# How do you find vertical, horizontal and oblique asymptotes for  Q(x)= (3x^2-2x-8)/(2x^3+x^2-3x)?

Jul 15, 2016

vertical asymptotes x$= - \frac{3}{2}$ , x = 0 , x = 1
horizontal asymptote y = 0

#### Explanation:

The denominator of Q(x) 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: $2 {x}^{3} + {x}^{2} - 3 x = 0 \Rightarrow x \left(2 x + 3\right) \left(x - 1\right) = 0$

$\Rightarrow x = - \frac{3}{2} , x = 0 , x = 1 \text{ are the asymptotes}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest power of x , that is ${x}^{3}$

$\frac{\frac{3 {x}^{2}}{x} ^ 3 - \frac{2 x}{x} ^ 3 - \frac{8}{x} ^ 3}{\frac{2 {x}^{3}}{x} ^ 3 + {x}^{2} / {x}^{3} - \frac{3 x}{x} ^ 3} = \frac{\frac{3}{x} - \frac{2}{x} ^ 2 - \frac{8}{x} ^ 3}{2 + \frac{1}{x} - \frac{3}{x} ^ 2}$

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 2 ,denominator-degree 3) Hence there are no oblique asymptotes.
graph{(3x^2-2x-8)/(2x^3+x^2-3x) [-10, 10, -5, 5]}