Chapter 3.7, Problem 25E

To determine

To find the vertical and horizontal asymptote of the rational function.

Answer to Problem 25E

The rational function has vertical asymptote $x=\frac{1}{2}$ , $x=-1$ and has horizontal asymptote $y=3$ .

Explanation of Solution

Given information:

The rational function is $s\left(x\right)=\frac{6x^2+1}{2x^2+x-1}$

Calculation:

Vertical asymptote of $r\left(x\right)$ are the lines $x=a$ , where $a$ is the zero of the denominator.

Therefore,

The zero of the denominators is obtained as:

  $2x^2+x-1=0$

Use Quadratic formula to solve the above equation,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, $a=2,b=1\text{ and }c=-1$

  $\begin{array}{l}x=\frac{-\left(1\right)\pm \sqrt{{\left(1\right)}^2-4\left(2\right)\left(-1\right)}}{2\left(2\right)}\\ x=\frac{-1\pm \sqrt{1+8}}{4}\\ x=\frac{-1\pm \sqrt{9}}{4}\\ x=\frac{-1\pm 3}{4}\end{array}$

Either,

  $\begin{array}{l}x=\frac{-1+3}{4}\\ x=\frac{2}{4}\\ x=\frac{1}{2}\end{array}$

Or,

  $\begin{array}{l}x=\frac{-1-3}{4}\\ x=-\frac{4}{4}\\ x=-1\end{array}$

Hence,

The rational function has vertical asymptote $x=\frac{1}{2}$ and $x=-1$ .

Horizontal asymptote:

The degree of the numerator is $n=2$ and denominator is $m=2$ .

Since, $n=m$ ,

Therefore,

  $\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}=\frac{6}{2}=3$

So, the horizontal asymptote is the line $y=3$ .

Hence,

The rational function hasvertical asymptote $x=\frac{1}{2}$ , $x=-1$ and has horizontal asymptote $y=3$ .