Chapter 4.1, Problem 98AYU

To determine

To analyze: the givenpolynomial function

Answer to Problem 98AYU

Figure 1(a):

  

Figure 1(b):

  

Explanation of Solution

Given:

  $f\left(x\right)=x^3+2x^2-8x$

Calculation:

Let us consider the function $f\left(x\right)=x^3+2x^2-8x$

This polynomial function $f\left(x\right)$ can be factored as

  $\begin{array}{l}f\left(x\right)=x\left(x^2+2x-8\right)\hfill \\ =x\left(x^2+4x-2x-8\right)\hfill \\ \begin{array}{l}=x\left\{x\left(x+4\right)-2\left(x+4\right)\right\}\\ =x\left(x-2\right)\left(x+4\right)\end{array}\hfill \end{array}$

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

Expand the polynomial to write it in the form

  $\begin{array}{l}\begin{array}{l}f\left(x\right)=x\left(x-2\right)\left(x+4\right)\\ =x\left\{x\left(x+4\right)-2\left(x+4\right)\right\}\end{array}\hfill \\ =x\left(x^2+4x-2x-8\right)\hfill \\ \begin{array}{l}=x\left(x^2+2x-8\right)\\ =x^3+2x^2-8x\end{array}\hfill \end{array}$

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

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

The y -intercept is $f\left(0\right)=0$ .

To find the x -intercepts, solve $f\left(x\right)=0$

  $x\left(x+4\right)\left(x-2\right)=0$ if and only if any of the factors is zero.

So, $x=-4,0,2$ are the x -intercepts of the given polynomial.

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 or roots of fare -4, 0,2.

The root-4is a zero of multiplicity 1, so the graph off crosses the x -axis at $x=-4$ .

The root 0 is a zero of multiplicity1. So the graph of f crosses thex -axis at $x=0$ .

The root 2 is a zero of multiplicity 1. So the graph of f crosses the x -axis at $x=2$ .

Step 4: Determine the maximum number of turning points on the graph of the function.

While the polynomial function is of degree 3 (step1), the graph of the function will have at most $3-1=2$ points.

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

Thex -intercepts are - 4,0, and 2.

So, the behavior of the function near -4 is

  $\begin{array}{l}f\left(x\right)=x\left(x+4\right)\left(x-2\right)\\ \approx \left(-4\right)\left(x+4\right)\left(-4-2\right)\\ \approx 24\left(x+4\right)\\ \approx 24x+96\end{array}$

This is a straight line with a positive slope 24.

Near $x=0$ :

  $\begin{array}{l}f\left(x\right)=x\left(x+4\right)\left(x-2\right)\\ \approx x\left(0+4\right)\left(0-2\right)\\ \approx -8x\end{array}$

This is a straight line with a negative slope -8.

Near $x=2$ :

  $\begin{array}{l}f\left(x\right)=x\left(x+4\right)\left(x-2\right)\\ \approx 2\left(2+4\right)\left(x-2\right)\\ \approx 12\left(x-2\right)\\ \approx 12x-24\end{array}$

This is a straight line with a positive slope 16.

Step 6: Put all the information from Steps 1 through 5 together to obtain the graph of f Figure 1(a) illustrates the information obtained from Steps 1 through 5. Evaluate f at -2.5, l to help establish the scale on the y -axis. The graph of f is given in Figure 1(b).

Figure 1(a):

  

Figure 1(b):

  

Conclusion:

Therefore, the polynomial function $f\left(x\right)=x^3+2x^2-8x$ is analysed with graph.