# How do you find the vertical, horizontal or slant asymptotes for ((x-1)(x-3))/(x(x-2) )?

Apr 15, 2016

vertical asymptotes x = 0 , x = 2
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(x - 2) = 0 → x = 0 , x = 2

$\Rightarrow x = 0 \text{ and " x = 2" are the asymptotes }$

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} f \left(x\right) \to 0$

now $\frac{\left(x - 1\right) \left(x - 3\right)}{x \left(x - 2\right)} = \frac{{x}^{2} - 4 x + 3}{{x}^{2} - 2 x}$

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

$\frac{{x}^{2} / {x}^{2} - \frac{4 x}{x} ^ 2 + \frac{3}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{2 x}{x} ^ 2} = \frac{1 - \frac{4}{x} + \frac{3}{x} ^ 2}{1 - \frac{2}{x}}$

as $x \to \pm \infty , \frac{4}{x} , \frac{2}{x} \text{ and } \frac{3}{x} ^ 2 \to 0$

$y = \frac{1 - 0 + 0}{1 - 0} = \frac{1}{1} = 1$

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.

This is the graph of the function.
graph{((x-1)(x-3))/(x(x-2)) [-10, 10, -5, 5]}