How do you find the vertical, horizontal or slant asymptotes for y=(3x+1)/(2-x)?

Dec 17, 2016

vertical asymptote at x = 2
horizontal asymptote at y = - 3

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

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

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

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

Slant 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 slant asymptotes.
graph{(3x+1)/(2-x) [-20, 20, -10, 10]}