# How do you find the asymptotes for Y=(3x)/(x^2-x-6) + 3?

Mar 8, 2016

Horizontal asymptote is $y = 3$
I have taken you to a point where you should be able to take over for the vertical ones.

#### Explanation:

As $\left\mid x \right\mid$ increases then $\left\mid {x}^{2} \right\mid$becomes significantly larger and the other values of the fraction become more an more insignificant

$\textcolor{b l u e}{\text{Horizontal asymptote}}$

${\lim}_{\left\mid x \right\mid \to \infty} \left[\frac{3 x}{{x}^{2} - x - 6} + 3\right] = {\lim}_{\left\mid x \right\mid \to \infty} \left[\frac{3 x}{{x}^{2} - x - 6}\right] + 3 = 3$

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The equation becomes undefined as the fractions denominator approaches 0. So it is a matter of solving
$\text{ } {x}^{2} - x - 6 = 0$

$\text{ } \left(x + 2\right) \left(x - 3\right)$

$\textcolor{b l u e}{\text{Vertical asymptotes at "x=-2" and } x = + 3}$