Chapter 2.7, Problem 1E

To determine

For the rational function $f(x)=\frac{N(x)}{D(x)}$ , if the degree of $N(x)$ is exactly one more than the degree of $D(x)$ , then the graph of f has a ______ (or oblique) ________.

Answer to Problem 1E

Slant asymptote

Explanation of Solution

Given:

Function: $f(x)=\frac{N(x)}{D(x)}$

For the rational function $f(x)=\frac{N(x)}{D(x)}$ , if the degree of $N(x)$ is exactly one more than the degree of $D(x)$ , then the graph of f has a slant (or oblique) asymptote.

Example:

Let $f(x)=\frac{2x^2-5x+5}{x-2}.$

Here, the degree of numerator is exactly one more than the degree of denominator.

Calculation of asymptotes:

Vertical asymptotes:

To find vertical asymptotes, put denominator of the given function equal to zero.

  $\begin{array}{l}\Rightarrow x-2=0\\ \Rightarrow x=2\end{array}$

Horizontal asymptotes:

As the degree of numerator is larger than the degree of the denominator, the horizontal asymptote does not exist for given function.

Slant asymptotes:

To find slant asymptote, divide given function numerator with denominator using long division.

Dividend: $2x^2-5x+5$

Divisor: $x-2$

  $\begin{array}{l}x-2\begin{array}{c}\hfill 2x-1\\ \hfill \overline{)\begin{array}{l}2x^2-5x+5\\ \begin{array}{c}2x^2-4x\end{array}\\ -x+5\\ \begin{array}{c}-x+2\end{array}\end{array}}\end{array}\\ 3\end{array}$  $\begin{array}{l}\frac{2x^2}{x}=2x\\ \text{Multiply}2x(x-2)\text{and subtract}\text{.}\\ \\ \frac{-x}{x}=-1\\ \text{Multiply}-1(x-2)\text{and subtract}\text{.}\end{array}$

Here, quotient is $2x-1$ and remainder is $3.$

So, the slant asymptote is $y=2x-1.$