Question 134844
Asymptotes occur when the denominator is zero. 

So what values of x make the denominator 0? 
{{{(x^2-x-12) = 0}}}
{{{(x-4)(x+3) = 0}}}
x = 4, x =-3

asymptotes are the lines  x=4 and x=-3

Holes occur when both the numerator and denominator are 0.
{{{((2x^3+6x^2)/(x^2-x-12))}}}
{{{ (2x^2(x+3))/ ((x-4)(x+3))}}}
So we have a hole at x = -3