Question

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

$F\left(x\right)=\frac{x^2+6x+5}{2x^2+7x+5}$

Short answer

$y=\frac{1}{2}$will be the horizontal asymptote for given rational expression.

Step-by-step solution

Step 1. Given information

Given rational expression:

$F\left(x\right)=\frac{x^2+6x+5}{2x^2+7x+5}$

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, $$\begin{gathered} n=2,m=2 \\ thus,n=m \end{gathered}$$

Now

$$\begin{gathered} F\left(x\right)=\frac{x^2+6x+5}{2x^2+7x+5} \\ Taking{x}^{2}commonfromnumeratoranddenominatorweget \\ =li{m}_{x\to \infty }\frac{x^2(1+{\displaystyle \raisebox{1ex}{$6x$}\!\left/ \!\raisebox{-1ex}{$x^2+{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.})$}\right.}}{x^2(2+{\displaystyle \raisebox{1ex}{$7x$}\!\left/ \!\raisebox{-1ex}{$x^2+{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.})$}\right.}} \\ =\frac{1}{2} \\ i.e\frac{a_n}{b_m}=\frac{1}{2} \end{gathered}$$