#### Answer

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

#### Work Step by Step

$$y=\frac{x^3+1}{x^2}$$
- As $x\to0$, $x^2$ approaches $0$ and $x^3+1$ approaches $(0)^3+1=1\gt0$ , so $(x^3+1)/x^2$ will approach $\infty$. In other words, $$\lim_{x\to0}\frac{x^3+1}{x^2}=\infty$$
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: $$y=\frac{(x^3)+1}{x^2}$$ $$y=\frac{x^3}{x^2}+\frac{1}{x^2}$$ $$y=x+\frac{1}{x^2}$$
As $x\to\pm\infty$, $1/x^2$ 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.