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

1 Answer
Mar 28, 2016

Answer:

First, factor the numerator: #y = ((x+3)(x-3))/(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).