# How do you graph f(x)=(x^2-1)/x using holes, vertical and horizontal asymptotes, x and y intercepts?

Sep 5, 2017

You have a vertical asymptote at x=0 because that would make the denominator equal zero. There's a slant asymptote at $y = x$ because ${x}^{2} / x = x$

#### Explanation:

You know this graph can't exist at $x = 0$, since that would make the denominator equal $0$. Because the polynomial on the top is of a bigger degree, there is a slant asymptote. Dividing the initial terms, we get ${x}^{2} / x = x$, so there is a slant asymptote at $y = x$.
Since we can factor the top into $\left(x - 1\right) \left(x + 1\right)$, we know the function has two solutions at $x = \pm 1$.

Plotting these solutions and following the asymptotes makes this a straightforward graph to sketch: graph{(x^2-1)/x [-10, 10, -5, 5]}

Also, not all rational functions are so easy to predict the behavior of, so creating a table of x and y values is always a good idea! And if you need more information about how to find the asymptotes, look here.