# How do you find the asymptotes for (9x^2 – 36) /( x^2 - 9)?

Jul 28, 2018

$\text{vertical asymptotes at } x = \pm 3$
$\text{horizontal asymptote at } y = 9$

#### Explanation:

$\text{let } f \left(x\right) = \frac{9 {x}^{2} - 36}{{x}^{2} - 9}$

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve } {x}^{2} - 9 = 0 \Rightarrow \left(x - 3\right) \left(x + 3\right) = 0$

$x = \pm 3 \text{ are the asymptotes}$

$\text{Horizontal asymptotes occur as }$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

$\text{divide terms on numerator/denominator by the highest}$
$\text{power of "x" that is } {x}^{2}$

f(x)=((9x^2)/x^2-(36)/x^2)/(x^2/x^2-9/x^2)=(9-(36)/x^2)/(1-9/x^2

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{9 - 0}{1 - 0}$

$y = 9 \text{ is the asymptote}$
graph{(9x^2-36)/(x^2-9) [-40, 40, -20, 20]}

vertical asymptotes: $x = \setminus \pm 3$

horizontal asymptote: $y = 9$

#### Explanation:

The given function:

$f \left(x\right) = \setminus \frac{9 {x}^{2} - 36}{{x}^{2} - 9}$

Vertical asymptotes:

The above function will have vertical asymptote where denominator becomes zero i.e.

$\setminus \therefore {x}^{2} - 9 = 0$

${x}^{2} = 9$

$x = \setminus \pm 3$

Horizontal asymptotes:

The above function will have horizontal where

$y = \setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{9 {x}^{2} - 36}{{x}^{2} - 9}$

$y = \setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{9 - \frac{36}{x} ^ 2}{1 - \frac{9}{x} ^ 2}$

$y = \setminus \frac{9 - 0}{1 - 0}$

$y = 9$

Hence, vertical asymptotes: $x = \setminus \pm 3$

horizontal asymptote: $y = 9$