Chapter 4.1, Problem 101AYU

To determine

To analyze: the given polynomial function

Answer to Problem 101AYU

Figure 1(a)

  

Figure 1(b)

  

Explanation of Solution

Given:

  $f\left(x\right)=-x^5-x^4+x^3+x^2$

Calculation:

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

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

  $\begin{array}{c}f\left(x\right)=x^2\left(x^3+x^2+x+1\right)\\ =x^2\left(-x^2\left(x+1\right)+1\left(x+1\right)\right)\\ =\left(x-0\right)\left(x-0\right)\left(1-x^2\right)\left(x+1\right)\\ =\left(x-0\right)\left(x-0\right)\left(1-x\right)\left(1+x\right)\left(x+1\right)\end{array}$

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

We expand the polynomial to write it in the form

  $\begin{array}{c}f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_{1}x+a_0\\ f\left(x\right)=\left(x-0\right)\left(x-0\right)\left(1-x\right)\left(1+x\right)\left(x+1\right)\\ =x^2\left(1-x\right){\left(x+1\right)}^2\\ =\left(x^2-x^3\right)\left(x^2+2x+1\right)\\ =-x^5-2x^4-x^3+x^4+2x^3+x^2\\ =-x^5-x^4+x^3+x^2\end{array}$

The polynomial function f is of degree 5. The graph of f behaves like $y=-x^5$ for large positive values of x and like $y=x^5$ for the negative values less than $-1$ .

Step2: we 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 $f\left(x\right)=0$

  $\begin{array}{c}-x^2\left(1-x\right){\left(1+x\right)}^2=0\\ x=-1,-1,0,0,1\end{array}$

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.

The zeroes or roots of fare $-1,0,1$ . The root $-1$ has multiplicity 2.

So, the graph f intersects x-axis at $x=-1$ two times.

The root $x=0$ has multiplicity 2. So, the curve f intersects x-axis at $x=0$ two times

The root $x=1$ has the multiplicity 1. So, the curve f intersects x-axis at $x=1$ one time.

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

Near the x-intercept 0, the behavior of the given polynomial is

  $\begin{array}{c}f\left(x\right)=-x^2\left(1-x^2\right)\left(1+x\right)\\ \approx -x^2\left(1-0\right)\left(1+0\right)\\ \approx -x^2\end{array}$

This is a parabola.

So, the function behaves like a downward parabola near $x=0$

Near the x-intercept $-1$ , the behavior of the polynomial is

  $\begin{array}{c}f\left(x\right)=-x^2\left(1-x\right){\left(1+x\right)}^2\\ \approx -{\left(-1\right)}^2\left(1-\left(-1\right)\right){\left(1+x\right)}^2\\ \approx -2{\left(1+x\right)}^2\end{array}$

This is a parabola

So, the function looks like a parabola near $x=-1$ .

Near the x-intercept 1, the behavior of the polynomial is

  $\begin{array}{c}f\left(x\right)=-x^2\left(1-x^2\right)\left(1+x\right)\\ \approx -{\left(1\right)}^2\left(1-x\right){\left(1+1\right)}^2\\ \approx -4\left(1-x\right)\\ \approx 4x-4\end{array}$

This is a straight line with slope 4.

So, at the x-intercept 1, the function behaves like a straight line.

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 $-2$ 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.