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AP Precalculus -1.8 Rational Functions and Zeros- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.8 Rational Functions and Zeros- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.8 Rational Functions and Zeros- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question 

The functions \( g \) and \( f \) are given by \( g(x) = 3x^2 – 2x \) and \( f(x) = 6x^4 + 5x^3 + 3x – 5 \). Which of the following statements is true about the remainder when \( f(x) \) is divided by \( g(x) \)?
(A) The remainder is 0, so \( g(x) \) is a factor of \( f(x) \).
(B) The remainder is 0, so \( f(x) \) is a factor of \( g(x) \).
(C) The remainder is \( (7x – 5) \), so \( g(x) \) is not a factor of \( f(x) \), and the graph of \( y = \frac{f(x)}{g(x)} \) has a slant asymptote.
(D) The remainder is \( (7x – 5) \), so \( g(x) \) is not a factor of \( f(x) \), and the graph of \( y = \frac{f(x)}{g(x)} \) does not have a slant asymptote.
▶️ Answer/Explanation
Detailed solution

Degree of \( g \) is 2, degree of \( f \) is 4. After division, quotient degree = 2, remainder degree < 2.
Given answer says remainder is \( 7x – 5 \). Then \( g(x) \) is not a factor. Slant asymptote occurs when degree of numerator is exactly one more than degree of denominator. Here degree of \( f \) is 4, degree of \( g \) is 2 ⇒ degree difference = 2 ⇒ quotient is quadratic, not linear ⇒ no slant asymptote.
✅ Answer: (D)

Question 

The function \( g \) is given by \( g(x) = x^3 – 3x^2 – 18x \), and the function \( h \) is given by \( h(x) = x^2 – 2x – 35 \). Let \( k \) be the function given by \( k(x) = \frac{h(x)}{g(x)} \). What is the domain of \( k \)?
(A) all real numbers \( x \) where \( x \neq 0 \)
(B) all real numbers \( x \) where \( x \neq -5, x \neq 7 \)
(C) all real numbers \( x \) where \( x \neq -3, x \neq 0, x \neq 6 \)
(D) all real numbers \( x \) where \( x \neq -5, x \neq -3, x \neq 0, x \neq 6, x \neq 7 \)
▶️ Answer/Explanation
Detailed solution

Factor \( g(x) \) and \( h(x) \):
\( g(x) = x^3 – 3x^2 – 18x = x(x^2 – 3x – 18) = x(x – 6)(x + 3) \)
Zeros of \( g \): \( x = 0, 6, -3 \)
\( h(x) = x^2 – 2x – 35 = (x – 7)(x + 5) \)
Thus \( k(x) = \frac{(x-7)(x+5)}{x(x-6)(x+3)} \)
Domain excludes zeros of denominator unless canceled by numerator. None of the numerator factors cancel with denominator factors.
So domain: all real \( x \) except \( x = 0, 6, -3 \).
✅ Answer: (C)

Question 

The rational function \( g \) is given by \( g(x) = \frac{(x^2+3x)(x^2-4x-5)}{(x+3)(x-1)(x-2)} \). For what input values of \( g \) are the output values of \( g \) equal to 0?
(A) 0 only
(B) \(-1, 0\), and 5 only
(C) \(-3, 1\), and 2
(D) \(-3, -1, 0\), and 5
▶️ Answer/Explanation
Detailed solution

Zeros of \( g \) occur where numerator = 0 and denominator ≠ 0.
Numerator: \( (x^2+3x)(x^2-4x-5) = x(x+3)(x-5)(x+1) \).
Zeros: \( x = 0, -3, 5, -1 \).
Denominator zeros: \( x = -3, 1, 2 \) (excluded from domain).
Thus zero \( x = -3 \) is not in domain (makes denominator zero).
So valid zeros: \( -1, 0, 5 \).
✅ Answer: (B)

Question 

The rational function \( r \) is given by \( r(x) = \frac{x^3 + 4x^2 + 4x}{x^2 – 9} \). On what intervals of \( x \) is \( r(x) \geq 0 \)?
(A) \( x \geq 0 \)
(B) \(-3 < x < 3\)
(C) \(-3 < x < -2, -2 < x < 0\), and \( x > 3 \) only
(D) \(-3 < x \leq 0\) and \( x > 3 \)
▶️ Answer/Explanation
Detailed solution

Factor: \( r(x) = \frac{x(x^2+4x+4)}{x^2-9} = \frac{x(x+2)^2}{(x-3)(x+3)} \).
Zeros of numerator: \( x = 0 \) (odd multiplicity 1), \( x = -2 \) (even multiplicity 2).
Vertical asymptotes at \( x = 3, -3 \).
Sign chart with test points in intervals:
\( (-\infty, -3) \): test \( x=-4 \) ⇒ \( r(-4) = (-)(+)/(-)(-) = (+)/(-) = (-) \) ⇒ negative.
\( (-3, -2) \): test \( x=-2.5 \) ⇒ \( (-)(+)/(-)(+) = (-)/(-) = (+) \) ⇒ positive, include \( x=-2 \) because zero but multiplicity even doesn’t change sign, check \( r(-2)=0 \) ⇒ include.
\( (-2, 0) \): test \( x=-1 \) ⇒ \( (-)(+)/(-)(+) = (-)/(-) = (+) \) ⇒ positive.
\( (0, 3) \): test \( x=1 \) ⇒ \( (+)(+)/(-)(+) = (+)/(-) = (-) \) ⇒ negative.
\( (3, \infty) \): test \( x=4 \) ⇒ \( (+)(+)/(+)(+) = (+) \) ⇒ positive.
We need \( r(x) \geq 0 \), so include zeros \( x=0, x=-2 \) and positive intervals.
Thus solution: \( -3 < x \leq 0 \) (since zero at 0, zero at -2 included, and -3 not included due to VA) and \( x > 3 \).
✅ Answer: (D)

Question 

A polynomial function \( p \) is given by \( p(x) = -x(x – 4)(x + 2) \). What are all intervals on which \( p(x) \geq 0 \)?
(A) \([-2, 4]\)
(B) \([-2, 0] \cup [4, \infty)\)
(C) \((-\infty, -4] \cup [0, 2]\)
(D) \((-\infty, -2] \cup [0, 4]\)
▶️ Answer/Explanation
Detailed solution

Zeros: \( x = -2, 0, 4 \).
Leading term: \( -x \cdot x \cdot x = -x^3 \), so as \( x \to -\infty \), \( p(x) \to +\infty \) (since \(-(\text{negative}) = \text{positive}\)).
Sign analysis:
Intervals: \( (-\infty, -2), (-2, 0), (0, 4), (4, \infty) \).
Test values:
\( x = -3 \): \( p(-3) = -(-3)(-7)(-1) = -(\text{negative})(\text{negative})(\text{negative}) = -(\text{negative}) = \text{positive} \) ⇒ \( + \).
\( x = -1 \): \( -(-1)(-5)(1) = -(\text{negative})(\text{negative})(\text{positive}) = -(\text{positive}) = \text{negative} \) ⇒ \( – \).
\( x = 1 \): \( -(1)(-3)(3) = -(\text{positive})(\text{negative})(\text{positive}) = -(\text{negative}) = \text{positive} \) ⇒ \( + \).
\( x = 5 \): \( -(5)(1)(7) = \text{negative} \) ⇒ \( – \).
We want \( p(x) \geq 0 \): intervals where sign is positive or zero.
Include zeros: \( x=-2,0,4 \).
So solution: \( (-\infty, -2] \cup [0, 4] \).
✅ Answer: (D)

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