# How do you find the vertical, horizontal or slant asymptotes for (2x^2+x-7)/(x^2-1)?

Nov 2, 2016

The vertical asymptotes are $x = - 1$ and $x = 1$
The horizontal asymptote is $y = 2$

#### Explanation:

The denominator $\left({x}^{2} - 1\right) = \left(x + 1\right) \left(x - 1\right)$
As we cannot divide by zero, the vertical asymptotes are $x = - 1$ and $x = 1$
As the degree of the numerator and denominator are the same, we don't have slant asymptotes
limit $\frac{2 {x}^{2} + x - 7}{{x}^{2} - 1} = \frac{2 {x}^{2}}{x} ^ 2 = 2$
$x \to \pm \infty$
So $y = 2$ is a horizontal asymptote
graph{(2x^2+x-7)/(x^2-1) [-10, 10, -5, 5]}