Home / AP Calculus AB 1.14 Connecting Infinite Limits and Vertical Asymptotes – MCQs

AP Calculus AB 1.14 Connecting Infinite Limits and Vertical Asymptotes - MCQs - Exam Style Questions

AP Calculus AB-MCQs and Answers - All Topics

No-Calc Question

How many vertical asymptotes does the graph of \(y=\dfrac{x-2}{x^{4}-16}\) have?
(A) One   
(B) Two   
(C) Three   
(D) Four
▶️ Answer/Explanation
Factor: \(x^{4}-16=(x-2)(x+2)(x^{2}+4)\).
Simplify the function: \(\dfrac{x-2}{x^{4}-16}=\dfrac{1}{(x+2)(x^{2}+4)}\) with a removable hole at \(x=2\).
Vertical asymptotes occur where the simplified denominator is \(0\): \(x=-2\) (real), \(x^{2}+4=0\) (not real).
So there is exactly one vertical asymptote at \(x=-2\).
✅ Answer: (A)

No-Calc Question


The function \(f\) is given by \(\displaystyle f(x)=\frac{a x^{2}+12}{x^{2}+b}\). The figure above shows a portion of the graph of \(f\). Which of the following could be the values of the constants \(a\) and \(b\)?

(A) \(a=-3,\; b=2\)
(B) \(a=2,\; b=-3\)
(C) \(a=2,\; b=-2\)
(D) \(a=3,\; b=-4\)
(E) \(a=3,\; b=4\)

▶️ Answer/Explanation

From the graph, the horizontal asymptote is \(y=3\).
For \(\displaystyle f(x)=\frac{a x^{2}+12}{x^{2}+b}\) with equal degrees, the horizontal asymptote is the ratio of leading coefficients \(\Rightarrow y=a\). Hence \(a=3\).

The vertical asymptote in the figure is at \(x=2\).
Vertical asymptotes occur where the denominator is \(0\): \(x^{2}+b=0 \Rightarrow x=\pm\sqrt{-b}\).
Setting \(\sqrt{-b}=2\) gives \(-b=4 \Rightarrow b=-4\).

Therefore \((a,b)=(3,-4)\).
✅ Answer: (D)

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