# How do you find the vertical, horizontal or slant asymptotes for  ((3x-2)(x+5))/((2x-1)(x+6))?

Jul 14, 2016

vertical asymptotes x = -6 , x$= \frac{1}{2}$
horizontal asymptote y$= \frac{3}{2}$

#### Explanation:

The denominator of the rational function cannot be zero as this is undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: (2x-1)(x+6)=0$\Rightarrow x = - 6 , x = \frac{1}{2} \text{ are the asymptotes}$

Horizontal asymptotes occur as

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

$f \left(x\right) = \frac{\left(3 x - 2\right) \left(x + 5\right)}{\left(2 x - 1\right) \left(x + 6\right)} = \frac{3 {x}^{2} + 13 x - 10}{2 {x}^{2} + 11 x - 6}$

divide terms on numerator/denominator by the highest exponent of x , that is ${x}^{2}$

$\frac{\frac{3 {x}^{2}}{x} ^ 2 + \frac{13 x}{x} ^ 2 - \frac{10}{x} ^ 2}{\frac{2 {x}^{2}}{x} ^ 2 + \frac{11 x}{x} ^ 2 - \frac{6}{x} ^ 2} = \frac{3 + \frac{13}{x} - \frac{10}{x} ^ 2}{2 + \frac{11}{x} - \frac{6}{x} ^ 2}$

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

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2) Hence there are no slant asymptotes.
graph{((3x-2)(x+5))/((2x-1)(x+6)) [-10, 10, -5, 5]}