Answer
See graph and explanations.
Work Step by Step
See graph; when $x\to-2^+, y\to\infty$ or $\lim_{x\to -2^+}\frac{x^2-1}{2x+4}=\infty$;
while when $x\to-2^-, y\to-\infty$ or $\lim_{x\to -2^-}\frac{x^2-1}{2x+4}=-\infty$;
the asymptote of the function is $x=-2$