Question 48399
f(x) = 2/(x^2 - 9) = 2/((x - 3)(x + 3))
If {{{x = 3}}} or {{{x = -3}}}, you would be dividing by zero. Since that is described as undefined, those are your two vertical asymptotes.
{{{graph(600,600,-10,10,-10,10,2/(x^2-9))}}}