# What are the vertical and horizontal asymptotes of f(x)=5/((x+1)(x-3))?

Mar 5, 2018

$\text{vertical asymptotes at "x=-1" and } x = 3$
$\text{horizontal asymptote at } y = 0$

#### Explanation:

$\text{the denominator of f(x) cannot be zero as this}$
$\text{would make f(x) undefined. Equating the denominator}$
$\text{to zero and solving gives the values that x cannot be}$
$\text{and if the numerator is non-zero for these values then}$
$\text{they are vertical asymptotes}$

$\text{solve } \left(x + 1\right) \left(x - 3\right) = 0$

$\Rightarrow x = - 1 \text{ and "x=3" are the asymptotes}$

$\text{Horizontal asymptotes occur as}$

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

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

$f \left(x\right) = \frac{\frac{5}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{2 x}{x} ^ 2 - \frac{3}{x} ^ 2} = \frac{\frac{5}{x} ^ 2}{1 - \frac{2}{x} - \frac{3}{x} ^ 2}$

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

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{5/((x+1)(x-3)) [-10, 10, -5, 5]}