}$.
Next, calculate the total change in height for Drone B by summing its respective changes: $5 + 3 + 3 + 2 + 3 = 16 \text{ feet}$.
Both drones experienced the same total change in height ($16 \text{ ft}$) over the same total time interval ($30 \text{ seconds}$).
Since both the numerator (change in height) and denominator (time) are identical for both drones, their average rates of change are equal ($\frac{16}{30} \text{ ft/s}$).
Therefore, the correct statement is that the average rates of change are equal.
Correct Option: (A)

Question

The table gives the average rates of change of a function $ f $ over different intervals. On which of the intervals does the function increase the most?

(A) $ 0 \le x \le 1 $

(B) $ 1 \le x \le 4 $

(D) $ 8 \le x \le 10 $

▶️ Answer/Explanation

Answer:

(C) $ 4 \le x \le 8 $

Explanation:

The total increase over an interval equals: \[ \text{Average Rate of Change} \times \text{Length of Interval}. \]

Compute each increase:

$0 \le x \le 1:\; 10 \times (1-0) = 10$

$1 \le x \le 4:\; (-5) \times (4-1) = -15$

$4 \le x \le 8:\; 2 \times (8-4) = 8$

$8 \le x \le 10:\; 6 \times (10-8) = 12$

The greatest positive increase is $12$, which occurs on $8 \le x \le 10$.

Detailed solution

1. Increase = (average rate of change) × (interval length).
2. For $0 \le x \le 1$: $10 \times 1 = 10$.
3. For $1 \le x \le 4$: $(-5) \times 3 = -15$.
4. For $4 \le x \le 8$: $2 \times 4 = 8$.
5. For $8 \le x \le 10$: $6 \times 2 = 12$.
6. Largest increase is $12$, so the correct interval is $8 \le x \le 10$.

Question

The function $f$ is defined for all real values of $x$. For a constant $a$, the average rate of change of $f$ from $x=a$ to $x=a+1$ is given by $2a+1$. Which of the following statements is true?

(A) The average rate of change of $f$ over consecutive equal-length input-value intervals is positive, so the graph of $f$ could be a line with a positive slope.

(B) The average rate of change of $f$ over consecutive equal-length input-value intervals is positive, so the graph of $f$ could be a parabola that opens up.

(C) The average rate of change of $f$ over consecutive equal-length input-value intervals is increasing at a constant rate, so the graph of $f$ could be a line with a positive slope.

(D) The average rate of change of $f$ over consecutive equal-length input-value intervals is increasing at a constant rate, so the graph of $f$ could be a parabola that opens up.

▶️ Answer/Explanation

Answer: (D)

The average rate of change from $x=a$ to $x=a+1$ is $2a+1$, which is a linear expression in $a$. Since $2a+1$ increases at a constant rate (slope $2$), the average rates of change over equal intervals increase linearly. A function whose average rate of change increases linearly is quadratic, meaning its graph is a parabola that opens upward. Therefore, statement (D) is correct.

Detailed solution

1. The average rate of change on $[a,a+1]$ is $2a+1$.
2. This expression is linear in $a$ with constant slope $2$.
3. Therefore, consecutive average rates increase at a constant rate.
4. A constant first difference implies a quadratic function.
5. The leading coefficient is positive since the rate increases.
6. Hence, the graph is a parabola opening upward.
7. Therefore, the correct statement is (D).

Question

The graph of the function $ y = g(x) $ is given. On which interval is the average rate of change of $ g $ least?

(A) $ -3 \le x \le -2 $

(B) $ -1 \le x \le 0 $

(D) $ 3 \le x \le 4 $

▶️ Answer/Explanation

Answer:

(D) $ 3 \le x \le 4 $

Explanation:

The average rate of change on an interval $ [a,b] $ is \[ \frac{g(b)-g(a)}{b-a}. \] We estimate values from the graph and compute slopes. The least average rate of change corresponds to the most negative slope.

Detailed solution

1. For $ -3 \le x \le -2 $: $ \frac{g(-2)-g(-3)}{1} \approx \frac{4.5-0}{1} = 4.5 $.
2. For $ -1 \le x \le 0 $: $ \frac{g(0)-g(-1)}{1} \approx \frac{0-4}{1} = -4 $.
3. For $ 1 \le x \le 2 $: $ \frac{g(2)-g(1)}{1} \approx \frac{-1-1}{1} = -2 $.
4. For $ 3 \le x \le 4 $: $ \frac{g(4)-g(3)}{1} \approx \frac{-4-0.5}{1} = -4.5 $.
5. The smallest (most negative) value is $ -4.5 $.
6. Therefore, the least average rate of change occurs on $ 3 \le x \le 4 $.

Question (Calc Allowed)

The function $ f $ is given by $ f(x) = x^2 + 3x – 5 $. Which of the following describes $ f $?

(A) For any interval of $ x $, the function always has a positive rate of change.

(B) For any interval of $ x $, the function always has a negative rate of change.

(C) For any interval of $ x < -1.5 $, the function has a positive rate of change, and for any interval of $ x > -1.5 $, the function has a negative rate of change.

(D) For any interval of $ x < -1.5 $, the function has a negative rate of change, and for any interval of $ x > -1.5 $, the function has a positive rate of change.

▶️ Answer/Explanation

Correct Answer: (D)

Explanation:
The rate of change of a function is given by its derivative. $ f(x) = x^2 + 3x – 5 $ is a quadratic function that opens upward since the coefficient of $ x^2 $ is positive. Its derivative is $ f'(x) = 2x + 3 $. Setting $ f'(x) = 0 $ gives $ 2x + 3 = 0 \Rightarrow x = -\dfrac{3}{2} = -1.5 $. For $ x < -1.5 $, $ f'(x) < 0 $, so the function is decreasing. For $ x > -1.5 $, $ f'(x) > 0 $, so the function is increasing. Therefore, option (D) correctly describes the function.

Detailed solution

1. Given $ f(x) = x^2 + 3x – 5 $.
2. Compute derivative: $ f'(x) = 2x + 3 $.
3. Set derivative equal to zero: $ 2x + 3 = 0 $.
4. Solve: $ x = -\dfrac{3}{2} = -1.5 $.
5. If $ x < -1.5 $, then $ f'(x) < 0 $ → decreasing.
6. If $ x > -1.5 $, then $ f'(x) > 0 $ → increasing.
7. Hence, the correct answer is (D).

Question

The average rate of change of the quadratic function $p$ is $-4$ on the interval $0 \le x \le 2$ and $-1$ on the interval $2 \le x \le 4$. What is the average rate of change of $p$ on the interval $6 \le x \le 8$?

(A) $2$

(B) $3$

(D) The average rate of change on the interval $6 \le x \le 8$ cannot be determined from the information given.

▶️ Answer/Explanation

Answer: (C) $5$

Since $p(x)$ is quadratic, its average rate of change over equal-length intervals changes linearly. The intervals $0 \le x \le 2$, $2 \le x \le 4$, $4 \le x \le 6$, and $6 \le x \le 8$ each have length $2$. The given average rates are $-4$ and $-1$, increasing by $3$. Therefore, the sequence continues increasing by $3$: $2$, then $5$. Hence, the average rate of change on $6 \le x \le 8$ is $5$.

Detailed solution

1. For a quadratic function, average rates over equal intervals form an arithmetic sequence.
2. Interval length is $2$ for all given intervals.
3. Given rates: $-4$ and $-1$.
4. Common difference $= -1 – (-4) = 3$.
5. Next interval $4 \le x \le 6$: $ -1 + 3 = 2$.
6. Next interval $6 \le x \le 8$: $ 2 + 3 = 5$.
7. Therefore, required average rate of change $= 5$.

Question

(A) The rate of change of the rates of change is $ 0 $ meters per second, and the object is neither speeding up nor slowing down.

(B) The rate of change of the rates of change is $ 0 $ meters per second per second, and the object is neither speeding up nor slowing down.

(C) The rate of change of the rates of change is $ 4 $ meters per second, and the object is neither speeding up nor slowing down.

(D) The rate of change of the rates of change is $ 4 $ meters per second per second, and the object is speeding up.

▶️ Answer/Explanation

Detailed solution

1. Calculate the rate of change of distance (velocity) for the intervals: $ \frac{9-1}{3-1} = 4 $, $ \frac{21-9}{6-3} = 4 $, and $ \frac{41-21}{11-6} = 4 $.
2. The velocity is constant at $ 4 \, \text{m/s} $ for all time intervals shown.
3. The “rate of change of the rates of change” is the acceleration (derivative of velocity).
4. Since the velocity is constant, the change in velocity is zero, so acceleration is $ 0 \, \text{m/s}^2 $.
5. The correct unit for acceleration is meters per second per second ($ \text{m/s}^2 $).
6. Since acceleration is $ 0 $, the object is moving at a steady speed, neither speeding up nor slowing down.
7. Therefore, option (B) matches both the value and the description of the motion.

Question

The table gives values of a function $ f $ for selected values of $ x $. Which of the following conclusions with reason is consistent with the data in the table?

(A) $ f $ could be a linear function because the rates of change over consecutive equal-length intervals in the table can be described by $ y = 2x $.

(B) $ f $ could be a linear function because the rates of change over consecutive equal-length intervals in the table can be described by $ y = 2x + 1 $.

(C) $ f $ could be a quadratic function because the rates of change over consecutive equal-length intervals in the table can be described by $ y = 2x $.

(D) $ f $ could be a quadratic function because the rates of change over consecutive equal-length intervals in the table can be described by $ y = 2x + 1 $.

▶️ Answer/Explanation

Detailed solution

The correct answer is (D).
First, calculate the first differences (rates of change) between consecutive $ f(x) $ values:
$ f(-1) – f(-2) = 2 – 5 = -3 $
$ f(0) – f(-1) = 1 – 2 = -1 $
$ f(1) – f(0) = 2 – 1 = 1 $
Since the first differences ($-3, -1, 1$) are not constant, $ f $ is not linear.
Next, calculate the second differences: $ -1 – (-3) = 2 $ and $ 1 – (-1) = 2 $. Since the second differences are constant, $ f $ could be a quadratic function.
Finally, check which linear equation describes the rates of change ($-3, -1, 1$) based on the starting $ x $ values ($-2, -1, 0$):
Using $ y = 2x + 1 $: for $ x = -2 $, $ y = -3 $; for $ x = -1 $, $ y = -1 $; for $ x = 0 $, $ y = 1 $.
This matches the calculated rates perfectly.

Question

The table gives values for a polynomial function $ f $ at selected values of $ x $ . What is the average rate of change of $ f $ over the closed interval $ [1,4] $ ?

(A) $ -2 $

(B) $ -\dfrac{1}{2} $

(D) $ 13 $

▶️ Answer/Explanation

Correct Answer: (A) $ -2 $

Detailed solution

1. The average rate of change on $ [1,4] $ is $ \dfrac{f(4)-f(1)}{4-1} $.

2. From the table, $ f(4)=10 $ and $ f(1)=16 $.

3. Substitute values: $ \dfrac{10-16}{4-1} $.

4. Simplify numerator: $ 10-16=-6 $.

5. Denominator: $ 4-1=3 $.

6. Compute: $ \dfrac{-6}{3}=-2 $.

7. Therefore, the average rate of change is $ -2 $.

Question

The table shows values for a function $ g $ at selected values of $ x $. Which of the following claim and explanation statements best fits these data?

(A) $ g $ is best modeled by a linear function, because the rate of change over consecutive equal-length input-value intervals is constant.

(B) $ g $ is best modeled by a linear function, because the change in the average rates of change over consecutive equal-length input-value intervals is constant.

(C) $ g $ is best modeled by a quadratic function, because the rate of change over consecutive equal-length input-value intervals is constant.

(D) $ g $ is best modeled by a quadratic function, because the change in the average rates of change over consecutive equal-length input-value intervals is constant.

▶️ Answer/Explanation

Correct Answer: (D)

Explanation:
The first differences (average rates of change) are:
$ 78-53 = 25 $,
$ 97-78 = 19 $,
$ 110-97 = 13 $,
$ 117-110 = 7 $.
These are not constant, so the function is not linear.
The second differences are:
$ 19-25 = -6 $,
$ 13-19 = -6 $,
$ 7-13 = -6 $.
Since the second differences are constant ($-6$), the function is quadratic.

Detailed solution

1. Compute first differences: $25, 19, 13, 7$.
2. These values are not constant ⇒ not linear.
3. Compute second differences: $19-25=-6$.
4. $13-19=-6$.
5. $7-13=-6$.
6. Second differences are constant ⇒ quadratic model.
7. Therefore, option (D) is correct.

Question

The table gives values of a polynomial function $ g $ at selected values of $ x $. If $ a < b $, then $ g(a) > g(b) $ for all $ a $ and $ b $ in the interval $ 3 < x < 7 $. Which of the following could be true about the graph of $ g $ on the interval $ 3 < x < 7 $?

(A) The graph of $ g $ is concave down because the function is decreasing, and the average rate of change over equal-length input-value intervals is increasing.

(B) The graph of $ g $ is concave up because the function is decreasing, and the average rate of change over equal-length input-value intervals is increasing.

(C) The graph of $ g $ is concave down because the function is decreasing, and the average rate of change over equal-length input-value intervals is decreasing.

(D) The graph of $ g $ is concave up because the function is decreasing, and the average rate of change over equal-length input-value intervals is decreasing.

▶️ Answer/Explanation

Answer: (C)

Detailed solution

1. Since $ a < b \Rightarrow g(a) > g(b) $, the function is decreasing on $ 3 < x < 7 $.
2. Compute average rates of change over intervals of length $ 1 $:
3. $ \frac{-19 – (-11)}{4-3} = -8 $, $ \frac{-29 – (-19)}{5-4} = -10 $.
4. $ \frac{-41 – (-29)}{6-5} = -12 $, $ \frac{-55 – (-41)}{7-6} = -14 $.
5. The average rates of change are $ -8, -10, -12, -14 $, which are decreasing (becoming more negative).
6. A decreasing function with decreasing slopes indicates the graph is concave down.
7. Therefore, the correct statement is (C).

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- MCQ Exam Style Questions - Effective Fall 2023