Chapter 3.7, Problem 30E

To determine

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

Answer to Problem 30E

The rational function has vertical asymptote $x=0$ and has horizontal asymptote $y=5$ .

Explanation of Solution

Given information:

The rational function is $r\left(x\right)=\frac{5x^3}{x^3+2x^2+5x}$

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:

  $\begin{array}{l}{x}^3+2x^2+5x=0\\ x\left(x^2+2x+5\right)=0\end{array}$

Either,

  $x=0$

Or,

  $x^2+2x+5=0$

Use Quadratic formula to solve the above equation,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, $a=1,b=2\text{ and }c=5$

  $\begin{array}{l}x=\frac{-\left(2\right)\pm \sqrt{{\left(2\right)}^2-4\left(1\right)\left(5\right)}}{2\left(1\right)}\\ x=\frac{-2\pm \sqrt{4-20}}{2}\\ x=\frac{-2\pm \sqrt{-16}}{2}\\ x=\frac{-2\pm 4i}{2}\\ x=-1\pm 2i\end{array}$

So, there is only one real zero of the denominator that is $x=0$ .

Hence,

The rational function has vertical asymptote $x=0$ .

Horizontal asymptote:

Here, the degree of the numerator is $n=3$ and denominator is $m=3$ .

Since, $n=m$ ,

Therefore,

  $\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}=\frac{5}{1}=5$

So, the horizontal asymptote is the line $y=5$ .

Hence,

The rational function hasvertical asymptote $x=0$ and has horizontal asymptote $y=5$ .