How do you find vertical, horizontal and oblique asymptotes for (x^2+x-6)/(x^2-x-6)?

Jul 28, 2018

$\text{vertical asymptotes at "x=-2" and } x = 3$
$\text{horizontal asymptote at } y = 1$

Explanation:

$\text{let } f \left(x\right) = \frac{{x}^{2} + x - 6}{{x}^{2} - x - 6}$

The denominator of f(x) cannot be zero as this would make f(x) 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.

$\text{solve } \left({x}^{2} - x - 6\right) = 0 \Rightarrow \left(x - 3\right) \left(x + 2\right) = 0$

$x = - 2 \text{ and "x=3" are the asymptotes}$

$\text{Horizontal asymptotes occur as}$

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

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

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} + \frac{x}{x} ^ 2 - \frac{6}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{x}{x} ^ 2 - \frac{6}{x} ^ 2} = \frac{1 + \frac{1}{x} - \frac{6}{x} ^ 2}{1 - \frac{1}{x} - \frac{6}{x} ^ 2}$

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{1 + 0 - 0}{1 - 0 - 0}$

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

$\text{Oblique asymptotes occur when the degree of the }$
$\text{numerator is greater than the degree of the denominator.}$
$\text{This is not the case here hence there are no oblique}$
$\text{asymptotes}$
graph{(x^2+x-6)/(x^2-x-6) [-10, 10, -5, 5]}