How do you graph #f(x)=1/x^2# using holes, vertical and horizontal asymptotes, x and y intercepts?

Let's go one by one. Here, I take the numerator as #n# and the denominator as #m#. Here, #n=1# and #m=x^2#

Vertical asymptotes:

Vertical asymptotes are found when #m=0#. Since #m=x^2#, we can say, the asymptote is at:

#x^2=0#

#x=sqrt(0)#

#x=0#

There is a vertical asymptote at #x=0#

Holes:

There are no holes here. Holes are found only if you can simplify #m#.

Horizontal asymptotes:

There are a few rules to remember when finding a horizontal asymptote:

We must look at the degree of #m# and #n#. Say the degree of #m# is #gamma#, and the degree of #n# is #delta#. Then, when:

#gamma>delta#, the asymptote is at nowhere. There is a slant asymptote, however.
#gamma=delta#, the asymptote is at #y=a/b#, where #a# and #b# are the leading coefficients of #n# and #m# respectively.
#gamma<##delta#, the asymptote is at #y=0#

Here, #delta=2# and #gamma=0#. So #gamma<##delta#, and there is an asymptote at #y=0#.

Intercepts:

The #x#-intercept is found when you set #y=0# and solve for #x#.

So:

#1/x^2=0#

This is impossible, as #1!=0#.

The #y#-intercept is found when you set #x=0#.

So:

#1/0^2#

However, the above is undefined.

So there are no #x# or #y# intercepts.

The graph looks like so:

graph{1/x^2 [-10, 10, -3, 7]}