# How do you find the vertical, horizontal or slant asymptotes for (4x^2+11) /( x^2+8x−9)?

Dec 27, 2016

The vertical asymptotes are $x = - 9$ and $x = 1$
No slant asymptote.
The horizontal asymptote is $y = 4$

#### Explanation:

Let's factorise the denominator

${x}^{2} + 8 x - 9 = \left(x - 1\right) \left(x + 9\right)$

Let, $f \left(x\right) = \frac{4 {x}^{2} + 11}{{x}^{2} + 8 x - 9}$

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

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

So, the vertical asymptotes are $x = - 9$ and $x = 1$

The degree of the numerator is $=$ to the degree of the denominator, we don't have a slant asymptote.

To find the horizontal asymptote, we compute the limits of $f \left(x\right)$ as $x \to \pm \infty$

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x - \pm \infty} 4 {x}^{2} / {x}^{2} = 4$

The horizontal asymptote is $y = 4$

graph{(y-(4x^2+11)/(x^2+8x-9))(y-4)(x-1)(x+9)=0 [-52, 52, -26, 26]}