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

##### 1 Answer
May 26, 2016

Horizontal asymptote is $y = 3$

Vertical asymptotes are $x = \pm 3$

#### Explanation:

Write as $y = \frac{3 {x}^{2}}{{x}^{2} \left(1 - \frac{9}{x} ^ 2\right)} = \frac{3}{1 - \frac{9}{x} ^ 2}$

$\textcolor{b l u e}{\text{Determine horizontal asymptote}}$

As $x$ becomes increasingly large then $\frac{9}{x} ^ 2$ becomes increasingly small.

$\textcolor{b l u e}{y = {\lim}_{x \to \pm \infty} \frac{3}{1 - \frac{9}{x} ^ 2} \to \frac{3}{1} \leftarrow \text{horizontal asymptote}}$

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$\textcolor{b l u e}{\text{Determine vertical asymptote}}$

$\textcolor{b r o w n}{\text{As "x^2" gets closer and closer to 9 then "1-9/x^2" becomes }}$$\textcolor{b r o w n}{\text{smaller and smaller. So "3/(1-9/x^2)" becomes larger and larger.}}$

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

color(brown)("However; if "x=color(white)()^+3" then "9/(color(white)()^+3)<1

$\text{ } y = {\lim}_{x \to {\textcolor{w h i t e}{}}^{+} 3} \frac{3}{1 - \frac{9}{x} ^ 2} \to + \infty$

color(brown)("However; if "x=color(white)()^-3" then "9/(color(white)()^+3)>1

$\text{ } y = {\lim}_{x \to {\textcolor{w h i t e}{}}^{+} 3} \frac{3}{1 - \frac{9}{x} ^ 2} \to - \infty$

So $\textcolor{b l u e}{y = {\lim}_{x \to 3} \frac{3}{1 - \frac{9}{x} ^ 2} \to \pm \infty \leftarrow \text{ vertical asymptotes}}$ 