Chapter 2.6, Problem 82E

To determine

To write: a rational function f whose graph has vertical asymptote at $x=2$ and has slant asymptote at $y=x+1$ and zero of the function at $x=-2$ .

Answer to Problem 82E

  $f\left(x\right)=\frac{x^2-x-6}{x-2}$

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 $D\left(x\right)$ .”

Slant Asymptote:

“For the given rational function $f\left(x\right)=\frac{N\left(x\right)}{D\left(x\right)}$ where $N\left(x\right)$ and $D\left(x\right)$ are polynomials and the degree of $N\left(x\right)$ is 1 more than the degree of $D\left(x\right)$ , then $f\left(x\right)$ can be expressed in the form:

  $f\left(x\right)=mx+b+\frac{R\left(x\right)}{D\left(x\right)}$

Where the degree of $R\left(x\right)$ is less than the degree of $D\left(x\right)$ . The line $y=mx+b$ is an oblique asymptote for the graph of f .

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

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

Also, the function has slant asymptote at $y=x+1$ , this implies the degree of the numerator of the function is 1 greater than the degree of its denominator.

Since degree of denominator is 1, so the degree of numerator is 2.

Also, the function has a zero at $x=-2$ , so its numerator must have a zero at $x=-2$

So, let $N\left(x\right)=x^2-x-6$

Then, the given function may be written as:

  $\begin{array}{c}f\left(x\right)=\frac{x^2-x-6}{x-2}\\ =\left(x+1\right)+\frac{\left(-4\right)}{x-2}\end{array}$

Then, the required rational function is $f\left(x\right)=\frac{x^2-x-6}{x-2}$ .