# How do you find the vertical, horizontal and slant asymptotes of: y = (x^2 + 2x - 3)/( x^2 - 5x - 6) ?

Jun 10, 2016

vertical asymptotes x = -1 , x = 6
horizontal asymptote y = 1

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve : x^2-5x-6)=0rArr(x-6)(x+1)=0

$\Rightarrow x = - 1 , x = 6 \text{ are the asymptotes}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by ${x}^{2}$

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

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

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

Slant 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 slant asymptotes.
graph{(x^2+2x-3)/(x^2-5x-6) [-10, 10, -5, 5]}