Answer
- Vertical asymptote: $x=-2$
- Oblique asymptote: $y=\frac{x}{2}-1$
Work Step by Step
$$y=\frac{x^2-1}{2x+4}$$
- As $x\to-2$, $(2x+4)$ approaches $0$ and $x^2-1$ approaches $(-2)^2-1=3\gt0$ , so $(x^2-1)/(2x+4)$ will approach $\infty$. In other words, $$\lim_{x\to-2}\frac{x^2-1}{2x+4}=\infty$$
Therefore, the line $x=-2$ is the vertical asymptote of the graph.
- There is one more oblique asymptote, but to find it, we need to change the form of the function: $$y=\frac{(x^2-4)+3}{2x+4}$$ $$y=\frac{x^2-4}{2x+4}+\frac{3}{2x+4}$$ $$y=\frac{(x-2)(x+2)}{2(x+2)}+\frac{3}{2x+4}$$ $$y=\Big(\frac{x}{2}-1\Big)+\frac{3}{2x+4}$$
As $x\to\pm\infty$, $3/(2x+4)$ approaches $0$, making the function turn into
$$y=\frac{x}{2}-1$$ which is the oblique asymptote of the graph.
The graph, along with 2 asymptotes, is shown below.