# How do you find the asymptotes for y=(3/(x-2))+1?

Jun 10, 2016

vertical asymptote x = 2
horizontal asymptote y = 1

#### Explanation:

Begin by expressing y as a single rational function.

$\Rightarrow y = \frac{3}{x - 2} + 1 = \frac{3}{x - 2} + \frac{x - 2}{x - 2} = \frac{3 + x - 2}{x - 2}$

Thus y simplifies to. $y = \frac{x + 1}{x - 2}$

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve: x-2 = 0 → x = 2 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

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

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

$\Rightarrow y = 1 \text{ is the asymptote}$
graph{(x+1)/(x-2) [-10, 10, -5, 5]}