#### Answer

- Vertical asymptote: $x=0$
- Oblique asymptote: $y=x$

#### Work Step by Step

$$y=\frac{x^2-1}{x}$$
- As $x\to0$, $x^2-1$ approaches $(0)^2-1=-1\lt0$ , so $(x^2-1)/x$ will approach $-\infty$. In other words, $$\lim_{x\to0}\frac{x^2-1}{x}=-\infty$$
Therefore, the line $x=0$ 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: $$y=\frac{(x^2)-1}{x}$$ $$y=\frac{x^2}{x}-\frac{1}{x}$$ $$y=x-\frac{1}{x}$$
As $x\to\pm\infty$, $1/x$ approaches $0$, making the function turn into
$$y=x$$ which is the oblique asymptote of the graph.
The graph, along with 2 asymptotes, is shown below.