# How do you find the vertical, horizontal or slant asymptotes for f(x)={ x^2 - 81 }/ {x^2 - 4x}?

Jun 4, 2016

vertical asymptotes x = 0 , x = 4
horizontal asymptote y = 1

#### 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 : ${x}^{2} - 4 x = 0 \Rightarrow x \left(x - 4\right) = 0 \Rightarrow x = 0 , x = 4$

$\Rightarrow x = 0 , x = 4 \text{ are the asymptotes}$

Horizontal asymptotes occur as

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

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

$\frac{\frac{{x}^{2}}{x} ^ 2 - \frac{81}{x} ^ 2}{\frac{{x}^{2}}{x} ^ 2 - \frac{4 x}{x} ^ 2} = \frac{1 - \frac{81}{x} ^ 2}{1 - \frac{4}{x}}$

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

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

Slant 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 slant asymptotes.
graph{(x^2-81)/(x^2-4x) [-20, 20, -10, 10]}