Chapter 2.6, Problem 81E

To determine

To write: a rational function f whose graph has vertical asymptotes at $x=-2$ and $x=1$ and no horizontal asymptote.

Answer to Problem 81E

  $f\left(x\right)=\frac{x^3}{\left(x+2\right)\left(x-1\right)}$

Explanation of Solution

Concept Used:

Vertical Asymptote:

“The line $x=a$ is a vertical asymptote of the graph of a rational function $f\left(x\right)=\frac{N\left(x\right)}{D\left(x\right)}$ , only if ‘ a’ is a zero of $D\left(x\right)$ .”

Horizontal Asymptote:

“For the given rational function $f\left(x\right)=\frac{N\left(x\right)}{D\left(x\right)}$ where degree of $N\left(x\right)$ is m and degree of $D\left(x\right)$ is n .

1. The line $y=0$ is a horizontal asymptote if $m<n$ .

2. The line $y=\frac{a_m}{b_n}$ (the ratio of leading coefficients) is a horizontal asymptote if $m=n$ .

3. There is no horizontal asymptote if $m>n$ .”

Since, the required function has vertical asymptote at $x=-2$ and $x=1$ , it means that the denominator of this function contains the term $\left(x+2\right)\left(x-1\right)$ .

So, let $D\left(x\right)=\left(x+2\right)\left(x-1\right)$

Also, the function has no horizontal asymptote, this implies the degree of the numerator of the function must be greater than the degree of its denominator.

Now, as denominator is $\left(x+2\right)\left(x-1\right)$ which is of degree 2. So, any function of degree 3 or more can be its numerator.

So, let $N\left(x\right)=x^3$

Then, the required rational function is $f\left(x\right)=\frac{x^3}{\left(x+2\right)\left(x-1\right)}$ .