# How do you find vertical, horizontal and oblique asymptotes for (6x-7)/(11x+8)?

Jun 5, 2016

vertical asymptote $x = - \frac{8}{11}$
horizontal asymptote $y = \frac{6}{11}$

#### Explanation:

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

solve : 11x + 8 = 0 $\Rightarrow x = - \frac{8}{11} \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

$\frac{\frac{6 x}{x} - \frac{7}{x}}{\frac{11 x}{x} + \frac{8}{x}} = \frac{6 - \frac{7}{x}}{11 + \frac{8}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{6 - 0}{11 + 0}$

$\Rightarrow y = \frac{6}{11} \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here ( both degree 1 ). Hence there are no oblique asymptotes.
graph{(6x-7)/(11x+8) [-10, 10, -5, 5]}