# How do you find the vertical, horizontal or slant asymptotes for f(x)=( x-4)/ (x^2-1)?

Jun 16, 2016

vertical asymptotes x = ± 1
horizontal asymptote y = 0

#### 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^2-1=0rArrx^2=1rArrx=±1

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

Horizontal asymptotes occur as

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

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

$\frac{\frac{x}{x} ^ 2 - \frac{4}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{1}{x} ^ 2} = \frac{\frac{1}{x} - \frac{4}{x} ^ 2}{1 - \frac{1}{x} ^ 2}$

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1, denominator-degree 2 ) Hence there are no slant asymptotes.
graph{(x-4)/(x^2-1) [-10, 10, -5, 5]}