Chapter 4.4, Problem 52AYU

In Problems 45-56, find the vertical, horizontal, and oblique asymptotes, if any, of each rational function.

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

To determine

To find: Vertical, horizontal, or oblique asymptotes.

Answer to Problem 52AYU

Vertical asymptote is $x=-\frac{5}{2}$, horizontal asymptote is $y=\frac{1}{2}$.

Explanation of Solution

Given:

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

$2x^2+7x+5=0$

$2x^2+5x+2x+5=0$

$x\left( 2x+5\right)+1\left( 2x+5\right)=0$

$\left( 2x+5\right)\left( x+1\right)=0$

$x^2+6x+5=0$

$\left( x+1\right)\left( x+5\right)=0$

$F\left( x\right)=\frac{ x^2 +6x+5}{ 2x^2 +7x+5}=\frac{ \left( x+1 \right)\left( x+5 \right)}{ \left( 2x+5 \right)\left( x+1 \right)}=\frac{ \left( x+5 \right)}{ \left( 2x+5 \right)}$

$F$ is in lowest terms, and the zero of the denominator is $2x+5=0$. The line $x=-\frac{5}{2}$ is the vertical asymptote of the graph of $F$ .

Therefore vertical asymptote is $x=-\frac{5}{2}$ .

Since the degree of the numerator, 1, is equal to the degree of the denominator, 1, the horizontal asymptote equals the ratio of the leading coefficients.

Therefore horizontal asymptote is $y=\frac{1}{2}$ .

Oblique asymptotes: None.