# How do you graph y^2(x^2-1)=1 ?

Jan 13, 2017

See explanation...

#### Explanation:

Interesting!

Let's start by considering:

$y \left({x}^{2} - 1\right) = 1$

Dividing both sides by ${x}^{2} - 1$ we get:

$y = \frac{1}{{x}^{2} - 1} = \frac{1}{\left(x - 1\right) \left(x + 1\right)}$

This has vertical asymptotes at $x = \pm 1$ and a local maximum at $x = 0$, where $y = - 1$.

It is only positive when $\left\mid x \right\mid > 1$

It has a horizontal asymptote $y = 0$ since $\frac{1}{{x}^{2} - 1} \to 0$ as $x \to \pm \infty$.

So it looks like this...
graph{1/(x^2-1) [-10, 10, -5, 5]}

Going back to our original equation, we can divide both sides by ${x}^{2} - 1$ to get:

${y}^{2} = \frac{1}{{x}^{2} - 1}$

and hence:

$y = \pm \sqrt{\frac{1}{{x}^{2} - 1}}$

This does not describe a function, but has a graph which will be symmetrical about the $x$ axis.

Considering just the positive values, note that square roots are monotonically increasing, approaching vertical for small values of the radicand...
graph{(y^2-x) = 0 [-10, 10, -5, 5]}

So combined with the function $\frac{1}{{x}^{2} - 1}$ this results in a graph which is less 'steep' near the vertical asymptotes, undefined for $\left\mid x \right\mid \le 1$ and slower to approach the asymptote $y = 0$, with a reflected copy...
graph{y^2(x^2-1)=1 [-10, 10, -5, 5]}