# How do you find vertical, horizontal and oblique asymptotes for y= (x^2 - 9)/(x-3)?

##### 1 Answer
Mar 28, 2016

First, factor the numerator: $y = \frac{\left(x + 3\right) \left(x - 3\right)}{x - 3}$

#### Explanation:

Cancel the factor $x - 3$ from both the numerator and denominator. You are left with $y = x + 3$, a linear function!

The factor that was cancelled leaves a "hole" or removable discontinuity in your graph at x = 3.

If you plot the line $y = x + 3$, there will be NO asymptotes at all, but just a little hole at the point (3,6).