Question 1157587
{{{T(x)=(x^3)/(x^4-16)}}}


Horizontal asymptote:

{{{(x^3)/(x^4-16)}}}->{{{0}}} as {{{x}}}-> ±{{{infinity}}}

so, horizontal asymptote is {{{y=0}}}

Vertical asymptote:

{{{x^4-16=0}}}
{{{x^4=16}}}
{{{x^4=2^4}}}

{{{x}}}=±{{{2}}}=> vertical asymptote: {{{x=2}}} and {{{x=-2}}}


no oblique​ asymptote

{{{drawing( 600, 600, -10, 10, -10, 10, 
green(line(-2,10,-2,-10)),green(line(2,10,2,-10)),
 graph( 600, 600, -10, 10, -10, 10,0,0 ,x^3/(x^4-16))) }}}