# How do you find the Vertical, Horizontal, and Oblique Asymptote given y = (x^2 + 2x - 3)/( x^2 - 5x - 6) ?

Jan 23, 2017

vertical asymptotes at x = -1 and x = 6
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.

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

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

Horizontal asymptotes occur as

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

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

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

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

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

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