# How do you find the vertical, horizontal or slant asymptotes for y=(x^2-1)/(2x^2 + 3x-2)?

Oct 23, 2016

The vertical asymptotes are $x = \frac{1}{2}$ and $x = - 2$

The horizontal asymptote is $y = \frac{1}{2}$

#### Explanation:

We facorise $y = \frac{\left(x + 1\right) \left(x - 1\right)}{\left(2 x - 1\right) \left(x + 2\right)}$
As we cannot divide by $0$
So for $x = \frac{1}{2}$ and $x = - 2$ we have vertical asymptotes
To look for horizontal asymptotes, we rewrite
$y$ as $y = \frac{1 - \frac{1}{x} ^ 2}{2 + \frac{3}{x} - \frac{2}{x} ^ 2}$

the limit $y = \frac{1 - 0}{2 + 0 + 0} = \frac{1}{2}$
$y \to - \infty$

the limit $y = \frac{1 - 0}{2 + 0 + 0} = \frac{1}{2}$
$y \to + \infty$
So $y = \frac{1}{2}$ is a horizontal asymptote