# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)=((3x-2)(x+5))/((2x-1)(x+6))?

Jun 29, 2016

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

#### Explanation:

For this rational function the denominator cannot be zero. This would lead to division by zero which is undefined.By setting the denominator equal to zero and solving for x we can find the values that x cannot be and if the numerator is also non-zero for these values of x then they must be vertical asymptotes.

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

$\Rightarrow x = - 6 \text{ and " x=1/2" are the asymptotes}$

Horizontal asymptotes occur as

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

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

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

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

Oblique 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 oblique asymptotes.
graph{((3x-2)(x+5))/((2x-1)(x+6) [-10, 10, -5, 5]}