Question 1176662

{{{f(x)=(2x^2-32)/(6x^2+13x-5)}}}

{{{f(x)=(2x^2-32)/((3x-1)(2x+ 5))}}}

asymptotes

Vertical: 
{{{(3 x - 1) (2 x + 5)=0}}}
{{{3 x - 1=0}}}
{{{3 x =1}}}
{{{x =1/3}}}

{{{(2 x + 5)=0}}}
{{{x =-5}}}
{{{2 x =-5/2}}}

Vertical asymptotes are:

{{{x=-5/2}}} and {{{x=1/3}}}


Horizontal: take coefficients of variable with highest degree , {{{2/6}}}
{{{y=2/6}}}
{{{y=1/3 }}}


intercepts:

to find x-intercepts, set numerator to zero

{{{2x^2-32=0}}}
{{{2(x^2-16)=0}}}
{{{2(x^2-4^2)=0}}}
{{{2(x-4)(x+4)=0}}}

=>{{{x=4}}} or {{{x=-4}}}

{{{x}}}-intercepts are:  ({{{4}}},{{{0}}}),({{{-4}}},{{{0}}})


to find y-intercepts, set {{{x=0}}}

{{{f(x)=(2*0^2-32)/(6*0^2+13*0-5)=-32/-5=32/5}}}

 {{{y}}}-intercept is ({{{0}}},{{{32/5}}})