# What are the vertical and horizontal asymptotes of y = (x+3)/(x^2-9)?

Jun 20, 2018

vertical asymptote at $x = 3$

horizontal asymptote at $y = 0$

hole at $x = - 3$

#### Explanation:

$y = \frac{x + 3}{{x}^{2} - 9}$

First factor:

$y = \frac{\left(x + 3\right)}{\left(x + 3\right) \left(x - 3\right)}$

Since the factor $x + 3$ cancels that is a discontinuity or hole, the factor $x - 3$ does not cancel so it is a asymptote:

$x - 3 = 0$

vertical asymptote at $x = 3$

Now let's cancel out the factors and see what the functions does as x gets really big in the positive or negative:

x -> +-oo, y ->?

$y = \frac{\cancel{\left(x + 3\right)}}{\cancel{\left(x + 3\right)} \left(x - 3\right)} = \frac{1}{x - 3}$

As you can see the reduced form is just $1$ over some number $x$, we can ignore the $- 3$ because when $x$ is huge it is insignificant.

We know that: $x \to \pm \infty , \frac{1}{x} \to 0$ hence, our original function has the same behavior:

$x \to \pm \infty , \frac{\left(x + 3\right)}{\left(x + 3\right) \left(x - 3\right)} \to 0$

Therefore, the function has a horizontal asymptote at $y = 0$

graph{y = (x+3)/(x^2-9) [-10, 10, -5, 5]}