Chapter 9.4, Problem 4CYU

To determine

To graph: The given function.

Explanation of Solution

Given information:

The following function:

  $f\left(x\right)=\frac{6x^2-3x+2}{x}$ .

Formula used:

Discriminant is given by:

  $D=b^2-4ac$

Calculation:

Step 1:

Find the zeroes.

  $\begin{array}{l}6x^2-3x+2=0[\text{Set }a(x)=0]\\ D=b^2-4ac[\text{Formula for discriminant]}\\ D={\left(-3\right)}^2-4\left(6\right)\left(2\right)[\text{Substitute]}\\ D=9-48[\text{Simplify]}\\ D=-39[\text{Subtract]}\end{array}$

Since discriminant is negative, the equation has no real solutions. Hence, the function has no zeroes

Step 2:

Find the asymptotes.

  $x=0[\text{Set }b(x)=0]$

There is a vertical asymptote at $x=0$ .

The degree of numerator is greater than the degree of the denominator.

So, there is no horizontal asymptote.

The difference between the degree of the numerator and the degree of the

denominator is $1$ , so there is an oblique asymptote.

Divide the numerator by the denominator to determine the equation of the oblique asymptote.

The equation of the oblique asymptote is the quotient excluding any remainder.

  $x\begin{array}{c}\hfill 6x-3\\ \hfill \overline{)\begin{array}{l}\text{ 6}{x}^2-3x+2\\ \underset{\_}{(-)6x^2}\\ -3x\\ \underset{\_}{(+)-3x}\\ \text{ 2}\end{array}}\end{array}$

Thus, the oblique asymptote is the line $f(x)=6x-3$ .

Step 3:

Draw the asymptotes, and then use a table of values to graph the function.