# How do you find the asymptote(s) or hole(s) of f(x) = (x+3)/(x^2-9)?

Jun 6, 2018

asymptote at $x = 3$

hole at $x = - 3$

#### Explanation:

$f \left(x\right) = \frac{x + 3}{{x}^{2} - 9}$

Factor the denominator, it is the difference of squares:

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

The factor(s) in the denominator that will cancel out, in this case $x + 3$ are holes.

The factor(s) in the denominator that will not cancel out, in this case $x - 3$ are asymptotes.

Here is the graph, it shows the asymptote, but not the hole at $x = - 3$

graph{(x+3)/(x^2-9) [-6.67, 13.33, -4.44, 5.56]}

Jun 6, 2018

VA: $x = 3$

HA: $y = 0$

#### Explanation:

We have the following expression

$\frac{x + 3}{\textcolor{b l u e}{\left({x}^{2} - 9\right)}}$

What I have in blue is a Difference of Squares of the form

${a}^{2} - {b}^{2}$, where $a = x$ and $b = 3$, which is factored as

$\left(a + b\right) \left(a - b\right)$

Now, our new expression is

$\frac{x + 3}{\textcolor{b l u e}{\left(x + 3\right) \left(x - 3\right)}}$

which simplifies to

$\frac{1}{x - 3}$

Vertical asymptotes are found by setting the denominator equal to zero, because this is the value for which the function is undefined.

$x - 3 = 0$

$\implies \textcolor{red}{x = 3}$ (Vertical Asymptote)

Horizontal asymptotes are found by comparing the degrees. Since there is a higher degree in the denominator, we have a horizontal asymptote at

$\implies \textcolor{\mathrm{da} r k v i o \le t}{y = 0}$ (Horizontal Asymptote)

We have a vertical asymptote at $x = 3$, and a horizontal asymptote at $y = 0$.

Hope this helps!