# What are the vertical asymptotes of f(x) = (2)/(x^2 - 1)?

Vertical asymptotes for this function will occur when the denominator is negative. Therefore, it behooves us to find the zeroes of the function in the denominator, namely ${x}^{2} - 1$. Factoring, we get ${x}^{2} - 1 = \left(x - 1\right) \left(x + 1\right)$ and thus our zeroes occur at $x = \pm 1$. Ergo, the locations of our vertical asymptotes are also $x = \pm 1$.