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

1 Answer
Jan 13, 2017

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]}