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

Question

1711 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-28T22:29:21

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

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).

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