# How do you find vertical, horizontal and oblique asymptotes for  f(x) = (x+1) / (x+2)?

Jul 14, 2016

vertical asymptote x = -2
horizontal asymptote y = 1

#### Explanation:

The denominator of f(x) cannot be zero as this is 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 + 2 = 0 → x = -2 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

$\frac{\frac{x}{x} + \frac{1}{x}}{\frac{x}{x} + \frac{2}{x}} = \frac{1 + \frac{1}{x}}{1 + \frac{2}{x}}$

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

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

Oblique 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 oblique asymptotes.
graph{(x+1)/(x+2) [-10, 10, -5, 5]}