Chapter 4.1, Problem 83AYU

To determine

To analyze: the given polynomial

Answer to Problem 83AYU

  

  

Explanation of Solution

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

Calculation:

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

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

Expand the polynomial to write it in the form

  $f\left(x\right)=\text{}\text{}{a}_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$

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

  $=\left(x^2+4x+4\right)\left(x^2-8x+16\right)$ Multiply

  $=\left(x^4-8x^3+16x^2+4x^3-32x^2+64x+4x^2-32x+64\right)$

  $=x^4-4x^3+12x^2+32x+64$ Combine like terms

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

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

The $y$ -intercepts is $f\left(0\right)=64$ . To find the $x$ intercepts, we solve

  

  ${\left(x+2\right)}^2{\left(x-4\right)}^2=0$

So

  ${\left(x+2\right)}^2=0$ or ${\left(x-4\right)}^2=0$

  $x+2=0$ or $x-4=0$

  $x=-2$

or $x=4$

The $x$ -intercepts are $-2$ and $4$ .

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

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

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 haveatmost $4-1=3$ points.

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

The three $x$ -intercept are $-2$ and $4$ .

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

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

  $={\left(x+2\right)}^2{\left(36\right)}^2$ A parabola that open up

  $=36{\left(x+2\right)}^2$

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

  $\approx {\left(4+2\right)}^2{\left(x-4\right)}^2$ A parabola that opens down

  $=36{\left(x-4\right)}^2$

Step 6: Put all the information form Steps 1 through 5 together to obtain the graph of $f$

Figure 1(a) illustrates the information obtained from Steps 1 through 5. We evaluate $f$ at $-3,0,1$ , to help establish the scale on the $y-axis$ . The graph of $f$ is given in Figure 1(b).

Figure 1

  

  

Conclusion:

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