# How do you identify all asymptotes for f(x)=(x+1)/(x-3)?

May 14, 2017

$\text{vertical asymptote at } x = 3$
$\text{horizontal asymptote at } y = 1$

#### 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 value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

$\text{solve " x-3=0rArrx=3" is the asymptote}$

Horizontal asymptotes occur as

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

$\text{divide terms on numerator/denominator by x}$

$f \left(x\right) = \frac{\frac{x}{x} + \frac{1}{x}}{\frac{x}{x} - \frac{3}{x}} = \frac{1 + \frac{1}{x}}{1 - \frac{3}{x}}$

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

$\Rightarrow y = 1 \text{ is the asymptote}$
graph{(x+1)/(x-3) [-11.25, 11.25, -5.625, 5.625]}