Chapter 4, Problem 8RE

To determine

To analyze : The polynomial function $f(x)=x(x+2)(x+4)$ .

Answer to Problem 8RE

The polynomial function $f(x)=x^3+6x^2+8x$ .

The y -intercept is zero.

The x -intercept is zero, -2 and -4.

Explanation of Solution

Given information:

  $f(x)=x(x+2)(x+4)$ .

Formula used:

The graph of a polynomial function $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_{1}x+a_0;a_n\ne 0$

Degree of a polynomial function $f:n$

Maximum number of turning points : $n-1$

At a zero even multiplicity: The graph of f touches the x -axis.

At a zero odd multiplicity : The graph of f crosses the x -axis.

For large $\left|x\right|$ , the graph of f behaves like the graph of $y=a_n{x}^n$ .

Calculation:

Consider ,

  $f(x)=x(x+2)(x+4)$

Step 1.

Re-writing the polynomial we have,

  $=f(x)=x\left[(x+2)(x+4)\right]$

  $\begin{array}{l}=x\left[x(x+4)+2(x+4)\right]\\ =x\left[x^2+4x+2x+8\right]\\ =x\left[x^2+6x+8\right]\\ =x^3+6x^2+8x\end{array}$

The polynomial function f is of degree 3. The graph of f behaves like $x^3$ .

Step 2.

The y −intercept is zero.

  $\begin{array}{l}=f(x)=x^3+6x^2+8x\\ =f(0)=0\end{array}$

The x -intercept :-

  $\begin{array}{l}=f(x)=0\\ =x^3+6x^2+8x=0\\ =x(x+2)(x+4)=0\end{array}$

  $x=0$ or $\begin{array}{l}x+2=0\\ x=-2\end{array}$ or $\begin{array}{l}x+4=0\\ x=-4\end{array}$

Step3.

The zeros of f are 0,-2,-4. The zero 0 is a zero of multiplicity 2, so the graph of f touches the x - axis at $x=0$ . The zero -2 is a zero of multiplicity 1, so the graph of f crosses the x -axis at $x=-2$ . The zero -4 is a zero of multiplicity 1, so the graph of f crosses the x -axis at $x=-4$ .

Step4.

Since the degree of polynomial function is 3. Therefore, the graph of the function will have at most $=3-1=2$ turning points.

Step5.

The x -intercept are 0,-2,-4.

The behavior of the graph of f near each x -intercept are as follows:-

Near $x=0$ ;

  $f(x)=x(x+2)(x+4)$

  $\begin{array}{l}=x(0+2)(0+4)\\ =x(2)(4)\\ =8x\end{array}$

A line with slope 8.

Near $x=-2$

  $f(x)=x(x+2)(x+4)$

  $\begin{array}{l}=(-2)(x+2)(-2+4)\\ =(-2)(x+2)(2)\\ =-4(x+2)\\ =-4x-8\end{array}$

A line with slope -4.

Near $x=-4$

  $f(x)=x(x+2)(x+4)$

  $\begin{array}{l}=(-4)(-4+2)(x+4)\\ =(-4)(-2)(x+4)\\ =8(x+4)\\ =8x+32\end{array}$

The line with slope 8.

Step6.

Using Step1- Step5 graph is drawn showing the x - intercept and y -intercept. With the help of graph the behavior of each x −intercept can be observed.

  

Figure1.

Hence , the polynomial function f has been analyzed.