# How do you find the vertical, horizontal and slant asymptotes of: y= (3x-2) /(2x+5)?

##### 1 Answer
Aug 30, 2016

vertical asymptote at $x = - \frac{5}{2}$
horizontal asymptote at $y = \frac{3}{2}$

#### 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: $2 x + 5 = 0 \Rightarrow x = - \frac{5}{2} \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

$y = \frac{\frac{3 x}{x} - \frac{2}{x}}{\frac{2 x}{x} + \frac{5}{x}} = \frac{3 - \frac{2}{x}}{2 + \frac{5}{x}}$

as $x \to \pm \infty , y \to \frac{3 - 0}{2 + 0}$

$\Rightarrow y = \frac{3}{2} \text{ is the asymptote}$

Slant 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 slant asymptotes.
graph{(3x-2)/(2x+5) [-20, 20, -10, 10]}