Answer
not continuous at $x=-2$ or $x=2$
Work Step by Step
1. See graph for $R(x)=\frac{3x+6}{x^2-4}=\frac{3(x+2)}{(x+2)(x-2)}=\frac{3}{x-2}, x\ne-2$ with V.A. $x=2$ and a hole at $(-2, -\frac{3}{4})$.
2. Although $\lim_{x\to-2}R(x)=-\frac{3}{4}$, the function is not continuous at $x=-2$ because of the hole.
3. We have $\lim_{x\to 2^-}R(x)=-\infty$ and $\lim_{x\to 2^+}R(x)=\infty$, thus the function is not continuous at $x=2$.
