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

##### 1 Answer
Sep 16, 2016

vertical asymptote at x = 3
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.

solve: $x - 3 = 0 \Rightarrow x = 3 \text{ is the asymptote}$

Horizontal asymptotes occur as

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

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

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes.
graph{(x+1)/(x-3) [-10, 10, -5, 5]}