# How do you identify all asymptotes or holes for f(x)=(2x)/(x^2-1)?

Sep 26, 2016

VA: 1,-1
HA: 0
SA: N/A

#### Explanation:

Vertical asymptote: Set denominator = 0 but factor first
${x}^{2} - 1$=$\left(x - 1\right) \left(x + 1\right)$
$\left(0 - 1\right) = - 1$
$\left(0 + 1\right) = 1$

$1$ and $- 1$ are your vertical asymptotes

Horizontal asymptote: When the x with the biggest power in the numerator is less than the x with the biggest power in the numerator, the horizontal asymptote is $0$.

Since a horizontal asymptote exists, then a slanted asymptote does not exist. It would only exist if the x with the biggest power in the numerator is biggest than the x with the biggest power in the denominator.