Chapter 4.1, Problem 102AYU

To determine

To analyze: the given polynomial function

Answer to Problem 102AYU

Figure 1(a)

  

Figure 1(b)

  

Explanation of Solution

Given:

  $f\left(x\right)=-x^5+5x^4+4x^3-20x^2$

Calculation:

Let us consider $f\left(x\right)=-x^5+5x^4+4x^3-20x^2$

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

  $\begin{array}{c}f\left(x\right)=-x^2\left(x3-5x^2-4x+20\right)\\ =-x^2\left(x^2\left(x-5\right)-4\left(x-5\right)\right)\\ =-x^2\left(x^2-4\right)\left(x-5\right)\\ =-x^2\left(x-2\right)\left(x+2\right)\left(x-5\right)\end{array}$

Step1: we determine the end behavior of the graph of the function

We expand the polynomial to write it in the form

  $\begin{array}{c}f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_{1}x+a_0\\ f\left(x\right)=-x^2\left(x-2\right)\left(x+2\right)\left(x-5\right)\\ =-x^2\left(x^2-4\right)\left(x-5\right)\\ =-x^2\left(x^2\left(x-5\right)-4\left(x-5\right)\right)\\ =-x^2\left(x^3-5x^2-4x+20\right)\\ =-x^5+5x^4+4x^3-20x^2\end{array}$

The polynomial function f is of degree 5. The graph of f behaves like $y=-x^5$ 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, we solve $f\left(x\right)=0$

  $-x^2\left(x-2\right)\left(x+2\right)\left(x-5\right)=0$

  $x=0,0,-2,2,5$ are the x-intercepts.

Step3: we 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.

We see that the zeroes or roots of f is 0. So, at $x=0$ , the function touches x-axis.

The root $x=-2$ has multiplicity 1. So, f crosses x-axis at $x=-1$ .

The root $x=2$ has multiplicity 1. So, f crosses x-axis at $x=2$ .

The root $x=5$ has multiplicity 1. So, f crosses x-axis at $x=5$ .

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

While the polynomial function is of degree 5 (step1), the graph of the function will have at most $5-1=4$ turning points.

Step5: we determine the behavior of the graph of f near each x-intercept

The x-intercepts are, $-2,0,2$ and 5.

Near $x=-2$ ,

  $\begin{array}{c}f\left(x\right)=\left(x-2\right)\left(x+2\right)\left(x-5\right)\\ \approx {\left(-2\right)}^2\left(-2-2\right)\left(x+2\right)\left(-2-5\right)\\ \approx -112\left(x+2\right)\\ \approx -112x-224\end{array}$

This is a straight line with slope $-112$ .

Near $x=0$

  $\begin{array}{c}f\left(x\right)=-x^2\left(x-2\right)\left(x+2\right)\left(x-5\right)\\ \approx -x^2\left(0-2\right)\left(0+2\right)\left(0-5\right)\\ \approx -20x^2\end{array}$

This is a parabola opens down wards.

Near $x=2$

  $\begin{array}{c}f\left(x\right)=-x^2\left(x-2\right)\left(x+2\right)\left(x-5\right)\\ \approx -{\left(2\right)}^2\left(x-2\right)\left(2+2\right)\left(2-5\right)\\ \approx 48\left(x-2\right)\\ \approx 48x-96\end{array}$

This is a straight line with slope 48.

Near $x=5$ ,

  $\begin{array}{c}f\left(x\right)=-x^2\left(x-2\right)\left(x+2\right)\left(x-5\right)\\ \approx -{\left(5\right)}^2\left(5-2\right)\left(5+2\right)\left(x-5\right)\\ \approx -525\left(x-5\right)\\ \approx -525x+2625\end{array}$

This is a straight line with slope $-525$ .

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. We evaluate f at 4 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 given polynomial function is analyzed.