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

1 Answer
Aug 24, 2014

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 = (x-1)(x+1)# and thus our zeroes occur at #x=+-1#. Ergo, the locations of our vertical asymptotes are also #x=+-1#.

For a more thorough explanation of the concept of vertical asymptotes, go here: http://socratic.org/questions/what-is-a-vertical-asymptote-in-calculus?