Question

How do you find the asymptotes for `f(x)=(3x^2+2) / (x^2 -1)`?

Answer 1

The vertical asymptotes are `x=1` and `x=-1`

The horizontal asymptote is `y=3`

Explanation:

The asymptotes occur where the denominator approaches zero, and where `x` becomes very large, either positively or negatively.

`f(x) = (3x^2+2)/(x^2-1)`

In this case the denominator is the difference of two squares so the function can be rewritten as
`f(x)= (3x^2+2)/((x-1)(x+1))`

The denominator is zero when either `x=1` or `x=-1` and these are therefore the vertical asymptotes.

AS `x` becomes very large, either positively or negatively,
`lim_(x->oo) =(3cancel(x^2) )/cancel(x^2)`

The horizontal asymptote is therefore `y=3`