Question 47800
{{{ f(x) = 2/(x^2-9) }}} can be written as 
{{{ f(x) = 2/((x+3)(x-3)) }}}


Now, in ANY fraction, if the denominator is zero you are screwed!. So we want to avoid that at all costs. If x was either 3 or -3, then one of the brackets would be zero..and we would be screwed.


So (vertical) asymptotes are x=3 and x=-3.


Now, to find any horizontal asymptotes, we need to re-write the function in terms of y...


{{{ f(x) = 2/(x^2-9) }}}
{{{ y = 2/(x^2-9) }}}
{{{ y(x^2-9) = 2 }}}
{{{ x^2y-9y = 2 }}}
{{{ x^2y = 2+9y }}}
{{{ x^2 = ((2+9y)/y) }}}
{{{ x = sqrt((2+9y)/y) }}} taking just the positive version (taking square roots gives you 2 versions: a positive and a negative version.)


and so we get in function notation:
{{{ f(y) = sqrt((2+9y)/y) }}} which is just the same as writing {{{ f(x) = sqrt((2+9x)/x) }}}.


And so again we ask the question: at what value(s) of y does the denominator become zero? answer is y=0.


So vertical asymptote is at y=0


These are your 3 asymptotes in this question.


jon