# How do you find the asymptotes for f(x) = (x^2+4x-2)/(x^2-x-7)?

Jun 9, 2015

$y = 1$
$x = \frac{- 1 - \sqrt{29}}{2}$
$x = \frac{- 1 + \sqrt{29}}{2}$

#### Explanation:

1. Calculate Horizontal Asymptotes
${\lim}_{x \to \pm \infty} \left(f \left(x\right)\right) = 1$
So we have a horizontal asymptote that is $y = 1$.
2. Calculate Vertical Asymptotes
x^2-x-7=0 ?
${\Delta}_{x} = 29$
x_1= (-1+sqrt(29))/2 ; x_2= (-1-sqrt(29))/2
${\lim}_{x \to {x}_{1}} \left(f \left(x\right)\right) = + \infty$
${\lim}_{x \to {x}_{2}} \left(f \left(x\right)\right) = - \infty$
So we have two vertical asymptotes in $x = {x}_{1}$ and in $x = {x}_{2}$
3. There aren't Oblique Asymptotes because there is a horizontal asymptote.
4. For sure... Let's see the graph of $f \left(x\right) = \frac{{x}^{2} - 4 x - 2}{{x}^{2} - x - 7}$

graph{(x^2-4x-2)/(x^2-x-7) [-10, 10, -5, 5]}