How do you find the vertical, horizontal or slant asymptotes for ((x^2)+3)/((9x^2)-80x-9)?

Dec 31, 2016

The vertical asymptotes are $x = - \frac{1}{9}$ and $x = 9$
No slant asymptote
The horizontal asymptote is $y = \frac{1}{9}$

Explanation:

We can factorise the denominator

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

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

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- \frac{1}{9} , 9\right\}$

As we cannot divide by $0$, $x \ne - \frac{1}{9}$ and $x \ne 9$

The vertical asymptotes are $x = - \frac{1}{9}$ and $x = 9$

As the degree of the numerator is $=$ to the degree of the denominator, there is no slant asymptote

Now, we calculate the limits of $f \left(x\right)$ as $x \to \pm \infty$

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

The horizontal asymptote is $y = \frac{1}{9}$

graph{(y-(x^2+3)/(9x^2-80x-9))(y-1/9)=0 [-52.1, 51.93, -26, 25.94]}