# How do you find vertical, horizontal and oblique asymptotes for y= ((x+2)(x+1))/((3x-1)(x+2))?

Sep 21, 2016

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

#### Explanation:

The first step here is to simplify the function y.

$y = \frac{\cancel{\left(x + 2\right)} \left(x + 1\right)}{\left(3 x - 1\right) \cancel{\left(x + 2\right)}} = \frac{x + 1}{3 x - 1}$

The denominator of y cannot be zero as this would make y 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:$3 x - 1 = 0 \Rightarrow x = \frac{1}{3} \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

$y = \frac{\frac{x}{x} + \frac{1}{x}}{\frac{3 x}{x} - \frac{1}{x}} = \frac{1 + \frac{1}{x}}{3 - \frac{1}{x}}$

as $x \to \pm \infty , y \to \frac{1 + 0}{3 - 0}$

$\Rightarrow y = \frac{1}{3} \text{ is the asymptote}$

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