Chapter 4.1, Problem 73AYU

To determine

To analyze: the given polynomial function

Answer to Problem 73AYU

  

  

Explanation of Solution

Given:

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

Formula Used:

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

Calculation:

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

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

Expand the polynomial to write it in the form

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

  $\begin{array}{l}f\left(x\right)=-2\left(x+2\right){\left(x-2\right)}^2\\ =-2\left(x+2\right)\left(x^3-6x^2+12x-8\right)\\ =-2\left(x^4-6x^3+12x^2-8x+2x^3-12x^2+24x-16\right) Multiply\\ =-2x^4+12x^3-24x^2+16x-4x^3+24x^2-48x+32\\ =-2x^4+8x^3-32x+32 Combine like terms\end{array}$

The polynomial function $f$ is of degree $4$ . The graph of $f$ behaves like $y=-2x^4$ for large values of $|x|$ .

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

The $y-intercept$ is $f\left(0\right)=32$ . To find the $x-intercepts$ . we solve $f\left(x\right)=0$ .

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

The $x-intercepts$ are $-2and2$ .

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 $-2and2$ . The zero $-2$ is azero of multiplicity $1$ , so the graph of $f$ crosses the $x-axis$ at $x=-2$ . The zero $2$ is a zero of multiplicity $3$ , so the graph of $f$ crosses 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 $4$ (step1), the graph of the function will have

atmost $4-1=3$ points.

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

The two $x-intercept$ are $-2and2$ .

  $\begin{array}{l}Near-2:f\left(x\right)=-2\left(x+2\right){\left(x-2\right)}^3\hfill \\ \approx -2\left(x+2\right){\left(-2-2\right)}^2\hfill \\ =-2\left(x+2\right)\left(-64\right)\text{ A }tinewithslope128\hfill \\ =128\left(x+2\right)\hfill \\ =128x+256\hfill \\ Near2:f\left(x\right)=-2\left(x42\right){\left(x-2\right)}^2\hfill \\ =-2\left(2+2\right){\left(x-2\right)}^2\hfill \\ =-8{\left(x-2\right)}^2\hfill \end{array}$

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 $-1,0,3$ 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.