Chapter 4, Problem 28RE

To determine

To discuss: the given polynomial function.

Answer to Problem 28RE

  

Explanation of Solution

Given: $R\left(x\right)=\frac{x^2-6x+9}{x^2}$

Calculation:

Let us consider the following polynomial function,

  $R\left(x\right)=\frac{x^2-6x+9}{x^2}$

apply $8$ steps to analyze the rational function.

Step $1$ :

factor the numerator and denominator of the function and find the domain of the function.

  $\begin{array}{l}R\left(x\right)=\frac{x^2-6x+9}{x^2}\\ =\frac{\left(x-3\right)\left(x-3\right)}{x^2}\\ =\frac{{\left(x-3\right)}^2}{x^2}\end{array}$

Thus, the domain of the function is $\left\{x|x\ne 0\right\}$

Step $2$ :

Now, write $R$ in lowest terms.

Since, there are no more common factors between numerator and denominator therefore $R$ is in lowest terms.

Step $3$ :

The $x-\mathrm{int}ercepts$ are found by determining real zeros of the numerator of $R$ written in lowest terms, equating numerator to zero,

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

So, $x-\mathrm{int}ercept$ of $R$ is $3$ .

There is no $y-\mathrm{int}ercept$ , since $x^2$ can’t equal to zero.

Step $4$ :

Now, test for symmetry if $R\left(-x\right)=R\left(x\right)$ , the function is even and its graph will be symmetric with respect to $y-axis$

If $R\left(-x\right)=-R\left(x\right)$ , the function is odd and its graph will be symmetric with respect to origin.

So,

  $\begin{array}{l}R\left(-x\right)=\frac{\left(-x-3\right)\left(-x-3\right)}{{\left(-x\right)}^2}\\ =\frac{\left(-x-3\right)\left(-x-3\right)}{x^2}\end{array}$

Conclude that, neither $R\left(-x\right)=R\left(x\right)$ nor $R\left(-x\right)=-R\left(x\right)$ so, no symmetry with $y-axis$ or origin.

Step $5$ :

Locate the vertical asymptotes by finding the zeros of the denominator with the rational function in lowest terms. With $R$ written in lowest terms, find that the graph of $R$ has a vertical asymptote $x=0$ .

Step $6$ :

Since the degree of numerator equals degree of numerator, therefore the graph is horizontal.

To determine the graph of $R$ intersects the horizontal asymptote, solve the equation

  $\begin{array}{l}R\left(x\right)=0\\ \frac{{\left(x-3\right)}^2}{x^2}=0\\ {\left(x-3\right)}^2=0\\ x=3\end{array}$

The only solution $x=3$ , so the graph intersects at $\left(3,0\right)$ .

Step $7$ :

From the graph, there is no $y-\mathrm{int}er\sec t$ and there is one $x-\mathrm{int}er\sec t$ , $3$ .

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

  

Step $8$ :

The graph of this function is as follows,

  

Conclusion:

Therefore,the given polynomial function is analyzed.