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

##### 1 Answer

See explanation...

#### Explanation:

Interesting!

Let's start by considering:

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

Dividing both sides by

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

This has vertical asymptotes at

It is only positive when

It has a horizontal asymptote

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

#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

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

graph{y^2(x^2-1)=1 [-10, 10, -5, 5]}