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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?
\( x \)-11357
\( f(x) \)-3604012
(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 \( Q \)?
\( x \)-4-3-2-101234
\( Q(x) \)3891393551-1-125125
(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)

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