AP Precalculus -1.6 Polynomial Functions and End Behavior- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.6 Polynomial Functions and End Behavior- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.6 Polynomial Functions and End Behavior- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

For a polynomial function $ f $, $ \lim_{x \to -\infty} f(x) = \infty $ and $ \lim_{x \to \infty} f(x) = -\infty $. Which of the following must be true about $ f $?

(A) The degree of $ f $ is even, and the leading coefficient is negative.
(B) The degree of $ f $ is even, and the leading coefficient is positive.
(C) The degree of $ f $ is odd, and the leading coefficient is negative.
(D) The degree of $ f $ is odd, and the leading coefficient is positive.

▶️ Answer/Explanation

Detailed solution

End behavior: as $ x \to -\infty $, $ f(x) \to \infty $ (up to the left); as $ x \to \infty $, $ f(x) \to -\infty $ (down to the right).
For odd degree polynomials: ends go in opposite directions.
Leading coefficient negative ⇒ up-left, down-right. Leading coefficient positive ⇒ down-left, up-right.
Given up-left, down-right ⇒ odd degree, negative leading coefficient.
✅ Answer: (C)

Question

The polynomial function $ p $ is given by $ p(x) = -4x^5 + 3x^2 + 1 $. Which of the following statements about the end behavior of $ p $ is true?

(A) The sign of the leading term of $ p $ is positive, and the degree of the leading term of $ p $ is even; therefore, $ \lim_{x \to -\infty} p(x) = \infty $ and $ \lim_{x \to \infty} p(x) = \infty $.
(B) The sign of the leading term of $ p $ is negative, and the degree of the leading term of $ p $ is odd; therefore, $ \lim_{x \to -\infty} p(x) = \infty $ and $ \lim_{x \to \infty} p(x) = -\infty $.
(C) The sign of the leading term of $ p $ is positive, and the degree of the leading term of $ p $ is odd; therefore, $ \lim_{x \to -\infty} p(x) = -\infty $ and $ \lim_{x \to \infty} p(x) = \infty $.
(D) The sign of the leading term of $ p $ is negative, and the degree of the leading term of $ p $ is odd; therefore, $ \lim_{x \to -\infty} p(x) = -\infty $ and $ \lim_{x \to \infty} p(x) = \infty $.

▶️ Answer/Explanation

Detailed solution

Leading term: $ -4x^5 $ → negative coefficient, degree 5 (odd).
For odd degree, negative leading coefficient: up-left, down-right.
So $ \lim_{x \to -\infty} p(x) = \infty $ and $ \lim_{x \to \infty} p(x) = -\infty $.
This matches option (B).
✅ Answer: (B)

Question

The polynomial function $ f $ is given by $ f(x) = (x – 4)(3x – 1)^2 $. Which of the following descriptions of $ f $ is true?

(A) $ f $ is a polynomial of degree 2 with a leading coefficient of 3.
(B) $ f $ is a polynomial of degree 2 with a leading coefficient of 9.
(C) $ f $ is a polynomial of degree 3 with a leading coefficient of 3.
(D) $ f $ is a polynomial of degree 3 with a leading coefficient of 9.

▶️ Answer/Explanation

Detailed solution

Factor $ (3x – 1)^2 $ has degree 2, factor $ (x – 4) $ has degree 1 ⇒ total degree = 3.
Leading term from expanding: $ (3x)^2 \cdot x = 9x^3 $ ⇒ leading coefficient 9.
✅ Answer: (D)

Question

A polynomial function has the form $ p(x) = ax^j + bx^k $, where $ a $ and $ b $ are nonzero constants, $ j $ and $ k $ are nonnegative integers. Which of the following conditions guarantees that $ p $ is an even function?

(A) $ k = 0 $
(B) $ j = 2k $ and $ k $ is even.
(C) $ j + k $ is even.
(D) $ j \cdot k $ is even.

▶️ Answer/Explanation

Detailed solution

Even function means $ p(-x) = p(x) $.
This requires $ j $ and $ k $ both even, because $ (-x)^n = x^n $ only if $ n $ is even.
Check options:
(A) $ k=0 $ (0 is even) but $ j $ could be odd ⇒ not guaranteed even.
(B) $ j = 2k $ and $ k $ even ⇒ $ j $ even (since even × integer = even). Both exponents even ⇒ $ p $ even.
(C) $ j+k $ even: e.g., $ j=1, k=3 $ ⇒ both odd ⇒ $ p $ not even.
(D) $ j \cdot k $ even: e.g., $ j=1, k=2 $ ⇒ $ j $ odd ⇒ not even.
Thus only (B) guarantees both exponents even.
✅ Answer: (B)

Question

The polynomial function $ P $ is given by $ P(x) = 3x(x + 1)^2(x – a) $, where $ a $ is a real number. Which of the following could be the graph of $ y = P(x) $?

▶️ Answer/Explanation

Detailed solution

Given $ P(x) = 3x(x + 1)^2(x – a) $.
For both $ x = -1 $ and $ x = 0 $ to have even multiplicity (graph tangent to x-axis at both), $ x = 0 $ must be multiplicity 2 ⇒ $ a = 0 $ makes $ P(x) = 3x^2(x + 1)^2 $, so both zeros multiplicity 2.
The correct graph must show tangency at $ x = 0 $ and $ x = -1 $, no crossing there, and degree 4, positive leading coefficient 3 ⇒ ends up.
Match to the option with that shape.
✅ Answer: (D)

Question

The polynomial function $ f $ is given by $ f(x) = ax^4 + bx^3 + cx^2 + dx + k $, where $ a \neq 0 $ and $ b, c, d, $ and $ k $ are constants. Which of the following statements about $ f $ is true?

(A) $ f $ has both a global maximum and a global minimum.
(B) $ f $ has either a global maximum or a global minimum, but not both.
(C) $ f $ has neither a global maximum nor a global minimum.
(D) The nature of a global maximum or a global minimum for $ f $ cannot be determined without more information about $ b, c, d, $ and $ k $.

▶️ Answer/Explanation

Detailed solution

$ f $ is a quartic polynomial (degree 4, even).
For even-degree polynomials with leading coefficient $ a \neq 0 $:
– If $ a > 0 $, as $ x \to \pm\infty $, $ f(x) \to +\infty $ ⇒ global minimum exists, no global maximum.
– If $ a < 0 $, as $ x \to \pm\infty $, $ f(x) \to -\infty $ ⇒ global maximum exists, no global minimum.
Thus $ f $ has either a global max or a global min, but not both.
✅ Answer: (B)

Question

The function $ f $ is given by $ f(x) = 5x^6 – 2x^3 – 3 $. Which of the following describes the end behavior of $ f $?

(A) $ \displaystyle \lim_{x \to -\infty} f(x) = -\infty $ and $ \displaystyle \lim_{x \to \infty} f(x) = -\infty $
(B) $ \displaystyle \lim_{x \to -\infty} f(x) = \infty $ and $ \displaystyle \lim_{x \to \infty} f(x) = \infty $
(C) $ \displaystyle \lim_{x \to -\infty} f(x) = -\infty $ and $ \displaystyle \lim_{x \to \infty} f(x) = \infty $
(D) $ \displaystyle \lim_{x \to -\infty} f(x) = \infty $ and $ \displaystyle \lim_{x \to \infty} f(x) = -\infty $

▶️ Answer/Explanation

Detailed solution

The end behavior of a polynomial is determined by its leading term when the degree is even or odd and the leading coefficient is positive or negative.
For $ f(x) = 5x^6 – 2x^3 – 3 $:
• Degree = 6 (even).
• Leading coefficient = 5 (positive).

For even degree and positive leading coefficient:
As $ x \to \pm\infty $, $ 5x^6 \to +\infty $.
Thus:
\[ \lim_{x \to -\infty} f(x) = \infty \] \[ \lim_{x \to \infty} f(x) = \infty \]
This matches option (B).
✅ Answer: (B)

Question

In the odd function $f(x)$, the $\lim_{x \to 3^-} f(x) = -\infty$. Which one of the following must also be true?

a. $\lim_{x \to -3^-} f(x) = -\infty$

b. $\lim_{x \to -3^-} f(x) = \infty$

c. $\lim_{x \to -3^+} f(x) = -\infty$

d. $\lim_{x \to -3^+} f(x) = \infty$

▶️ Answer/Explanation

Detailed solution

An odd function satisfies the condition $f(-x) = -f(x)$, which implies symmetry about the origin.
The given limit $\lim_{x \to 3^-} f(x) = -\infty$ describes the behavior as $x$ approaches $3$ from the left.
To find the behavior near $-3$, we substitute $x$ with $-t$, so as $x \to -3^+$, then $t \to 3^-$.
Using the odd property: $f(x) = f(-t) = -f(t)$.
Therefore, $\lim_{x \to -3^+} f(x) = \lim_{t \to 3^-} [-f(t)] = -(-\infty) = \infty$.
This matches option d.
Correct Option: d

Question

The function is a polynomial with the following end behavior:

$\lim_{x\to-\infty} g(x) = -\infty \quad \text{and} \quad \lim_{x\to\infty} g(x) = -\infty$

Which of the following could be an expression for $g(x)$?

(A) $-3x^2 + 4x^2 + 1$

(B) $-x^4 + 2x^3 + 6$

(D) $4x^6 + x^3 – 5$

▶️ Answer/Explanation

Detailed solution

The end behavior of a polynomial function is determined by its leading term (the term with the highest degree).
Since $\lim_{x\to\infty} g(x) = -\infty$ and $\lim_{x\to-\infty} g(x) = -\infty$, the graph points downwards on both ends.
This behavior indicates that the polynomial must have an even degree and a negative leading coefficient.
(A) simplifies to $x^2 + 1$: Even degree, positive coefficient (Ends $\to \infty$). Incorrect.
(B) $-x^4 + 2x^3 + 6$: Even degree (4), negative coefficient (-1). Ends $\to -\infty$. Correct.
(C) $2x^3 – 7x^2 + 4x$: Odd degree. Ends go in opposite directions. Incorrect.
(D) $4x^6 + x^3 – 5$: Even degree, positive coefficient (Ends $\to \infty$). Incorrect.

Question

The polynomial function $p$ is given by $p(x) = ax^n + x – 4$ where $a$ and $n$ are constants and $n > 1$. If the end behavior of $p$ is given by the statements $\lim_{x \to -\infty} p(x) = \infty$ and $\lim_{x \to \infty} p(x) = -\infty$, which of the following could be the values of $a$ and $n$?

(A) $a = -3$ and $n = 5$

(B) $a = -3$ and $n = 4$

(D) $a = 3$ and $n = 4$

▶️ Answer/Explanation

Detailed solution

The end behavior of the polynomial $p(x)$ is determined by its leading term, $ax^n$.
Since $\lim_{x \to \infty} p(x) = -\infty$, the leading term must be negative as $x$ gets large, which implies $a < 0$.
This condition eliminates options (C) and (D) because $a$ must be negative.
Since $\lim_{x \to -\infty} p(x) = \infty$, the function rises to the left. With $a < 0$, $x^n$ must be negative.
For $x^n$ to be negative when $x$ is negative, the degree $n$ must be an odd integer.
Looking at the remaining options, only (A) has an odd value for $n$ ($n=5$).
Therefore, the correct choice is $a = -3$ and $n = 5$.

Question

The polynomial function $f$ is an odd function with domain $-4 \leq x \leq 4$. The table gives information about values of $f(x)$ and the behavior of the function. What is the absolute maximum value of $f$ on its domain?

(A) $4$
(B) $3$
(C) $2$
(D) $0$

▶️ Answer/Explanation

Detailed solution

Since $f$ is an odd function, it satisfies the property $f(-x) = -f(x)$ for all $x$ in the domain.
From the table, the minimum value on the positive domain $[0, 4]$ is $f(2) = -3$.
By the odd property, $f(-2) = -f(2) = -(-3) = 3$.
The value $f(4) = 2$ corresponds to $f(-4) = -f(4) = -2$.
In the interval $2 < x < 3$, $f(x)$ increases from $-3$ to $0$, so on $-3 < x < -2$, $f(x)$ decreases from $0$ to $3$.
The maximum value in the positive interval is $f(4) = 2$, while the odd property reveals a higher value of $3$ at $x = -2$.
Comparing all critical points and boundaries, the absolute maximum value is $3$.
Therefore, the correct option is (B).

Question

The polynomial function $p$ is given by $p(x) = ax^n + b$, where $a$ and $b$ are nonzero constants and $n$ is a positive integer. As the input values of $p$ increase without bound, the output values of $p$ decrease without bound. Which of the following must be true?

A. $a$ is negative because $\lim_{x \to \infty} p(x) = -\infty$.

B. $a$ is positive because $\lim_{x \to \infty} p(x) = -\infty$.

C. $a$ is negative because $\lim_{x \to \infty} p(x) = \infty$.

D. $a$ is positive because $\lim_{x \to \infty} p(x) = \infty$.

▶️ Answer/Explanation

Detailed solution

The phrase “input values of $p$ increase without bound” corresponds to the limit as $x \to \infty$.
The phrase “output values of $p$ decrease without bound” corresponds to the limit equaling $-\infty$.
Therefore, the condition is defined as $\lim_{x \to \infty} p(x) = -\infty$.
Considering the leading term, $\lim_{x \to \infty} (ax^n + b) = \lim_{x \to \infty} ax^n$.
Since $n$ is a positive integer, $x^n \to \infty$ as $x$ increases.
For the product $ax^n$ to approach $-\infty$, the coefficient $a$ must be negative.
Thus, the correct statement is that $a$ is negative because $\lim_{x \to \infty} p(x) = -\infty$.

Question

Let $ k $ be a polynomial function with the following end behavior: \[ \lim_{x \to -\infty} k(x) = \infty \text{ and } \lim_{x \to \infty} k(x) = -\infty \] Which of the following could be an expression for $ k(x) $?

(A) $ 4x^5 + x^2 + 3 $

(B) $ 3x^4 – x^3 + 5 $

(D) $ -5x^6 + x^3 + 6 $

▶️ Answer/Explanation

Detailed solution

The end behavior of a polynomial function is determined by its leading term (the term with the highest degree).
The problem states that as $ x \to -\infty $, $ k(x) \to \infty $ (starts up) and as $ x \to \infty $, $ k(x) \to -\infty $ (ends down).
Since the ends point in opposite directions, the degree of the polynomial must be odd.
Since the function goes to $ -\infty $ as $ x $ approaches positive infinity (right side goes down), the leading coefficient must be negative.
Looking at the options, we analyze the leading terms:
(A) $ 4x^5 $: Odd degree, positive coefficient (Ends: Down/Up). Incorrect.
(B) $ 3x^4 $: Even degree, positive coefficient (Ends: Up/Up). Incorrect.
(C) $ -2x^3 $: Odd degree, negative coefficient (Ends: Up/Down). Correct.
(D) $ -5x^6 $: Even degree, negative coefficient (Ends: Down/Down). Incorrect.

Question

For a polynomial function $ f $, $ \displaystyle \lim_{x \to -\infty} f(x) = \infty $ and $ \displaystyle \lim_{x \to \infty} f(x) = -\infty $. Which of the following must be true about $ f $?

(A) The degree of $ f $ is even, and the leading coefficient is negative.

(B) The degree of $ f $ is even, and the leading coefficient is positive.

(D) The degree of $ f $ is odd, and the leading coefficient is positive.

▶️ Answer/Explanation

Answer:

(C) The degree of $ f $ is odd, and the leading coefficient is negative.

Explanation:

The end behavior of a polynomial is determined by its leading term $ a x^n $. If the limits at $ -\infty $ and $ \infty $ are opposite in sign, the degree must be odd. Since $ f(x) \to -\infty $ as $ x \to \infty $, the leading coefficient must be negative.

Detailed solution

Line 1: The end behavior of $ f(x) $ depends on its leading term $ a x^n $.
Line 2: If $ n $ is even, both ends of the graph go in the same direction.
Line 3: Here, $ \lim_{x \to -\infty} f(x) = \infty $ and $ \lim_{x \to \infty} f(x) = -\infty $, so the ends differ.
Line 4: Therefore, the degree $ n $ must be odd.
Line 5: For an odd degree polynomial, if $ a < 0 $, then $ f(x) \to -\infty $ as $ x \to \infty $.
Line 6: This matches the given condition, so the leading coefficient is negative.
Line 7: Hence, the correct answer is (C).

Question

For the polynomial function $ g $, $ \displaystyle \lim_{x \to -\infty} g(x) = -\infty $. Which of the following expressions could define $ g(x) $?

(A) $ -5x – 2x^4 $

(B) $ -5x^2 – 2x^7 $

(D) $ -x^3 + 2x^4 $

▶️ Answer/Explanation

Answer:

(A)

Detailed solution

1. The end behavior of a polynomial depends on its highest-degree term.
2. In (A), the leading term is $ -2x^4 $.
3. Since $ x^4 $ is even, $ x^4 \to +\infty $ as $ x \to -\infty $.
4. Multiplying by $ -2 $ gives $ -2x^4 \to -\infty $.
5. In (B) and (C), the leading terms are odd powers with negative coefficients, giving $ +\infty $ as $ x \to -\infty $.
6. In (D), the leading term is $ 2x^4 $, which gives $ +\infty $.
7. Therefore, only option (A) satisfies $ \lim_{x \to -\infty} g(x) = -\infty $.

Question

The function $ f $ is given by $ f(x) = -2x^{7} + 5x^{4} + 6x^{2} – 3$. Which of the following correctly describes the end behavior of $ f $ as the input values increase without bound?

(A) $\displaystyle \lim_{x \to \infty} f(x) = \infty$

(B) $\displaystyle \lim_{x \to \infty} f(x) = -\infty$

(D) $\displaystyle \lim_{x \to -\infty} f(x) = -\infty$

▶️ Answer/Explanation

Answer:

(B) $\displaystyle \lim_{x \to \infty} f(x) = -\infty$

Detailed solution

1. The highest-degree term is $ -2x^{7} $.
2. End behavior of a polynomial is determined by its leading term.
3. As $ x \to \infty $, $ x^{7} \to \infty $.
4. Therefore, $ -2x^{7} \to -\infty $.
5. Lower-degree terms $ 5x^{4}, 6x^{2}, -3 $ become insignificant compared to $ x^{7} $.
6. Hence, $ f(x) \to -\infty $ as $ x \to \infty $.
7. Therefore, the correct answer is option (B).

Question

The function $ f $ is given by $ f(x) = 5x^6 – 2x^3 – 3 $. Which of the following describes the end behavior of $ f $?

(A) $\displaystyle \lim_{x \to -\infty} f(x) = -\infty$ and $\displaystyle \lim_{x \to \infty} f(x) = -\infty$

(B) $\displaystyle \lim_{x \to -\infty} f(x) = \infty$ and $\displaystyle \lim_{x \to \infty} f(x) = \infty$

(D) $\displaystyle \lim_{x \to -\infty} f(x) = \infty$ and $\displaystyle \lim_{x \to \infty} f(x) = -\infty$

▶️ Answer/Explanation

Answer:

(B)

Explanation:

The leading term of the polynomial is $5x^6$. Since the degree is even ($6$) and the leading coefficient $5$ is positive, the function rises to positive infinity on both ends. Therefore, \[ \lim_{x \to -\infty} f(x) = \infty \quad \text{and} \quad \lim_{x \to \infty} f(x) = \infty. \]

Detailed solution

1. The highest-degree term is $5x^6$.
2. End behavior depends only on the leading term.
3. The degree $6$ is even.
4. The leading coefficient $5 > 0$.
5. For even degree with positive coefficient, both ends rise.
6. Thus, $f(x) \to \infty$ as $x \to \pm\infty$.
7. Therefore, the correct answer is (B).

Question

The polynomial function is given by $ p(x) = -4x^5 + 3x^2 + 1 $. Which of the following statements about the end behavior of $ p $ is true?

(A) The sign of the leading term of $ p $ is positive, and the degree of the leading term of $ p $ is even; therefore,
$ \displaystyle \lim_{x \to -\infty} p(x) = \infty $ and $ \displaystyle \lim_{x \to \infty} p(x) = \infty $

(B) The sign of the leading term of $ p $ is negative, and the degree of the leading term of $ p $ is odd; therefore,
$ \displaystyle \lim_{x \to -\infty} p(x) = \infty $ and $ \displaystyle \lim_{x \to \infty} p(x) = -\infty $

(C) The sign of the leading term of $ p $ is positive, and the degree of the leading term of $ p $ is odd; therefore,
$ \displaystyle \lim_{x \to -\infty} p(x) = -\infty $ and $ \displaystyle \lim_{x \to \infty} p(x) = \infty $

(D) The sign of the leading term of $ p $ is negative, and the degree of the leading term of $ p $ is odd; therefore,
$ \displaystyle \lim_{x \to -\infty} p(x) = -\infty $ and $ \displaystyle \lim_{x \to \infty} p(x) = \infty $

▶️ Answer/Explanation

Correct Answer: (B)

The leading term of the polynomial is $ -4x^5 $. Since the coefficient $ -4 $ is negative and the degree $ 5 $ is odd, the graph rises to $ \infty $ as $ x \to -\infty $ and falls to $ -\infty $ as $ x \to \infty $. Therefore, option (B) correctly describes the end behavior.

Detailed solution

1. The given polynomial is $ p(x) = -4x^5 + 3x^2 + 1 $.
2. The leading term is $ -4x^5 $.
3. The coefficient $ -4 $ is negative.
4. The degree $ 5 $ is odd.
5. For an odd degree with negative leading coefficient: left end rises, right end falls.
6. Thus, $ \lim_{x \to -\infty} p(x) = \infty $ and $ \lim_{x \to \infty} p(x) = -\infty $.
7. Hence, the correct statement is (B).

Question

The polynomial function $ f $ is given by $ f(x) = (x – 4)(3x – 1)^2$. Which of the following descriptions of $ f $ is true?

(A) $ f $ is a polynomial of degree $ 2 $ with a leading coefficient of $ 3 $.

(B) $ f $ is a polynomial of degree $ 2 $ with a leading coefficient of $ 9 $.

(D) $ f $ is a polynomial of degree $ 3 $ with a leading coefficient of $ 9 $.

▶️ Answer/Explanation

Answer:

(D) $ f $ is a polynomial of degree $ 3 $ with a leading coefficient of $ 9 $.

Explanation:

The factor $ (x – 4) $ has degree $ 1 $, and $ (3x – 1)^2 $ has degree $ 2 $. Thus, the total degree is $ 1 + 2 = 3 $. The leading term of $ (3x – 1)^2 $ is $ 9x^2 $. Multiplying by $ x $ from $ (x – 4) $ gives the leading term $ 9x^3 $. Therefore, the leading coefficient is $ 9 $.

Detailed solution

1. Given $ f(x) = (x – 4)(3x – 1)^2 $.
2. Degree of $ (x – 4) = 1 $.
3. Degree of $ (3x – 1)^2 = 2 $.
4. Total degree $ = 1 + 2 = 3 $.
5. Leading term of $ (3x – 1)^2 = 9x^2 $.
6. Multiply by leading term $ x $: $ x \cdot 9x^2 = 9x^3 $.
7. Hence, leading coefficient $ = 9 $.
8. Correct option is (D).

Question

A polynomial function has the form $ p(x) = ax^j + bx^k $, where $a$ and $b$ are nonzero constants, and $j$ and $k$ are nonnegative integers. Which of the following conditions guarantees that $p$ is an even function?

(A) $ k = 0 $

(B) $ j = 2k $ and $ k $ is even

(D) $ j \cdot k $ is even

▶️ Answer/Explanation

Answer:

(B)

Explanation:

A function is even if $ p(-x) = p(x) $. For $ p(x) = ax^j + bx^k $, we have \[ p(-x) = a(-x)^j + b(-x)^k. \] This equals $ ax^j + bx^k $ only when both exponents $j$ and $k$ are even. Option (B) guarantees this because if $k$ is even and $j = 2k$, then $j$ is also even. Hence both exponents are even, making $p(x)$ an even function.

Detailed solution

1. A function is even if $ p(-x) = p(x) $.
2. Compute $ p(-x) = a(-x)^j + b(-x)^k $.
3. For equality, both $ (-x)^j = x^j $ and $ (-x)^k = x^k $.
4. This occurs only when $ j $ and $ k $ are even integers.
5. In option (B), $ k $ is even and $ j = 2k \Rightarrow j $ is even.
6. Therefore, both exponents are even, guaranteeing an even function.

AP Precalculus -1.6 Polynomial Functions and End Behavior- MCQ Exam Style Questions - Effective Fall 2023