# How do you find the asymptotes for y = (x^2 + 2x - 3)/( x^2 - 5x - 6)?

Aug 8, 2018

$\text{vertical asymptotes at "x=-1" and } x = 6$
$\text{horizontal asymptote at } y = 1$

#### Explanation:

The denominator of y cannot be zero as this would make y 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.

$\text{solve } {x}^{2} - 5 x - 6 = 0 \Rightarrow \left(x - 6\right) \left(x + 1\right) = 0$

$x = - 1 \text{ and "x=6" are the asymptotes}$

$\text{Horizontal asymptotes occur as}$

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

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

$y = \frac{{x}^{2} / {x}^{2} + \frac{2 x}{x} ^ 2 - \frac{3}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{5 x}{x} ^ 2 - \frac{6}{x} ^ 2} = \frac{1 + \frac{2}{x} - \frac{3}{x} ^ 2}{1 - \frac{5}{x} - \frac{6}{x} ^ 2}$

$\text{ as } x \to \pm \infty , y \to \frac{1 + 0 - 0}{1 - 0 - 0}$

$y = 1 \text{ is the asymptote}$
graph{(x^2+2x-3)/(x^2-5x-6) [-20, 20, -10, 10]}