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]}