AP Precalculus -1.10 Rational Functions and Holes- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.10 Rational Functions and Holes- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.10 Rational Functions and Holes- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The rational function \( r \) is given by \( r(x) = \frac{x^{2}-4}{x^{2}-x-2} \). The table gives values of \( r(x) \) for selected values of \( x \).

Which of the following statements is true?

(A) \( \lim_{x \to 2} r(x) = \frac{4}{3}, \text{ so } r(2) = \frac{4}{3}. \)
(B) \( \lim_{x \to 2} r(x) = \frac{4}{3} \) and \( r \) is undefined at \( x = 2 \), so the graph of \( r \) has a hole at \( (2, \frac{4}{3}) \).
(C) \( \lim_{x \to \frac{4}{3}} r(x) = 2 \) and \( r \) is undefined at \( x = 2 \), so the graph of \( r \) has a hole at \( (2, \frac{4}{3}) \).
(D) \( \lim_{x \to 2^+} r(x) = \infty, \lim_{x \to 2^-} r(x) = \infty, \) and \( r \) is undefined at \( x = 2 \), so the graph of \( r \) has a vertical asymptote at \( x = 2 \).

▶️ Answer/Explanation

Detailed solution

Factor numerator and denominator:
\( r(x) = \frac{x^2-4}{x^2 – x – 2} = \frac{(x-2)(x+2)}{(x-2)(x+1)} \)
For \( x \neq 2 \), the function simplifies to \( r(x) = \frac{x+2}{x+1} \).
Thus \( r \) is undefined at \( x = 2 \) (original denominator zero) and at \( x = -1 \) (vertical asymptote from factor \( x+1 \) after cancellation).

Evaluate limit as \( x \to 2 \):
\( \lim_{x \to 2} r(x) = \frac{2+2}{2+1} = \frac{4}{3} \).
So \( r \) has a removable discontinuity (hole) at \( (2, \frac{4}{3}) \).

✅ Answer: (B)

Question

Let \( f \) be a rational function that is graphed in the \( xy \)-plane. Consider \( x = 1 \) and \( x = 7 \). The polynomial in the numerator of \( f \) has a zero at \( x = 1 \) and does not have a zero at \( x = 7 \). The polynomial in the denominator of \( f \) has zeros at both \( x = 1 \) and \( x = 7 \). The multiplicities of the zeros at \( x = 1 \) in the numerator and in the denominator are equal. Which of the following statements is true?

(A) The graph of \( f \) has holes at both \( x = 1 \) and \( x = 7 \).
(B) The graph of \( f \) has a vertical asymptote at \( x = 1 \) and a hole at \( x = 7 \).
(C) The graph of \( f \) has a hole at \( x = 1 \) and a vertical asymptote at \( x = 7 \).
(D) The graph of \( f \) has vertical asymptotes at both \( x = 1 \) and \( x = 7 \).

▶️ Answer/Explanation

Detailed solution

At \( x = 1 \): zero in numerator and denominator with equal multiplicities ⇒ factor cancels completely ⇒ hole (removable discontinuity).
At \( x = 7 \): zero in denominator but not in numerator ⇒ vertical asymptote.
Thus hole at \( x = 1 \), vertical asymptote at \( x = 7 \).
✅ Answer: (C)

Question

Which of the following functions has a zero at \( x = 3 \) and has a graph in the \( xy \)-plane with a vertical asymptote at \( x = 2 \) and a hole at \( x = 1 \)?

(A) \( h(x) = \frac{x^2 – 4x + 3}{x^2 – 3x + 2} \)
(B) \( j(x) = \frac{x^2 – 5x + 6}{x^2 – 3x + 2} \)
(C) \( k(x) = \frac{x – 3}{x^2 – 3x + 2} \)
(D) \( m(x) = \frac{x – 3}{x^2 – 4x + 3} \)

▶️ Answer/Explanation

Detailed solution

Factor each choice’s numerator and denominator:
(A) \( h(x) = \frac{x^2-4x+3}{x^2-3x+2} = \frac{(x-1)(x-3)}{(x-1)(x-2)} \) → hole at \( x=1 \), vertical asymptote at \( x=2 \), zero at \( x=3 \). ✔
(B) \( j(x) = \frac{x^2-5x+6}{x^2-3x+2} = \frac{(x-2)(x-3)}{(x-1)(x-2)} \) → hole at \( x=2 \), vertical asymptote at \( x=1 \), zero at \( x=3 \) → not matching.
(C) \( k(x) = \frac{x-3}{(x-1)(x-2)} \) → vertical asymptotes at \( x=1 \) and \( x=2 \), no hole.
(D) \( m(x) = \frac{x-3}{(x-1)(x-3)} \) → hole at \( x=3 \), vertical asymptote at \( x=1 \) → zero canceled.
Only (A) matches all conditions.
✅ Answer: (A)

Question

Which of the following names a function with a hole in its graph at \( x = 1 \) and provides correct reasoning related to the hole?

(A) The graph of \( f(x) = \frac{x^2 – 1}{x – 1} \) has a hole at \( (1, 2) \) because the values of \( \frac{x^2 – 1}{x – 1} \) get arbitrarily close to 2 for \( x \)-values sufficiently close to 1, but the function is undefined at \( x = 1 \).
(B) The graph of \( g(x) = \frac{x^2 + 1}{x – 1} \) has a hole at \( x = 1 \) because the values of \( \frac{x^2 + 1}{x – 1} \) increase without bound for \( x \)-values arbitrarily close to 1.
(C) The graph of \( h(x) = \frac{4x – 4}{x^2 + 1} \) has a hole at \( (1, 0) \) because the values of \( \frac{4x – 4}{x^2 + 1} \) are arbitrarily close to 0 for \( x \)-values sufficiently close to 1.
(D) The graph of \( k(x) = \frac{4x – 4}{(x – 1)^2} \) has a hole at \( x = 1 \) because the values of \( 4x – 4 \) and \( (x – 1)^2 \) are arbitrarily close to 0 for \( x \)-values sufficiently close to 1.

▶️ Answer/Explanation

Detailed solution

A hole occurs when a factor cancels in numerator and denominator, leaving the function undefined at that point but with a finite limit.
Check (A): \( f(x) = \frac{x^2 – 1}{x – 1} = \frac{(x-1)(x+1)}{x-1} = x+1 \) for \( x \neq 1 \). Limit as \( x \to 1 \) is 2, hole at \( (1, 2) \). ✔
(B): \( g(x) = \frac{x^2 + 1}{x – 1} \), denominator zero at \( x=1 \), numerator ≠ 0 ⇒ vertical asymptote, not a hole.
(C): \( h(x) \) defined everywhere (denominator \( x^2+1 > 0 \)), no hole.
(D): \( k(x) = \frac{4(x-1)}{(x-1)^2} = \frac{4}{x-1} \) for \( x \neq 1 \) ⇒ vertical asymptote at \( x=1 \), not a hole.
✅ Answer: (A)

Question

The rational function \( g \) is given by \( g(x) = \frac{x^3 + 1000}{x^2 – 100} = \frac{(x + 10)(x^2 – 10x + 100)}{x^2 – 100} \). Which of the following statements describes the behavior of the graph of \( g \)?

(A) The graph intersects the \( x \)-axis at \( x = -10 \) because \((-10)^3 + 1000 = 0\).
(B) The graph has a hole at \( x = -10 \) because \( (x + 10) \) appears exactly once in the numerator and exactly once in the denominator, when both the numerator and the denominator of \( g \) are factored.
(C) The graph has vertical asymptotes at \( x = 10 \) and at \( x = -10 \) because \( 10^2 – 100 = 0 \) and \( (-10)^2 – 100 = 0 \).
(D) The graph has no holes because the degree of the numerator is greater than the degree of the denominator.

▶️ Answer/Explanation

Detailed solution

Factor completely:
Numerator: \( x^3 + 1000 = (x+10)(x^2 – 10x + 100) \).
Denominator: \( x^2 – 100 = (x-10)(x+10) \).
So \( g(x) = \frac{(x+10)(x^2 – 10x + 100)}{(x-10)(x+10)} \).
Factor \( (x+10) \) cancels ⇒ hole at \( x = -10 \), not a vertical asymptote.
Denominator zero at \( x = 10 \) remains ⇒ vertical asymptote at \( x=10 \).
Check options:
(A) False: after cancellation, \( x=-10 \) not in domain (hole), no \( x \)-intercept there.
(B) True: \( x+10 \) cancels ⇒ hole at \( x=-10 \).
(C) False: No vertical asymptote at \( x=-10 \) because factor canceled.
(D) False: Hole exists regardless of degree comparison.
✅ Answer: (B)

Question

The graph of which of the following functions in the \( xy \)-plane has at least one \( x \)-intercept, at least one hole, at least one vertical asymptote, and a horizontal asymptote?

(A) \( f(x) = \frac{x^2 – 16}{x^2 – x – 6} \)
(B) \( f(x) = \frac{x^2 – 16}{x^2 -x- 30} \)
(C) \( f(x) = \frac{x^2 – 4}{x^2 – x – 30} \)
(D) \( f(x) = \frac{x^2 – 4}{x^2 -x- 6} \)

▶️ Answer/Explanation

Detailed solution

We want all four features on the same graph:

• at least one x-intercept
• at least one hole (removable discontinuity)
• at least one vertical asymptote
• a horizontal asymptote

Step 1: Horizontal asymptote

All choices are ratios of quadratics with the same degree, so every option has a horizontal asymptote
(specifically ( y = 1 )). Good so far.

Step 2: Factor everything

(A)
\(
f(x)=\dfrac{(x-4)(x+4)}{(x-3)(x+2)}
\)
No common factor
• vertical asymptotes: ( x=3,-2 )
• x-intercepts: ( x=\pm4 )
• no hole ❌

(B)
\(
f(x)=\dfrac{(x-4)(x+4)}{(x-6)(x+5)}
\)
No common factor
• vertical asymptotes: ( x=6,-5 )
• x-intercepts: ( x=\pm4 )
• no hole ❌

(C)
\(
f(x)=\dfrac{(x-2)(x+2)}{(x-6)(x+5)}
\)
No common factor
• vertical asymptotes: ( x=6,-5 )
• x-intercepts: ( x=\pm2 )
• no hole ❌

(D)
\(
f(x)=\dfrac{(x-2)(x+2)}{(x-3)(x+2)}
\)

Now cancel the common factor ( (x+2) ):

\(
f(x)=\dfrac{x-2}{x-3}, \quad x\neq -2
\)

• hole at ( x=-2 )
• vertical asymptote at ( x=3 )
• x-intercept at ( x=2 )
• horizontal asymptote ( y=1 )

✅ All four conditions satisfied.

AP Precalculus -1.10 Rational Functions and Holes- MCQ Exam Style Questions - Effective Fall 2023