Question 1084114
{{{(x^2+x-12)/(x^2-4)=((x+4)(x-3))/((x+2)(x-2))}}}
It is clear that the function is not defined for x=2 and for x=-2.
As x approaches those values the denominator approaches zero,
while the numerator approaches some non-zero value.
That means that as x approaches 2 or -2,
the absolute value of the function increases without bounds.
So, x=2, and x=z2 are vertical asymptotes.
 
Looking at the function a different way we see that
{{{(x^2+x-12)/(x^2-4)=((x^2-4)+(x-8))/(x^2-4)=(x^2-4)/(x^2-4)+(x-8)/(x^2-4)=1+(x-8)/(x^2-4)}}},
we see that as the absolute value of x increases,
the term {{{(x-8)/(x^2-4)}}} approaches zero,
meaning that the function's value approaches 1.
So, y=1 is the horizontal asymptote.