Question

Find vertical and horizontal asymtope for (x^2+2x-48)/x+8

Answer 1

There are none.

$y=\dfrac{x^{2}+2x-48}{x+8}=\dfrac{(x+8)(x-6)}{x+8}\\\\y=x-6$

The expression describes a straight line with a hole at x=-8 where it is undefined.