Question 927749
{{{f(x)= (8x+9)/(x^2-4)}}}

you can use the {{{domain}}} , because whichever values are not allowed in the domain will be vertical asymptotes on the graph

{{{(x^2-4)=0}}}=> for {{{x=-2}}} and {{{x=2}}}=>vertical asymptotes 

domain:
{ {{{x}}} element {{{R}}} : {{{x<>-2}}} and {{{x<>2}}} }


the {{{horizontal}}} asymptote is found by {{{dividing}}} the {{{leading}}} {{{terms}}}, so the asymptote is given by:

    {{{y }}}= (numerator's leading coefficient)/(denominator's leading coefficient)

 {{{y =(0*x^2+8x+9)/(x^2-4)=0/1=0}}}


the {{{horizontal}}} asymptote is  {{{y =0}}}


{{{drawing(600,600,-10,10,-10,10,rectangle(-2,-11,11,11),rectangle(2,-11,11,11),graph( 600, 600, -10,10, -10, 10, (8x+9)/(x^2-4),0)) }}}