# How do I find the asymptotes of y=1/((x-1)(x-3))?

Mar 29, 2018

Horizontal is when $\lim x \to \pm \infty \frac{1}{\left(x - 3\right) \left(x - 1\right)} = 0$

and vertical is when x is 1 or 3

#### Explanation:

The horizontal assymptotes are the assymptotes as x approaches infinity or negative infinity $\lim x \to \infty$ or $\lim x \to - \infty$

$\lim x \to \infty \frac{1}{{x}^{2} - 4 x + 3}$

Divide top and bottom by the highest power in the denominator
$\lim x \to \infty \frac{\frac{1}{x} ^ 2}{1 - \frac{4}{x} + \frac{3}{x} ^ 2}$

$\frac{0}{1 - 0 - 0} = \frac{0}{1} = 0$ so this is your horizontal assymptote negative infinty gives us the same result

For the vertical asymptote we are looking for when the denominator is equal to zero

$\left(x - 1\right) \left(x - 3\right) = 0$ so you have a vertical asymptote when

$x = 3 \mathmr{and} 1$