# How do you find the asymptotes for g(x) = (3x^2) / (x^2 - 9)?

Dec 12, 2015

Horizontal asymptotes: $y = 3$ when $x$ tends to both positive and negative infinity.

Vertical asymptote: $x = 3$ and $x = - 3$.

#### Explanation:

For the horizontal asymptote, compute the limit of this function as $x \setminus \to \setminus \pm \setminus \infty$. In both cases, you have

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{3 {x}^{2}}{{x}^{2} - 9} =$

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{3 {x}^{2}}{{x}^{2} \left(1 - \frac{9}{x} ^ 2\right)} =$

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{3}{1 - \frac{9}{x} ^ 2} = \frac{3}{1} = 3$

So, the horizontal asymptotes are $y = 3$ in both directions.

As for the vertical asymptotes, we need to cosider the points which are not in the domain of the function. These points are the ones that annihilate the denominator, in this case $\setminus \pm 3$, because they solve

${x}^{2} - 9 = 0 \setminus \iff {x}^{2} = 9 \setminus \iff x = \setminus \pm \setminus \sqrt{9} = \setminus \pm 3$

This means that $x = \setminus \pm 3$ are the two vertical lines which are vertical asymptotes of the function.