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

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 $y = \frac{a}{b}$, 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:

$\frac{1}{x} ^ 2 = 0$

This is impossible, as $1 \ne 0$.

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

So:

$\frac{1}{0} ^ 2$

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]}