Chapter 4.1, Problem 84AYU

  

  

To determine

To find:

Explanation of Solution

Given: To analyze the given polynomial

Calculation:

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

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

Expand the polynomial to write it in the form

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

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

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

  $=\left(x^2-4x+4\right)\left(x^2+6x+8\right)$

  $=\left(x^4+6x^3+8x^2-4x^3-24x^2-32x+4x^2+24x+32\right)$

  $=x^4+12x^3-12x^2-8x+32$ 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 $f\left(0\right)=32$ . To find the $x$ intercepts, we solve

  

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

So

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

  $x-2=0$ or $x=-2$ or $x=-4$

  $x=2$

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

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 $-4,-2$ and $2$ . 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 $1$ , so the graph $f$ touches the $x-axis$ at $x=-2$ . The zero $2$ is a zero of multiplicity $2$ , 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 $f$ near each $x$-intercept

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

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

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

  $=36\left(-2\right)\left(x+4\right)$

  $=-72\left(x+4\right)$ A line with slope -72

  $=-72x-288$

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

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

A line with slope 32 $=16\left(2\right)\left(x+2\right)$

  $=32\left(x+2\right)$

  $=32x+64$

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

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

  $=24{\left(x-2\right)}^2$

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 $-3,-1,0$ to help establish the scale on the $y-axis$ . The graph of $f$ is given in Figure 1(b).

Figure 1

  

  

Conclusion: