# How do you find the Vertical, Horizontal, and Oblique Asymptote given ((x^2)+3)/((9x^2)-80x-9)?

Sep 13, 2016

vertical asymptotes at $x = - \frac{1}{9} , x = 9$
horizontal asymptote at $y = \frac{1}{9}$

#### Explanation:

The denominator of the function cannot be zero as this would make the function 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.

solve: $9 {x}^{2} - 80 x - 9 = 0 \Rightarrow \left(9 x + 1\right) \left(x - 9\right) = 0$

$\Rightarrow x = - \frac{1}{9} \text{ and " x=9" are the asymptotes}$

Horizontal asymptotes occur as

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

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

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} + \frac{3}{x} ^ 2}{\frac{9 {x}^{2}}{x} ^ 2 - \frac{80 x}{x} ^ 2 - \frac{9}{x} ^ 2} = \frac{1 + \frac{3}{x} ^ 2}{9 - \frac{80}{x} - \frac{9}{x} ^ 2}$

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

$\Rightarrow y = \frac{1}{9} \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2 ) Hence there are no oblique asymptotes.
graph{(x^2+3)/(9x^2-80x-9) [-20, 20, -10, 10]}