Question

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

Answer 1

See explanation...

Explanation:

Interesting!

Let's start by considering:

`y(x^2-1) = 1`

Dividing both sides by `x^2-1` we get:

`y = 1/(x^2-1) = 1/((x-1)(x+1))`

This has vertical asymptotes at `x=+-1` and a local maximum at `x=0`, where `y = -1`.

It is only positive when `abs(x) > 1`

It has a horizontal asymptote `y = 0` since `1/(x^2-1) -> 0` as `x->+-oo`.

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 = 1/(x^2-1)`

and hence:

`y = +-sqrt(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 `1/(x^2-1)` this results in a graph which is less 'steep' near the vertical asymptotes, undefined for `abs(x) <= 1` and slower to approach the asymptote `y=0`, with a reflected copy...
graph{y^2(x^2-1)=1 [-10, 10, -5, 5]}