# How do you find the vertical, horizontal or slant asymptotes for y=1/(2-x)?

Dec 1, 2016

vertical asymptote at x = 2
horizontal asymptote at y = 0

#### Explanation:

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 : $2 - x = 0 \Rightarrow x = 2 \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x.

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

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator- degree 0 , denominator-degree 1 ) Hence there are no slant asymptotes.
graph{1/(2-x) [-10, 10, -5, 5]}