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

Jan 21, 2016

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.

$f \left(x\right) = \frac{3 {x}^{2} + 2}{{x}^{2} - 1}$

In this case the denominator is the difference of two squares so the function can be rewritten as
$f \left(x\right) = \frac{3 {x}^{2} + 2}{\left(x - 1\right) \left(x + 1\right)}$

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,
${\lim}_{x \to \infty} = \frac{3 \cancel{{x}^{2}}}{\cancel{{x}^{2}}}$

The horizontal asymptote is therefore $y = 3$