Chapter 4.1, Problem 71AYU

To determine

To analyze: the given polynomial function

Answer to Problem 71AYU

  

  

Explanation of Solution

Given:

  $f\left(x\right)=\left(x+4\right){\left(x-2\right)}^2$

Formula Used:

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

Calculation:

Given: $f\left(x\right)=\left(x+4\right){\left(x-2\right)}^2$

Step1: Determine the end behavior of the graph of the function

Expand the polynomial to write it in the form

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

  $\begin{array}{l}f\left(x\right)=\left(x+4\right){\left(x-2\right)}^2\hfill \\ =\left(x+4\right)\left(x^2-4x+4\right)\hfill \\ =x^3-4x^2+4x+4x^2-16x+16Multiply\hfill \\ =x^3-12x+16Combine liketerms\hfill \end{array}$

The polynomial function $f$ is of degree $3$ . The graph of $f$ behaves like $y=x^3$ for large values of $|x|$ .

Step2: find the $x-and\text{ y}-intercepts$ of the graph of the function

The $\text{y}-intercept$ is $f\left(0\right)=16$ . To find the $x-intercepts$ , we solve $f\left(x\right)=0$ .

  $\begin{array}{l}f\left(x\right)=0\\ \left(x+4\right){\left(x-2\right)}^2=0\\ x+4=0 or {\left(x-2\right)}^2=0\\ x=-4 or x-2=0\\ x=2\end{array}$

The $x-intercepts$ are $-4and2$ .

Step3: Determine the zeros of the function and their multiplicity Use this information to determine whether the graph crosses or touches the $x-axis$ ateach $x-intercept$ .

The zeros of $f$ are $-4and2$ . The zero $-4$ is a zero of multiplicity $1$ , so the graph of $f$ crosses the $x-axis$ at $x=-4$ . The zero $2$ is a zero of multiplicity $2$ , so the graph of $f$

touches the $x-axis$ at $x=2$ .

Step4: Determine the maximum number of turning points on the graph of the function.

Because the polynomial function is of degree $3$ (step1), the graph of the function will have

atmost $3-1=2$ points.

Step5: Determine the behavior of the graph of $f$ neareach $x-intercept$

The two $x-intercepts$ are $\text{-4 }and2$ .

  $\begin{array}{l}Near-4:f\left(x\right)=\left(x+4\right){\left(x-2\right)}^2\hfill \\ \approx \left(x+4\right){\left(-4-2\right)}^2\hfill \\ =\left(x+4\right)\left(36\right)Alinewithslope36\hfill \\ =36x+144\hfill \\ Near2:f\left(x\right)=\left(x+4\right){\left(x-2\right)}^2\hfill \\ \approx \left(2+4\right){\left(x-2\right)}^2\hfill \\ Aparabolathatopensup\hfill \\ =6{\left(x-2\right)}^2\hfill \end{array}$

Step 6: Put all the information from Steps 1 through 5 together to obtain the graph of $f$ Figure $1\left(a\right)$ illustrates the information obtained from Steps 1 through 5. We evaluate $f$ at $-5,-3,0,1,3$ to help establish the scale on the $y-axis$ . The graph of $f$ is given in Figure $1\left(b\right)$ .

Figure1

  

  

Conclusion:

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