# How do you find the vertical, horizontal or slant asymptotes for  [(9x-4) / (3x+2)] +2?

Oct 25, 2016

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

#### Explanation:

The denominator of the rational function cannot be zero as this would make the rational function 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: $3 x + 2 = 0 \Rightarrow x = - \frac{2}{3} \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

$f \left(x\right) = \frac{\frac{9 x}{x} - \frac{4}{x}}{\frac{3 x}{x} + \frac{2}{x}} + 2 = \frac{9 - \frac{4}{x}}{3 + \frac{2}{x}} + 2$

as $x \to \pm \infty , f \left(x\right) \to \frac{9 - 0}{3 + 0} + 2$

$\Rightarrow y = 5 \text{ is the asymptote}$

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{(9x-4)/(3x+2)+2 [-20, 20, -10, 10]}