Answer
vertical asymptotes $x=-2$ and $x=3$
horizontal asymptote $y=1$
no slant asymptote.
See graph.
Work Step by Step
Step 1. Factor the numerator and the denominator; we have $h(x)=\frac{x^2-3x-4}{x^2-x-6}=\frac{(x-4)(x+1)}{(x-3)(x+2)}$ and there are no common factors.
Step 2. The vertical asymptotes can be found as $x=-2$ and $x=3$
Step 3. The horizontal asymptote can be found when $x\pm\infty$ and we get $y=1$
Step 4. A slant asymptote exists if the numerator is one order higher than the denominator; for the given rational function, there is no slant asymptote.
Step 5. See graph.
