How do you find vertical, horizontal and oblique asymptotes for (x^2-x-8)/ (x+1)?

Feb 11, 2017

vertical asymptote at x = - 1
oblique asymptote y = x - 2

Explanation:

Let $f \left(x\right) = \frac{{x}^{2} - x - 8}{x + 1}$

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve $x + 1 = 0 \Rightarrow x = - 1 \text{ is the asymptote}$

Horizontal asymptotes occur as

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

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

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} - \frac{x}{x} ^ 2 - \frac{8}{x} ^ 2}{\frac{x}{x} ^ 2 + \frac{1}{x} ^ 2} = \frac{1 - \frac{1}{x} - \frac{8}{x} ^ 2}{\frac{1}{x} + \frac{1}{x} ^ 2}$

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

This is undefined, hence there are no horizontal asymptotes.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here hence there is an oblique asymptote. Expressing the numerator in terms of factors of the divisor.

$f \left(x\right) = \frac{\textcolor{red}{x} \left(x + 1\right) + \left(\textcolor{b l u e}{- x} - x\right) - 8}{x + 1}$

$= \frac{\textcolor{red}{x} \left(x + 1\right) \textcolor{red}{- 2} \left(x + 1\right) \textcolor{b l u e}{+ 2} - 8}{x + 1}$

$= x - 2 - \frac{6}{x + 1}$

as $x \to \pm \infty , f \left(x\right) \to x - 2$

$\Rightarrow y = x - 2 \text{ is the asymptote}$
graph{(x^2-x-8)/(x+1) [-10, 10, -5, 5]}