Chapter 4.1, Problem 69AYU

To determine

To analyze: the given polynomial function

Answer to Problem 69AYU

  

  

Explanation of Solution

Given:

  $f\left(x\right)=x^2\left(x-3\right)$

Formula Used:

  $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_{1}x+a_0$

Calculation:

Consider the following polynomial:

  $f\left(x\right)=x^2\left(x-3\right)$

The objective is to analyze each polynomial function using the analyzing the graph of a polynomial function.

Step1: To determine the end behavior of the graph of the function.

The graph of a polynomial function is.

  $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_{1}x+a_0$

Expand the given polynomial and its write to the above form.

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

The polynomial function has degree $3$ and the graph of $f$ behaves like $y=x^3$ for large values of $\left|x\right|$ .

Step2: Find $x - and y-$ intercepts of the graph of the function.

The $y-intercept$ values are obtained by substituting $x=0$ .

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

So, the $y-intercept$ is $0$ .

The $x-intercept$ values are obtained by substituting $y=0$ .

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

So, the $x-intercept$ are $0and3$ .

Steps3:

To determine the zeros of the function and their multiplicity using this information whether the graph crosses or touches the $x-axis$ ateach $x-intercept$ using the results.

The graph of $f$ touches the $x-axis$ at a zero of even multiplicity and the graph of $f$ crosses the $x-axis$ at a zero of odd multiplicity.

The zeros of the function $f$ are $0and3$ .

The intercept $0$ is a multiplicity of $\text{2}$ .

So, the graph of $f$ touches the $x-axis$ at $x=0$ .

The intercept $3$ is a multiplicity of $\text{1}$ .

So, the graph of $f$ crosses the $x-axis$ at $x=3$ .

Step4:

To determine the maximum number of points on the graph of the function use the result.

The maximum number of turning points is, $n-1$ ( $n$ is the degree of the given polynomial function).

Here, the degree of the given polynomial function is $3$ .

So, the maximum number of turning points is $3-1=2$ .

Hence, the graph of $f$ will contain at most two turning points.

Steps5:

To determine the behavior of the graph of $f$ neareach $x-intercept$ .

The two $x-intercept$ are $0and3$ .

The graph of $f$ nearthe $x-intercept$ at $0$ :

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

So, the behavior of the graph of $f$ nearthe $x-intercept$ at $0$ is a parabola that opens down.

The graph of $f$ nearthe $x-intercept$ at $3$ :

  $\begin{array}{l}f\left(x\right)={\left(x\right)}^2\left(x-3\right)\\ \approx 3^2\left(x-3\right)\\ =9\left(x-3\right) \left(Since9\left(x-3\right)isa\text{ line with }slope9.\right)\end{array}$

So, the behavior of the graph of $f$ nearthe $x-intercept$ at $3$ is a line with slope $9$ .

Step6: Use all the above information from step 1 to step 5 to draw a complete graph of the function.

First figure illustrates that the information obtained from step 1 to step 5. Notice that to evaluate that $f$ at $0and3$ to help the scale onthe $y-axis$ .

Draw the following figure1:

  

The graphs of the given polynomial function $f\left(x\right)=x^2\left(x-3\right)$ and draw the following figure2:

  

Conclusion:

Thus, the graph of $f$ is obtained which illustrates the information obtained from Steps 1 through 5.