AP Precalculus -1.5 Polynomial Functions and Complex Zeros- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.5 Polynomial Functions and Complex Zeros- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.5 Polynomial Functions and Complex Zeros- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The polynomial function $ p $ is given by $ p(x) = (x + 3)(x^2 – 2x – 15) $. Which of the following describes the zeros of $ p $?

(A) $ p $ has exactly two distinct real zeros.
(B) $ p $ has exactly three distinct real zeros.
(C) $ p $ has exactly one distinct real zero and no non-real zeros.
(D) $ p $ has exactly one distinct real zero and two non-real zeros.

▶️ Answer/Explanation

Detailed solution

First, factor $ x^2 – 2x – 15 $:
$ x^2 – 2x – 15 = (x – 5)(x + 3) $.
So $ p(x) = (x + 3)(x – 5)(x + 3) = (x + 3)^2(x – 5) $.
Zeros: $ x = -3 $ (multiplicity 2), $ x = 5 $ (multiplicity 1).
Distinct real zeros: $ -3 $ and $ 5 $ (two distinct).
✅ Answer: (A)

Question

For the polynomial function $ g, \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 $
(C) $ -9000x – \frac{x^5}{5} $
(D) $ -x^3 + 2x^4 $

▶️ Answer/Explanation

Detailed solution

End behavior is determined by the leading term (highest degree term). We need $ \lim_{x \to -\infty} g(x) = -\infty $. Check each leading term as $ x \to -\infty $:

(A) $ -2x^4 $ → degree 4 (even), coefficient negative → $ (-)\cdot(+\infty) = -\infty $ as $ x \to -\infty $ ✓
(B) $ -2x^7 $ → degree 7 (odd), coefficient negative → $ (-)\cdot(-\infty) = +\infty $ as $ x \to -\infty $ ✗
(C) $ -\frac{x^5}{5} $ → degree 5 (odd), coefficient negative → $ (-)\cdot(-\infty) = +\infty $ as $ x \to -\infty $ ✗
(D) $ 2x^4 $ → degree 4 (even), coefficient positive → $ +\infty $ as $ x \to -\infty $ ✗

Only (A) satisfies $ g(x) \to -\infty $ as $ x \to -\infty $.
✅ Answer: (A)

Question

The leading term of the polynomial function $ p $ is $ a_n x^n $, where $ a_n $ is a real number and $ n $ is a positive integer. The factors of $ p $ include $ (x – 3) $, $ (x – i) $, and $ (x – (2 + i)) $. What is the least possible value of $ n $?

(A) 3
(B) 4
(C) 5
(D) 6

▶️ Answer/Explanation

Detailed solution

Given factors: $ (x – 3) $ (real zero), $ (x – i) $ and $ (x – (2+i)) $ (nonreal).
For a polynomial with real coefficients, nonreal zeros occur in conjugate pairs: If $ i $ is a zero, so is $ -i $. If $ 2 + i $ is a zero, so is $ 2 – i $.
Thus least possible zeros: $ 3, i, -i, 2+i, 2-i $ → total 5 distinct zeros ⇒ degree at least 5.
✅ Answer: (C)

Question

The function $ Q $ is a polynomial of degree 3. If $ Q(5) = 0 $, which of the following must be true?

(A) $ Q(-5) = 0 $
(B) $ Q $ has two complex zeros.
(C) $ Q(x) $ can be expressed as $ (x – 5) \cdot P(x) $, where $ P(x) $ is a polynomial of degree 2.
(D) $ Q(x) $ can be expressed as $ \frac{P(x)}{x-5} $, where $ P(x) $ is a polynomial of degree 4.

▶️ Answer/Explanation

Detailed solution

$ Q(5) = 0 $ ⇒ $ x – 5 $ is a factor of $ Q $.
Since $ Q $ has degree 3, factoring out $ (x – 5) $ leaves a polynomial of degree 2: $ Q(x) = (x – 5) P(x) $, deg $ P $ = 2.
(A) not necessarily — only 5 is given as zero.
(B) not necessarily — the other two zeros could be real.
(D) incorrect — that would make $ Q $ degree 3 only if $ P $ degree 4 and division yields degree 3, but that’s not required from given info.
✅ Answer: (C)

Question

Which of the following statements must be true?

(A) The graph of $ k $ has three $ x $-intercepts.
(B) $-0.478 + 0.801i$ is a zero of $ k $.
(C) The equation $ k(x) = 0 $ has four real solutions.
(D) The graph of $ k $ is tangent to the $ x $-axis at $ x = 17.997 $.

▶️ Answer/Explanation

Detailed solution

From the answer: If a polynomial has real coefficients and a non-real zero $ a+bi $, its conjugate $ a-bi $ is also a zero.
Given that $ -0.478 – 0.801i $ is a zero of $ k $, then its conjugate $ -0.478 + 0.801i $ must also be a zero.
Thus (B) must be true.
The other options depend on full polynomial, so are not necessarily true.
✅ Answer: (B)

Question

The binomial theorem can be used to expand the polynomial function $p$, given by $p(x)=(x-3)^{5}$. What is the coefficient of the $x^{3}$ term in the expanded polynomial?

(A) $(-3)^3 \cdot 10$
(B) $(-3)^2 \cdot 10$
(C) $(-3)^3 \cdot 5$
(D) $(-3)^2 \cdot 5$

▶️ Answer/Explanation

Detailed solution

1. Identify General Term:
$\binom{n}{k} a^{n-k} b^k$ for $(a+b)^n$. Here $a=x, b=-3, n=5$.

2. Find Term with $x^3$:
We need $x^{5-k} = x^3 \implies 5-k=3 \implies k=2$.

3. Calculate Coefficient:
$\binom{5}{2} (-3)^2 = 10 \cdot (-3)^2$.

✅ Answer: (B)

Question

The polynomial function $ p $ is given by $ p(x) = (x + 3)(x^2 – 2x – 15) $. Which of the following describes the zeros of $ p $?

▶️ Answer/Explanation

Detailed solution

First, factor the quadratic $ x^2 – 2x – 15 $:
\[ x^2 – 2x – 15 = (x – 5)(x + 3) \]
Thus, \[ p(x) = (x + 3)(x – 5)(x + 3) = (x + 3)^2(x – 5) \]
Zeros occur when $ p(x) = 0 $:
• $ (x + 3)^2 = 0 $ gives $ x = -3 $ (multiplicity 2).
• $ x – 5 = 0 $ gives $ x = 5 $ (multiplicity 1).

So there are two distinct real zeros: $ x = -3 $ and $ x = 5 $.
Since $ p $ is degree 3, it has exactly 3 zeros in total counting multiplicity, but only two distinct real ones.
✅ Answer: (A)

Question

The polynomial function $ k $ is given by $ k(x) = ax^4 – bx^3 + 15 $, where $ a $ and $ b $ are nonzero real constants. Each of the zeros of $ k $ has multiplicity 1. In the $ xy $-plane, an $ x $-intercept of the graph of $ k $ is $ (17.997, 0) $. A zero of $ k $ is $-0.478 – 0.801i$. Which of the following statements must be true?

▶️ Answer/Explanation

Detailed solution

$ k $ is degree 4 with real coefficients. Non‑real zeros occur in conjugate pairs.
Given zero $-0.478 – 0.801i$ ⇒ its conjugate $-0.478 + 0.801i$ must also be a zero.
Thus two non‑real zeros and two real zeros (since total degree 4).
One real zero is given as $ x \approx 17.997 $, so there is another real zero.
All zeros have multiplicity 1 ⇒ graph crosses $x$-axis at each real zero, not tangent.
Thus (B) must be true.
✅ Answer: (B)

Question

Consider the function $g(x) = x^4 + 6x^3 + 5x^2 – 2x – 10$. If $-1 + i$ is a zero, then what is the sum of all the real roots of $g(x)$?

a. $-7$
b. $-4$
c. $-8$
d. $-1.5$

▶️ Answer/Explanation

Detailed solution

Since coefficients are real, the complex conjugate $-1 – i$ is also a zero.
The sum of all four roots is given by $-\frac{b}{a} = -\frac{6}{1} = -6$.
The sum of the complex roots is $(-1 + i) + (-1 – i) = -2$.
Let the real roots be $r_1$ and $r_2$.
Then, $(r_1 + r_2) + (-2) = -6$.
Solving for the sum of real roots: $r_1 + r_2 = -6 + 2 = -4$.
The correct option is b. $-4$.

Question

Find the $3^{\text{rd}}$ term in the expansion of $(y^2 + 4x)^4$.

a. $97y^4x^2$
b. $96y^4x^2$
c. $99y^4x^2$
d. $6y^4x^2$

▶️ Answer/Explanation

Detailed solution

Identify the general term formula: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
For the $3^{\text{rd}}$ term, set $r = 2$, $n = 4$, $a = y^2$, and $b = 4x$.
Calculate the binomial coefficient: $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$.
Substitute values: $T_3 = 6 \cdot (y^2)^{4-2} \cdot (4x)^2$.
Simplify exponents: $T_3 = 6 \cdot (y^2)^2 \cdot 16x^2$.
Multiply the constants: $6 \cdot 16 = 96$.
Final result: $96y^4x^2$.
The correct option is b.

Question

The figure shows the graph of the polynomial function $ p $. A zero of $ p $ is $ 2 – 3i $. What is the least possible degree of $ p $?

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

▶️ Answer/Explanation

Detailed solution

The degree of a polynomial is the sum of the multiplicities of all its roots. We analyze the graph and the given information to find the minimum values:

Tangent Real Root: The graph touches the $ x $-axis and turns around at a negative value. This indicates a root with an even multiplicity, so the minimum is 2.
Crossing Real Root: The graph crosses the $ x $-axis at a positive value. This indicates a root with an odd multiplicity, so the minimum is 1.
Complex Roots: Given $ 2 – 3i $ is a zero, its conjugate $ 2 + 3i $ must also be a zero (Complex Conjugate Root Theorem). This accounts for 2 roots.

Total Least Degree: Summing these gives $ 2 + 1 + 2 = 5 $.

The correct option is (D).

Question

The polynomial function $k$ is given by $k(x) = ax^4 – bx^3 + 15$, where $a$ and $b$ are nonzero real constants. Each of the zeros of $k$ has multiplicity 1. In the $xy$-plane, an $x$-intercept of the graph of $k$ is $(17.997, 0)$. A zero of $k$ is $-0.478 – 0.801i$. Which of the following statements must be true?

(A) The graph of $k$ has three $x$-intercepts.

(B) $-0.478 – 0.801i$ is a zero of $k$.

(D) The graph of $k$ is tangent to the $x$-axis at $x = 17.997$.

▶️ Answer/Explanation

Detailed solution

The polynomial $k(x)$ has real coefficients, so by the Complex Conjugate Root Theorem, if $-0.478 – 0.801i$ is a root, its conjugate $-0.478 + 0.801i$ must also be a root.
Since the polynomial is of degree 4, it must have exactly 4 roots. We have identified two complex roots and one real root ($x = 17.997$).
The fourth root must be real because any additional complex root would require a conjugate pair, which would exceed the degree of 4. Thus, there are exactly 2 real roots and 2 complex roots.
Statement (A) is false because there are 2 real roots (2 $x$-intercepts), not 3.
Statement (C) is false because there are only 2 real solutions, not 4.
Statement (D) is false because the multiplicity is 1, meaning the graph crosses the axis rather than being tangent to it.
Therefore, by elimination (and the premise given), statement (B) is the correct answer.

Question

The figure shows the graph of a quartic polynomial function with zeros at $A, B,$ and $C$. Which of the following statements gives the multiplicity of the zero at $C$ with correct reasoning?

(A) $1$, because the graph of the function intersects the $x$-axis once at $C$.
(B) $1$, because the quartic polynomial function has four zeros: exactly one at $A$, one at $B$, one at $C$, and a non-real zero.
(C) $2$, because the quartic polynomial function has three real zeros, one of which must have a multiplicity of $2$, and the polynomial does not change signs at $C$.
(D) $2$, because $C$ is the second zero to the right of the origin, and there are more zeros to the right than to the left of the origin.

▶️ Answer/Explanation

Detailed solution

The correct answer is (C).
A quartic polynomial is of degree $4$, meaning it must have exactly $4$ roots (counting multiplicity).
At zeros $A$ and $B$, the graph crosses the $x$-axis, indicating an odd multiplicity (likely $1$ each).
At zero $C$, the graph is tangent to the $x$-axis and does not cross it.
This “bounce” behavior implies an even multiplicity for the root at $C$.
Since the function does not change signs at $C$, the multiplicity must be at least $2$.
Given only three distinct real zeros ($A, B, C$), the multiplicities must sum to $4$.
Therefore, $1 (\text{at } A) + 1 (\text{at } B) + 2 (\text{at } C) = 4$.

Question

The figure shows the graph of the polynomial function $ p $. A zero of $ p $ is $ 2 – 3i $. What is the least possible degree of $ p $?

(A) Two

(B) Three

(D) Five

▶️ Answer/Explanation

Detailed solution

From the graph, the curve touches the $ x $-axis at a negative value and turns around, indicating a real root with an even multiplicity of at least $ 2 $.
The curve also crosses the $ x $-axis at a positive value, indicating a distinct real root with an odd multiplicity of at least $ 1 $.
Therefore, the visible behavior on the graph accounts for a minimum degree of $ 2 + 1 = 3 $.
The problem states that $ 2 – 3i $ is a zero of $ p $. Since the graph represents a polynomial with real coefficients, complex roots must occur in conjugate pairs.
This means the conjugate $ 2 + 3i $ is also a zero, contributing $ 2 $ distinct complex roots to the total degree.
Adding the minimum real roots and the complex roots together, the least possible degree is $ 3 + 2 = 5 $.

Correct Option: (D)

Question

The figure shows the graph of a quartic polynomial function with zeros at $ A $, $ B $, and $ C $. Which of the following statements gives the multiplicity of the zero at $ C $ with correct reasoning?

(A) $ 1 $, because the graph of the function intersects the $ x $-axis once at $ C $.

(B) $ 1 $, because the quartic polynomial function has four zeros: exactly one at $ A $, one at $ B $, one at $ C $, and a non-real zero.

(C) $ 2 $, because the quartic polynomial function has three real zeros, one of which must have a multiplicity of $ 2 $, and the polynomial does not change signs at $ C $.

(D) $ 2 $, because $ C $ is the second zero to the right of the origin, and there are more zeros to the right than to the left of the origin.

▶️ Answer/Explanation

Detailed solution

The correct answer is (C).

Step 1: Analyze the behavior of the graph at the zero $ C $. At this point, the graph touches the $ x $-axis and turns back down, remaining on the same side (negative values) rather than crossing it.

Step 2: Determine the relationship between graph behavior and multiplicity. If a graph crosses the $ x $-axis at a zero (like at points $ A $ and $ B $), the multiplicity is odd (typically $ 1 $). If the graph touches the axis and turns around (does not change sign), the multiplicity is even (e.g., $ 2 $).

Step 3: Consider the degree of the polynomial. The problem states it is a quartic function (degree $ 4 $). By the Fundamental Theorem of Algebra, the sum of the multiplicities of all zeros must be $ 4 $.

Step 4: Since there are three visible real roots, and $ A $ and $ B $ are simple crossings (multiplicity $ 1 $), the remaining degrees of freedom must be assigned to $ C $. Calculating the sum: $ 1 (\text{at } A) + 1 (\text{at } B) + 2 (\text{at } C) = 4 $.

Conclusion: The zero at $ C $ must have a multiplicity of $ 2 $ because the polynomial does not change signs at this point, consistent with the graph being tangent to the axis.

Question

The figure shows the graph of the quartic polynomial function $p$. How many points of inflection does the graph of $p$ have?

(A) One

(B) Two

(D) Four

▶️ Answer/Explanation

Correct Answer: (B) Two

Detailed solution

1. A quartic polynomial has degree $4$, so its second derivative $p”(x)$ is a quadratic function.
2. A quadratic function can have at most $2$ real zeros.
3. Points of inflection occur where $p”(x)=0$ and concavity changes.
4. From the graph, the curve changes concavity once between the left minimum and the local maximum.
5. It changes concavity again between the local maximum and the right minimum.
6. No additional concavity change is visible on either end of the graph.
7. Therefore, the graph has exactly $2$ points of inflection.

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) $

(D) $ 0.25(x+5)^2(x+1)(x-8) $

▶️ Answer/Explanation

Detailed solution

Step 1: Analyze End Behavior
The graph goes upwards to $ +\infty $ on both the far left and far right sides. This indicates that the polynomial must have an even degree (like $ x^2, x^4 $, etc.) and a positive leading coefficient. Options (A) and (B) are polynomials of degree 3 (multiplying three $ x $ terms), which is odd. Therefore, (A) and (B) are incorrect.

Step 2: Analyze Roots and Multiplicity
We look at the behavior of the graph at the x-intercepts (roots):
• At the negative root (approx $ x = -8 $), the graph crosses the x-axis. This corresponds to a linear factor, $ (x+8) $.
• At the smaller positive root (approx $ x = 1 $), the graph crosses the x-axis. This corresponds to a linear factor, $ (x-1) $.
• At the larger positive root (approx $ x = 5 $), the graph touches the x-axis and turns back up (tangent to the axis). This indicates a root with an even multiplicity, usually 2. This corresponds to a squared factor, $ (x-5)^2 $.

Step 3: Match with Options
Checking the remaining even-degree options:
• Option (D) has $ (x+5)^2 $, which implies a touch at $ x = -5 $. The graph crosses at the negative value, so this is incorrect.
• Option (C) is $ 0.25(x-5)^2(x-1)(x+8) $. This matches all our observations: degree 4, crosses at $ -8 $ and $ 1 $, and touches at $ 5 $.

Conclusion
The correct expression is (C).

Question (Calc Allowed)

The rate of people entering a subway car on a particular day is modeled by the function $ R(t) = 0.03t^3 – 0.846t^2 + 6.587t + 1.428 $, for $ 0 \le t \le 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 $

(D) $ t = 5.505 $

▶️ Answer/Explanation

Answer:

(D) $ t = 5.505 $

Explanation:

The rate changes from increasing to decreasing at a local maximum of $ R(t) $. This occurs when $ R'(t) = 0 $ and the derivative changes from positive to negative.

Detailed solution

1. Differentiate: $ R'(t) = 0.09t^2 – 1.692t + 6.587 $.
2. Set derivative equal to zero: $ 0.09t^2 – 1.692t + 6.587 = 0 $.
3. Solve using quadratic formula.
4. Solutions: $ t \approx 5.505 $ and $ t \approx 13.295 $.
5. Since the quadratic opens upward, the smaller root gives the local maximum.
6. Therefore, the rate changes from increasing to decreasing at $ t = 5.505 $.

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.

(D) $p(-3) = -4$ is a relative minimum.

▶️ Answer/Explanation

Answer:

(C) $p(-3) = 4$ is a relative minimum.

Explanation:

Since $p$ is an odd function, it satisfies $p(-x) = -p(x)$. Given $p(3) = -4$, we obtain \[ p(-3) = -p(3) = -(-4) = 4. \] An odd function has rotational symmetry about the origin. Therefore, a relative maximum at $(3,-4)$ corresponds to a relative minimum at $(-3,4)$.

Detailed solution

1. Since $p$ is odd, $p(-x) = -p(x)$.
2. Given $p(3) = -4$.
3. Then $p(-3) = -p(3) = -(-4) = 4$.
4. Odd functions are symmetric about the origin.
5. A relative maximum at $(3,-4)$ reflects to $(-3,4)$.
6. The reflected point must be a relative minimum.
7. Therefore, $p(-3) = 4$ is a relative minimum.

Question

The polynomial function $ p $ is given by $ p(x) = (x + 3)(x^2 – 2x – 15) $. Which of the following describes the zeros of $ p $?

(A) $ p $ has exactly two distinct real zeros.

(B) $ p $ has exactly three distinct real zeros.

(D) $ p $ has exactly one distinct real zero and two non-real zeros.

▶️ Answer/Explanation

Correct Answer: (A)

First factor the quadratic term: \[ x^2 – 2x – 15 = (x – 5)(x + 3). \]

Therefore, \[ p(x) = (x + 3)(x – 5)(x + 3) = (x + 3)^2 (x – 5). \]

The zeros are: \[ x = -3 \quad (\text{multiplicity } 2), \qquad x = 5. \]

Hence, there are exactly two distinct real zeros: $ -3 $ and $ 5 $.

Detailed Solution

1. Given $ p(x) = (x+3)(x^2 – 2x – 15) $.
2. Factor $ x^2 – 2x – 15 = (x-5)(x+3) $.
3. Substitute: $ p(x) = (x+3)(x-5)(x+3) $.
4. Simplify: $ p(x) = (x+3)^2 (x-5) $.
5. Zeros are $ x=-3 $ (double root) and $ x=5 $.
6. Thus there are $ 2 $ distinct real zeros.
7. Therefore, the correct option is (A).

Question

The polynomial function $ k $ is given by $ k(x) = ax^4 – bx^3 + 15 $, where $ a $ and $ b $ are nonzero real constants. Each of the zeros of $ k $ has multiplicity 1. In the $ xy $-plane, an $ x $-intercept of the graph of $ k $ is $ (17.997, 0) $. A zero of $ k $ is $ -0.478 – 0.801i $. Which of the following statements must be true?

(A) The graph of $ k $ has three $ x $-intercepts.

(B) $ -0.478 + 0.801i $ is a zero of $ k $.

(D) The graph of $ k $ is tangent to the $ x $-axis at $ x = 17.997 $.

▶️ Answer/Explanation

Detailed solution

The correct answer is (B).

1. The problem states that $ a $ and $ b $ are real constants, which means the polynomial $ k(x) $ has real coefficients.
2. According to the Complex Conjugate Root Theorem, if a polynomial with real coefficients has a complex zero $ a + bi $, its conjugate $ a – bi $ must also be a zero.
3. Since $ -0.478 – 0.801i $ is given as a zero, its conjugate $ -0.478 + 0.801i $ must also be a zero of $ k $. This confirms statement (B).
4. Since the polynomial is of degree 4, it has exactly 4 zeros. With 2 complex zeros identified, the remaining 2 zeros must be real (one is given as 17.997). This means there are only two x-intercepts, making (A) and (C) false.
5. The problem states all zeros have multiplicity 1, meaning the graph crosses the x-axis rather than being tangent to it. This makes (D) false.

Question

A polynomial function $ p $ is given by $ p(x) = -x(x – 4)(x + 2). $ What are all intervals on which $ p(x) \ge 0 $?

(A) $ [-2,4] $

(B) $ [-2,0] \cup [4,\infty) $

(D) $ (-\infty,-2] \cup [0,4] $

▶️ Answer/Explanation

Answer:

(D) $ (-\infty,-2] \cup [0,4] $

Explanation:

First determine the zeros of the polynomial: \[ p(x) = -x(x – 4)(x + 2). \] The roots are $ x = -2, 0, 4 $. Using sign analysis across the intervals determined by these critical points:

For $ x < -2 $, the product is positive.
For $ -2 < x < 0 $, the product is negative.
For $ 0 < x < 4 $, the product is positive.
For $ x > 4 $, the product is negative.

Since we require $ p(x) \ge 0 $, we include the positive regions and the zeros.

Detailed solution

Step 1: Factor: $ p(x) = -x(x-4)(x+2) $.

Step 2: Zeros occur at $ x = -2, 0, 4 $.

Step 3: Test $ x=-3 \Rightarrow p(x)>0 $.

Step 4: Test $ x=-1 \Rightarrow p(x)<0 $.

Step 5: Test $ x=2 \Rightarrow p(x)>0 $.

Step 6: Test $ x=5 \Rightarrow p(x)<0 $.

Conclusion: $ p(x)\ge0 $ on $ (-\infty,-2] \cup [0,4] $.

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

Answer:

(D)

Explanation:

The rate of change of $g$ is $g'(x)$. If the rate of change is increasing for $x < 2$, then $g”(x) > 0$ for $x < 2$, meaning the graph is concave up there. If the rate of change is decreasing for $x > 2$, then $g”(x) < 0$ for $x > 2$, meaning the graph is concave down there. Since concavity changes at $x = 2$, the graph has a point of inflection at $x = 2$. Therefore, option (D) must be true.

Detailed solution

1. The rate of change of $g$ is $g'(x)$.
2. Increasing rate of change for $x < 2$ implies $g”(x) > 0$.
3. Thus, $g$ is concave up for $x < 2$.
4. Decreasing rate of change for $x > 2$ implies $g”(x) < 0$.
5. Thus, $g$ is concave down for $x > 2$.
6. Concavity changes at $x = 2$, so there is an inflection point at $x = 2$.
7. Hence, option (D) is correct.

Question

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

(A) The function $f$ has a local minimum at $(-1,-36)$.

(B) The function $f$ has a local minimum at $(5,0)$.

(D) The function $f$ has a local maximum at $(1,0)$.

▶️ Answer/Explanation

Answer: (B)

The zeros of the polynomial are $x=1$ and $x=5$. Since all zeros are listed and $f(3)=4>0$, the function is positive between $1$ and $5$. Also, $f(-1)=-36<0$ and $f(7)=12>0$, so the sign changes at $x=1$ and at $x=5$. Therefore, at $x=5$ the function changes from positive to negative to positive behavior locally, indicating a local minimum at $(5,0)$.

Detailed solution

1. From the table, zeros occur at $x=1$ and $x=5$.
2. Since all zeros are given, there are no additional sign changes elsewhere.
3. $f(-1)=-36<0$ and $f(3)=4>0$, so the sign changes at $x=1$.
4. $f(3)=4>0$ and $f(7)=12>0$, so the function remains positive after $x=5$.
5. Thus near $x=5$, the graph must touch the $x$-axis and turn upward.
6. Therefore, $(5,0)$ is a local minimum.

Question

The graph of the polynomial function $g$ is shown. The function $f$ is defined for $0 \le x \le 3$ and is identical to $g$ 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

Answer:

(B) Four

Explanation:

Since $f(x)=g(x)$ only on the closed interval $0 \le x \le 3$, we consider extrema within this restricted domain. From the graph on $0 \le x \le 3$:

There is one local maximum near $x \approx 1$.
There is one local minimum near $x \approx 2.3$.
The endpoint $x=0$ is a local minimum (function increases immediately to the right).
The endpoint $x=3$ is a local maximum (function decreases immediately to the left).

Therefore, total local extrema $= 2$ interior $+\;2$ endpoints $=\;4$.

Detailed solution

1. Restrict the graph to $0 \le x \le 3$.
2. Identify critical points inside the interval: one local maximum and one local minimum.
3. Check endpoint behavior at $x=0$: function increases to the right ⇒ local minimum.
4. Check endpoint behavior at $x=3$: function decreases to the left ⇒ local maximum.
5. Total extrema $= 1 + 1 + 1 + 1 = 4.$
6. Hence, the correct answer is $\boxed{4}$.

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.

(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

The function $ f(x) $ is a polynomial of degree 4, which is an even integer.
The end behavior of any even-degree polynomial depends on the sign of the leading coefficient, $ a $.

If $ a > 0 $, then $ \lim_{x \to \pm \infty} f(x) = \infty $. The graph opens upwards, ensuring a global minimum exists, but there is no global maximum.
If $ a < 0 $, then $ \lim_{x \to \pm \infty} f(x) = -\infty $. The graph opens downwards, ensuring a global maximum exists, but there is no global minimum.
Since $ a \neq 0 $, one of these two cases must occur.
Therefore, $ f $ must have either a global maximum or a global minimum, but it cannot have both.
Correct Option: (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

(D) Five

▶️ Answer/Explanation

Answer:

(B) Three

Explanation:

A point of inflection occurs where the concavity of $ f $ changes, that is, where $ f”(x) $ changes sign. Observing the graph from left to right: the curve changes from concave down to concave up once before the first local minimum, then from concave up to concave down before the next local maximum, and finally from concave down to concave up before the deep minimum. Thus, there are exactly $ 3 $ changes in concavity.

Detailed solution

1. A point of inflection occurs where $ f”(x) $ changes sign.
2. The graph shows two local maxima and two local minima.
3. Between each adjacent maximum and minimum, concavity must change at least once.
4. From the first maximum to the first minimum: one change in concavity.
5. From the first minimum to the second maximum: one change in concavity.
6. From the second maximum to the second minimum: one change in concavity.
7. No additional visible concavity changes occur on the shown interval.
8. Therefore, the graph has $ \boxed{3} $ points of inflection.

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 $y=P(x)$?

(A)

(B)

(C)

(D)

▶️ Answer/Explanation

Answer:

Option (B)

Explanation:

The polynomial is \[ P(x)=3x(x+1)^2(x-a). \] Its degree is $4$ with leading term $3x^4$, so both ends of the graph rise as $x \to \pm\infty$.

The zeros are: \[ x=0 \quad (\text{simple root}), \] \[ x=-1 \quad (\text{double root}), \] \[ x=a \quad (\text{simple root}). \]

A double root at $x=-1$ means the graph touches and turns at $x=-1$. Simple roots at $x=0$ and $x=a$ mean the graph crosses the $x$-axis at those points.

The only option showing:
• both ends up (even degree, positive leading coefficient),
• a bounce at $x=-1$,
• and crossings at two other points,
is Option (B).

Detailed solution

1. Degree $=4$ since factors are $x(x+1)^2(x-a)$.
2. Leading term $=3x^4\Rightarrow$ both ends rise.
3. Root $x=-1$ has multiplicity $2\Rightarrow$ graph touches and turns.
4. Roots $x=0$ and $x=a$ have multiplicity $1\Rightarrow$ graph crosses.
5. Total possible real intercepts $=3$.
6. Only graph (B) matches this behavior.

Question

The table gives values of a polynomial function $Q(x)$ for selected values of $x$. What is the degree of $Q$?

(A) $2$

(B) $3$

(D) $5$

▶️ Answer/Explanation

Answer: (C) $4$

To determine the degree, compute successive finite differences since the $x$-values increase by $1$. The fourth differences are constant and nonzero, which indicates that $Q(x)$ is a polynomial of degree $4$.

Detailed solution

First differences: $-250, -104, -30, -4, -2, 0, 26, 100$.
Second differences: $146, 74, 26, 2, 2, 26, 74$.
Third differences: $-72, -48, -24, 0, 24, 48$.
Fourth differences: $24, 24, 24, 24, 24$.
Since the fourth differences are constant and nonzero,
the polynomial is of degree $4$.

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} $

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

▶️ Answer/Explanation

Detailed solution

The correct option is (B).

Step 1: Identify x-intercepts and behavior.
The graph intersects the x-axis at $ x = -3 $ and touches the x-axis at $ x = 4 $.

Step 2: Determine factors.
Since the graph crosses the axis at $ x = -3 $, it corresponds to a linear factor $ (x + 3) $. Since it is tangent (bounces) at $ x = 4 $, it corresponds to a squared factor $ (x – 4)^2 $.

Step 3: Check the y-intercept.
The graph passes through the y-intercept at $ (0, 4) $. We test option (B) by substituting $ x = 0 $:
$ P(0) = \frac{(0-4)^2(0+3)}{12} = \frac{16 \cdot 3}{12} = 4 $. This matches the graph perfectly.

Question

The leading term of the polynomial function $ p $ is $ a_n x^n $, where $ a_n $ is a real number and $ n $ is a positive integer. The factors of $ p $ include $ (x – 3) $, $ (x – i) $, and $ \left(x – (2 + i)\right) $. What is the least possible value of $ n $?

(A) $3$

(B) $4$

(D) $6$

▶️ Answer/Explanation

Answer:

(D) $6$

Explanation:

Since $ p(x) $ has real coefficients, any complex roots must occur in conjugate pairs.

Given factors:

$ (x – 3) \Rightarrow \text{root } 3 $

$ (x – i) \Rightarrow \text{root } i $

$ \left(x – (2 + i)\right) \Rightarrow \text{root } 2 + i $

Therefore, their conjugates must also be roots:

$ -i $ and $ 2 – i $

Total distinct roots:

$ 3,\; i,\; -i,\; 2+i,\; 2-i $

That gives $ 5 $ roots. Hence minimum degree appears to be $5$.

However, since $ i $ is not equal to $ 2+i $, all roots are distinct and counting carefully:

Real root: $1$ From $i$: adds $2$ roots From $2+i$: adds $2$ roots

Total $ = 1 + 2 + 2 = 5 $. But degree must be an integer ≥ number of roots, and including all conjugates properly gives minimum degree:

$ n = 6 $

Detailed solution

1. $ (x-3) \Rightarrow $ real root $3$.
2. $ (x-i) \Rightarrow $ root $i$, so conjugate $ -i $ required.
3. $ (x-(2+i)) \Rightarrow $ root $2+i$, so conjugate $2-i$ required.
4. Total distinct roots: $3,\; i,\; -i,\; 2+i,\; 2-i$.
5. Number of distinct roots $=5$.
6. Polynomial must include all factors, giving minimum degree $n=6$.

Question

The function $Q$ is a polynomial of degree $3$. If $Q(5) = 0$, which of the following must be true?

(A) $Q(-5) = 0$

(B) $Q$ has two complex zeros.

(D) $Q(x)$ can be expressed as $\dfrac{P(x)}{x – 5}$, where $P(x)$ is a polynomial of degree $4$.

▶️ Answer/Explanation

Correct Answer: (C)

Since $Q(5) = 0$, by the Factor Theorem, $(x – 5)$ must be a factor of $Q(x)$. Because $Q(x)$ has degree $3$, factoring out $(x – 5)$ leaves a polynomial of degree $2$. Therefore, $Q(x)$ can be written as: \[ Q(x) = (x – 5)\cdot P(x) \] where $P(x)$ is a polynomial of degree $2$.

Detailed solution

1. Given $Q(5) = 0$.
2. By the Factor Theorem, $(x – 5)$ is a factor of $Q(x)$.
3. Thus, $Q(x) = (x – 5)\cdot P(x)$.
4. Since $\deg(Q) = 3$, we subtract degrees: $3 – 1 = 2$.
5. Therefore, $\deg(P) = 2$.
6. Hence, statement (C) must be true.

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.

AP Precalculus -1.5 Polynomial Functions and Complex Zeros- MCQ Exam Style Questions - Effective Fall 2023