AP Calculus AB 1.14 Connecting Infinite Limits and Vertical Asymptotes - MCQs - Exam Style Questions
No-Calc Question
▶️ Answer/Explanation
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)