Question

5-7.
use the graph of $f(x) = -x^2 + 9$ to sketch the graph of $y = \frac{1}{f(x)}$ without using a graphing calculator. state the equations of the asymptotes in the graph of $y = \frac{1}{f(x)}$.
note: this problem was updated from its original version, which was a repeat of a problem in the reciprocal functions lesson.

Explanation:

Step1: Find zeros of $f(x)$

Set $-x^2 + 9 = 0$
$x^2 = 9$
$x = \pm 3$
These are vertical asymptotes for $y=\frac{1}{f(x)}$.

Step2: Find horizontal asymptote

As $x\to\pm\infty$, $f(x)=-x^2+9\to-\infty$, so $\frac{1}{f(x)}\to0$.
Horizontal asymptote is $y=0$.

Step3: Verify asymptote behavior

For $x\to3^+$, $f(x)\to0^-$, so $\frac{1}{f(x)}\to-\infty$; for $x\to3^-$, $f(x)\to0^+$, so $\frac{1}{f(x)}\to+\infty$.
For $x\to-3^+$, $f(x)\to0^+$, so $\frac{1}{f(x)}\to+\infty$; for $x\to-3^-$, $f(x)\to0^-$, so $\frac{1}{f(x)}\to-\infty$.
Confirms vertical asymptotes at $x=\pm3$, horizontal at $y=0$.

Answer:

Asymptotes: $x=3$, $x=-3$, $y=0$