#### Answer

- Vertical asymptote: $x=1$
- Oblique asymptote: $y=x+1$

#### Work Step by Step

$$y=\frac{x^2}{x-1}$$
- As $x\to1$, $(x-1)$ approaches $0$ and $x^2$ approaches $1\gt0$ , so $x^2/(x-1)$ will approach $\infty$. In other words, $$\lim_{x\to1}\frac{x^2}{x-1}=\infty$$
Therefore, the line $x=1$ 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: $$y=\frac{(x^2-1)+1}{x-1}$$ $$y=\frac{x^2-1}{x-1}+\frac{1}{x-1}$$ $$y=(x+1)+\frac{1}{x-1}$$
As $x\to\pm\infty$, $1/(x-1)$ approaches $0$, making the function turn to $$y=x+1$$ which is the oblique asymptote of the graph.
The graph, along with 2 asymptotes, are shown below.