Answer
vertical asymptote $x=4$,
horizontal asymptote $y=0$.
Work Step by Step
Factor the denominator, $R(x)=\frac{x+3}{x^2-x-12}=\frac{x+3}{(x+3)(x-4)}=\frac{1}{x-4}, x\ne-3$, thus there is a vertical asymptote $x=4$ and a horizontal asymptote $y=0$.
