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

1 Answer
Feb 18, 2018

There is a vertical asymptote at x=0
There are no holes here.
There is an asymptote at y=0.
There are no x or y intercepts.

Explanation:

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]}