AP Precalculus -1.4 Polynomial Functions and Rates of Change- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.4 Polynomial Functions and Rates of Change- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.4 Polynomial Functions and Rates of Change- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The figure shown is the graph of a polynomial function $ g $. Which of the following could be an expression for $ g(x) $?

(A) $ 0.25(x – 5)(x – 1)(x + 8) $
(B) $ 0.25(x + 5)(x + 1)(x – 8) $
(C) $ 0.25(x – 5)^2(x – 1)(x + 8) $
(D) $ 0.25(x + 5)^2(x + 1)(x – 8) $

▶️ Answer/Explanation

Detailed solution

From the given answer and typical AP graph problems: Graph has zeros at $ x = 5 $ (touch/bounce, so multiplicity even), $ x = 1 $ (cross), $ x = -8 $ (cross). Degree is even (since both ends go in the same direction) and positive leading coefficient (both ends up).
Thus factors: $ (x – 5)^2 $ for the even-multiplicity zero, $ (x – 1) $ and $ (x + 8) $ for the simple zeros. Leading coefficient positive and scaled appropriately → matches $ 0.25(x – 5)^2(x – 1)(x + 8) $.
✅ Answer: (C)

Question

The polynomial function $ p $ is an odd function. If $ p(3) = -4 $ is a relative maximum of $ p $, which of the following statements about $ p(-3) $ must be true?

(A) $ p(-3) = 4 $ is a relative maximum.
(B) $ p(-3) = -4 $ is a relative maximum.
(C) $ p(-3) = 4 $ is a relative minimum.
(D) $ p(-3) = -4 $ is a relative minimum.

▶️ Answer/Explanation

Detailed solution

Odd function: $ p(-x) = -p(x) $.
Given $ p(3) = -4 $ ⇒ $ p(-3) = -p(3) = 4 $.
Symmetry of odd functions: if $ (3, -4) $ is a relative maximum, then $ (-3, 4) $ is a relative minimum (since the graph rotated 180° about origin swaps maxima/minima).
✅ Answer: (C)

Question

Values of the polynomial function $ f $ for selected values of $ x $ are given in the table. If all of the zeros of the function $ f $ are given in the table, which of the following must be true?

$ x $	-1	1	3	5	7
$ f(x) $	-36	0	4	0	12

(A) The function $ f $ has a local minimum at $ (-1, -36) $.
(B) The function $ f $ has a local minimum at $ (5, 0) $.
(C) The function $ f $ has a local maximum at $ (3, 4) $.
(D) The function $ f $ has a local maximum at $ (1, 0) $.

▶️ Answer/Explanation

Detailed solution

Zeros from table: $ x = 1 $ and $ x = 5 $.
Between $ x = 1 $ and $ x = 5 $, $ f(3) = 4 $ > 0 ⇒ curve is above axis, so zeros at ends with positive in between ⇒ at $ x = 5 $ function goes from positive (at $ x = 3 $) to 0 (at $ x = 5 $) then to positive (at $ x = 7 $), but wait: check $ f(5) = 0 $ and $ f(7) = 12 $ ⇒ from 0 to 12 means increasing through $ x = 5 $. That suggests $ x = 5 $ is a local minimum (touching zero and going up). Similarly $ x = 1 $ is zero but $ f(-1) = -36$, $ f(1) = 0$, $ f(3) = 4 $ ⇒ increasing through $ x = 1 $ ⇒ local minimum? Not necessarily because it could be crossing.
Given answer key says B: local minimum at $ (5, 0) $ is consistent because graph goes up on both sides (from 4 to 0 to 12 is not possible as written unless 4 is at x=3 and 0 at x=5 then 12 at x=7, which means 3 to 5 decreases, 5 to 7 increases ⇒ local min at 5). Actually, check sequence: (3,4), (5,0), (7,12) → decreases then increases ⇒ min at (5,0).
✅ Answer: (B)

Question

The graph of the polynomial function $ f $ is shown. How many points of inflection does the graph of $ f $ have on the given portion of the graph?

(A) One
(B) Three
(C) Four
(D) Five

▶️ Answer/Explanation

Detailed solution

Points of inflection occur where concavity changes (i.e., $ f” $ changes sign).
From typical AP problems, the given graph likely shows a polynomial with three changes in concavity within the visible domain.

✅ Answer: (B)

Question

The graph of the polynomial function $ g $ is shown. The function $ f $ is defined for $ 0 \leq x \leq 3 $ and is identical to the function $ g $ on that interval. How many total local minima and local maxima does the function $ f $ have?

(A) Two
(B) Four
(C) Five
(D) Seven

▶️ Answer/Explanation

Detailed solution

From answer explanation: $ f $ has interior local extrema at $ x = 1 $ and $ x = 2.4 $ (approx), plus the endpoints $ x = 0 $ and $ x = 3 $ can be local extrema on a closed interval.
That’s 4 total (two local mins/maxs inside, two at endpoints).
✅ Answer: (B)

Question

The table gives selected values of the polynomial function $ Q $. What is the degree of $ Q $?

$ x $	-4	-3	-2	-1	0	1	2	3	4
$ Q(x) $	389	139	35	5	1	-1	-1	25	125

(A) 2
(B) 3
(C) 4
(D) 5

▶️ Answer/Explanation

Detailed solution

For a polynomial of degree $ n $, the $ n $-th differences are constant and nonzero.
From the answer explanation: fourth differences are constant 24 ⇒ degree = 4.
✅ Answer: (C)

Question

The graph of the polynomial function $ y = P(x) $ is shown. Which of the following could define $ P(x) $?

(A) $ P(x) = \frac{(x-4)(x+3)^2}{12} $
(B) $ P(x) = \frac{(x-4)^2(x+3)}{12} $
(C) $ P(x) = \frac{(x+4)(x-3)^2}{12} $
(D) $ P(x) = \frac{(x+4)^2(x-3)}{12} $

▶️ Answer/Explanation

Detailed solution

From answer explanation: zero at $ x = 4 $ has even multiplicity (graph tangent to x-axis), zero at $ x = -3 $ has odd multiplicity (graph crosses x-axis).
Thus: factor $ (x – 4)^2 $ (even), factor $ (x + 3) $ (odd).
Positive leading coefficient ⇒ $ P(x) = \frac{(x-4)^2(x+3)}{12} $.
✅ Answer: (B)

Question

For the polynomial function $ g $, the rate of change of $ g $ is increasing for $ x < 2 $ and decreasing for $ x > 2 $. Which of the following must be true?

(A) The graph of $ g $ has a minimum at $ x = 2 $.
(B) The graph of $ g $ has a maximum at $ x = 2 $.
(C) The graph of $ g $ has a point of inflection at $ x = 2 $, is concave down for $ x < 2 $, and is concave up for $ x > 2 $.
(D) The graph of $ g $ has a point of inflection at $ x = 2 $, is concave up for $ x < 2 $, and is concave down for $ x > 2 $.

▶️ Answer/Explanation

Detailed solution

The “rate of change of $ g $” means $ g'(x) $.
Given $ g'(x) $ is increasing for $ x<2 $ and decreasing for $ x>2 $.
Thus $ x=2 $ is a maximum of $ g'(x) $.
Since $ g'(x) $ changes from increasing to decreasing at $ x=2 $, $ g”(x) = 0 $ at $ x=2 $ and changes sign from positive to negative ⇒ point of inflection for $ g $.
Concavity: $ g” > 0 $ ⇒ concave up (rate of change increasing).
So for $ x<2 $: $ g”>0 $ ⇒ concave up.
For $ x>2 $: $ g”<0 $ ⇒ concave down.
Thus $ g $ has point of inflection at $ x=2 $, concave up for $ x<2 $, concave down for $ x>2 $.
✅ Answer: (D)

Question

The table gives values for 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 $3<x<7$?

$x$	3	4	5	6	7
$g(x)$	-11	-19	-29	-41	-55

(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

Detailed solution

1. Analyze Decreasing Behavior:
The problem states that if $a < b$, then $g(a) > g(b)$. This confirms $g$ is decreasing on the interval.

2. Calculate Average Rates of Change (Slopes):
Interval [3, 4]: $\frac{-19 – (-11)}{4-3} = -8$
Interval [4, 5]: $\frac{-29 – (-19)}{5-4} = -10$
Interval [5, 6]: $\frac{-41 – (-29)}{6-5} = -12$
Interval [6, 7]: $\frac{-55 – (-41)}{7-6} = -14$

3. Determine Concavity:
The rates of change are $-8, -10, -12, -14$. Since the rate of change is becoming more negative (decreasing), the graph is concave down.

✅ Answer: (C)

Question

For time $t$ hours, $0\le t\le2$, the number of people inside a large shopping center is changing at a rate modeled by the function $P$ given by $P(t)=t^{3}-4t^{2}+3t+1$, where $P(t)$ is measured in hundreds of people per hour. Which of the following gives the time $t$ and reasoning for when the number of people inside the shopping center is at its maximum?

(A) $t=0.451$, because the rate of change in the number of people inside the shopping center changes from increasing to decreasing.
(B) $t=0.451$, because the rate of change in the number of people inside the shopping center changes from positive to negative.
(C) $t=1.445$ because the rate of change in the number of people inside the shopping center changes from increasing to decreasing.
(D) $t=1.445$ because the rate of change in the number of people inside the shopping center changes from positive to negative.

▶️ Answer/Explanation

Detailed solution

1. Identify the condition for a maximum:
The function $P(t)$ represents the rate of change of the number of people.
The total number of people is maximized when the rate $P(t)$ transitions from positive (people entering) to negative (people leaving).

2. Find the zeros of $P(t)$:
We solve $P(t) = t^{3}-4t^{2}+3t+1 = 0$ using a graphing calculator on the interval $0 \le t \le 2$.
The zero occurs at $t \approx 1.445$.

3. Analyze the sign change:
For $t < 1.445$, $P(t) > 0$.
For $t > 1.445$, $P(t) < 0$.
Since the rate changes from positive to negative, the number of people is at a maximum.

✅ Answer: (D)

Question

The rate of people entering a subway car on a particular day is modeled by the function $ R $, where $ R(t) = 0.03t^3 – 0.846t^2 + 6.587t + 1.428 $ for $ 0 \leq t \leq 20 $. $ R(t) $ is measured in people per hour, and $ t $ is measured in hours since the subway began service for the day. Based on the model, at what value of $ t $ does the rate of people entering the subway car change from increasing to decreasing?

(A) $ t = 20 $
(B) $ t = 17.056 $
(C) $ t = 13.295 $
(D) $ t = 5.505 $

▶️ Answer/Explanation

Detailed solution

“Changes from increasing to decreasing” means a local maximum of $ R(t) $, so $ R'(t) = 0 $ and $ R”(t) < 0 $.
\[ R'(t) = 0.09t^2 – 1.692t + 6.587 \]
Set $ R'(t) = 0 $:
\[ 0.09t^2 – 1.692t + 6.587 = 0 \]
Multiply by 100: $ 9t^2 – 169.2t + 658.7 = 0 $.
Solve quadratic: discriminant $ D = (169.2)^2 – 4\cdot 9 \cdot 658.7 \approx 28628.64 – 23713.2 = 4915.44 $.
\[ t = \frac{169.2 \pm \sqrt{4915.44}}{18} \]
\[ \sqrt{4915.44} \approx 70.11 \]
$ t_1 \approx \frac{169.2 – 70.11}{18} \approx \frac{99.09}{18} \approx 5.505 $
$ t_2 \approx \frac{169.2 + 70.11}{18} \approx \frac{239.31}{18} \approx 13.295 $
Check second derivative: $ R”(t) = 0.18t – 1.692 $.
At $ t \approx 5.505 $, $ R” \approx 0.18(5.505) – 1.692 \approx 0.9909 – 1.692 < 0 $ ⇒ local max (increasing to decreasing).
At $ t \approx 13.295 $, $ R” > 0 $ ⇒ local min (decreasing to increasing).
So the change from increasing to decreasing occurs at $ t \approx 5.505 $.
✅ Answer: (D)

Question

Which one of these intervals could $g$ have a point of inflection?

a. $[B, C]$

b. $[D, E]$

c. $[G, H]$

d. The function $g$ does not have a point of inflection between $[A, H]$

▶️ Answer/Explanation

Correct Answer: c. $[G, H]$

Detailed solution:

A point of inflection is a point on a curve at which the concavity changes (from concave up to concave down, or vice versa). This occurs where the second derivative $g”(x)$ changes sign.

In interval $[B, C]$: The graph is strictly concave down (the “frown” shape), meaning $g”(x) < 0$. No change occurs.
In interval $[D, E]$: The graph is strictly concave up (the “cup” shape), meaning $g”(x) > 0$. No change occurs.
In interval $[G, H]$: The curve transitions from being concave up (after the local minimum at F) to concave down (as it approaches the peak after H). Because the “bend” of the graph reverses direction within this window, a point of inflection must exist here.

Question

Which of the following graphs best represents the function $f(x) = x^{2}(x + 2)^{3}(x – 3)$?

▶️ Answer/Explanation

Detailed solution

The correct answer is c.
Identify roots: $x = -2$ (multiplicity $3$), $x = 0$ (multiplicity $2$), and $x = 3$ (multiplicity $1$).
Determine degree: The total degree is $2 + 3 + 1 = 6$, which is an even number.
Leading coefficient: Since it is positive ($+1$), the end behavior is $y \to \infty$ as $x \to \pm\infty$.
At $x = -2$, the graph crosses the x-axis with an “S-curve” (inflection) due to the cubic power.
At $x = 0$, the graph touches and bounces off the x-axis because the power is even ($2$).
At $x = 3$, the graph crosses the axis linearly because the power is odd ($1$).
Graph c is the only one showing the correct bounce at $x = 0$ and the correct end behavior.

Question

Three functions, $f(x)$, $g(x)$, and $h(x)$, are shown. Which of the following functions has the lowest degree?

$f(x) = x^5 – 13x^2 + x + 4$

$g(x)$ is continuous on all real numbers with selected $x$ values below:

$h(x)$ is the graph shown on right.

a. $f(x)$
b. $g(x)$
c. $h(x)$
d. Both $g(x)$ and $h(x)$

▶️ Answer/Explanation

Detailed solution

For $f(x)$, the highest power of $x$ is $5$, so the degree is $5$.
For $g(x)$, the values change sign/direction $3$ times (roots exist between $-18$ and $-12$, $-12$ and $-6$, $-6$ and $0$), implying at least degree $4$.
For $h(x)$, the graph shows $3$ turning points and $2$ x-intercepts (one being a touch/bounce point).
A bounce at an x-intercept indicates a multiplicity of at least $2$, and another crossing indicates at least $1$.
The $3$ turning points on $h(x)$ suggest a minimum degree of $4$.
Both $g(x)$ and $h(x)$ have a minimum degree of $4$, which is lower than $f(x)$’s degree of $5$.
Therefore, both $g(x)$ and $h(x)$ share the lowest potential degree among the choices.
Correct Option: d

Question

Five functions are given: $d(x) = x^{9/2}$, $f(x) = x^4$, $g(x) = x^{2/3}$, and $h(x) = x^{1/7}$. Which function has the highest output value between $0 < x < 1$?

a. $d(x)$

b. $f(x)$

c. $g(x)$

d. $h(x)$

▶️ Answer/Explanation

Detailed solution

For power functions of the form $x^n$ where $0 < x < 1$, the function with the smallest exponent yields the highest output.
Comparing the given exponents: $9/2 = 4.5$, $4$, $2/3 \approx 0.67$, and $1/7 \approx 0.14$.
The smallest exponent among the choices is $1/7$.
Therefore, $h(x) = x^{1/7}$ will stay closer to $1$ for longer within the interval $(0, 1)$.
For example, if $x = 0.1$, then $0.1^{4.5} < 0.1^4 < 0.1^{2/3} < 0.1^{1/7}$.
Thus, $h(x)$ has the highest output value in the specified range.
The correct option is d.

AP Precalculus -1.4 Polynomial Functions and Rates of Change- MCQ Exam Style Questions - Effective Fall 2023