How Does the Function $f(x) = \frac{1}{g(x)}$ Behave?

[karush]

The graph of $y=(x-2)^2 - 1$ is given, then they say use this graph for problem 7

Next is
7) $f(x) =\frac{1} {g(x)} $ ?
Step one restrictions $y> \ge - 1$
Step two Equation Solution?
Step three: Domain:= $\left(-\infty, +\infty\right)$
Vertical Asymptopes at: none

What's with the?

 

[MarkFL]

I am assuming we have:

\(\displaystyle g(x)=y=(x-2)^2-1\)

and then:

\(\displaystyle f(x)=\frac{1}{g(x)}\)

If I was then going to look for the domain of $f$, I would observe that it will be all reals, with the exception of the roots of $g$ (where $g$ will have vertical asymptotes):

\(\displaystyle g(x)=(x-2)^2-1=(x-2+1)(x-2-1)=(x-1)(x-3)=0\)

Thus, the domain of $f$ is:

\(\displaystyle (-\infty,1)\,\cup\,(1,3)\,\cup\,(3,\infty)\)

We can also observe that:

\(\displaystyle \lim_{x\to\pm\infty}g=\infty\implies \lim_{x\to\pm\infty}f=0\)

And so there is a horizontal asymptote at $y=0$.

We also know that $f$ has the same sign as $g$.

I don't know what is meant by "Equation Solution."