How do you find the asymptotes for f(x) = ((3x-2)(x+5))/((2x-1)(x+6))?

Sep 8, 2016

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

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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: $\left(2 x - 1\right) \left(x + 6\right) = 0 \Rightarrow x = - 6 , x = \frac{1}{2}$

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

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

Horizontal asymptotes occur as

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

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

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

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

$\Rightarrow y = \frac{3}{2} \text{ is the asymptote}$
graph{((3x-2)(x+5))/((2x-1)(x+6)) [-10, 10, -5, 5]}