Question

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

Answer 1

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]}