Question

Find the vertical, horizontal, and oblique asymptotes, if any, of each rational function.

$Q\left(x\right)=\frac{2x^2-5x-12}{3x^2-11x-4}$

Short answer

$y=\frac{2}{3}$will be the horizontal asymptote for given rational expression.

Step-by-step solution

Step 1. Given information

Given rational expression:

$Q\left(x\right)=\frac{2x^2-5x-12}{3x^2-11x-4}$

Step 2. Write the concept

If n = m (the degree of the numerator equals the degree of the denominator), the quotient obtained will be

the number $\frac{a_n}{b_m}$, and the line $y=\frac{a_n}{b_m}$is a horizontal asymptote.

Step 3. Calculation

Here,$n=2,m=2$

Now

$$\begin{gathered} F\left(x\right)=\frac{2x^2-5x-12}{3x^2-11x-4} \\ Taking{x}^{2}commonfromnumeratoranddenominatorweget \\ =li{m}_{x\to \infty }\frac{x^2(2-{\displaystyle \raisebox{1ex}{$5x$}\!\left/ \!\raisebox{-1ex}{$x^2-{\displaystyle \raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.})$}\right.}}{x^2(3-{\displaystyle \raisebox{1ex}{$11x$}\!\left/ \!\raisebox{-1ex}{$x^2-{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.})$}\right.}} \\ =\frac{2}{3} \\ i.e\frac{a_n}{b_m}=\frac{2}{3} \end{gathered}$$