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

Aug 9, 2016

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

#### Explanation:

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.

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

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

Horizontal asymptotes occur as

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

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

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

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

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

Oblique 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 oblique asymptotes.
graph{(x^2-81)/(x^2-4x) [-50.63, 50.63, -25.3, 25.33]}