Chapter 4, Problem 11RE

To determine

To analyze : The polynomial function $f(x)={(x-1)}^2(x+3)(x+1)$ .

Answer to Problem 11RE

The polynomial function $f(x)=x^4+2x^3-4x^2-2x+3$ .

The y -intercept is 3.

The x -intercept is zero 1, -1 and -3.

Explanation of Solution

Given information:

  $f(x)={(x-1)}^2(x+3)(x+1)$ .

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 ,

  $f(x)={(x-1)}^2(x+3)(x+1)$

Step 1.

Re-writing the polynomial is,

  $=f(x)={(x-1)}^2(x+3)(x+1)$

  $\begin{array}{l}=x^2\left[x^2+4x+3\right]-2x\left[x^2+4x+3\right]+1\left[x^2+4x+3\right]\\ =x^4+4x^3+3x^2-2x^3-8x^2-6x+x^2+4x+3\\ =x^4+2x^3-4x^2-2x+3\end{array}$

The polynomial function f is of degree 4. The graph of f behaves like $x^4$ .

Step 2.

The y −intercept :-

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

The x -intercept :-

  $\begin{array}{l}=f(x)=0\\ =x^4+2x^3-4x^2-2x+3=0\\ ={(}^x(x+3)(x+1)=0\end{array}$

  $\begin{array}{l}{(}^x=0\\ (x-1)(x-1)=0\\ x=1,1\end{array}$ or $\begin{array}{l}x+1=0\\ x=-1\end{array}$ or $\begin{array}{l}x+3=0\\ x=-3\end{array}$

Step3.

The zeros of f are 1, -1 and -3. The zero 1 is a zero of multiplicity 3, so the graph of f crosses the x - axis at $x=1$ . The zero -1 is a zero of multiplicity 1, so the graph of f crosses the x -axis at $x=-1$ . The zero -3 is a zero of multiplicity 1, so the graph of f crosses the x -axis at $x=-3$ .

Step4.

Since the degree of polynomial function is 4. Therefore, the graph of the function will have at most $=4-1=3$ turning points.

Step5.

The x -intercept are 1, -1 and -3.

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

Near $x=1$ ;

  $f(x)=x^4+2x^3-4x^2-2x+3$

  $\begin{array}{l}={(}^x(x+3)(x+1)\\ ={(}^x(1+3)(1+1)\\ ={(}^x(4)(2)\\ =8{(}^x\end{array}$

A parabola that opens up .

Near $x=-1$

  $f(x)=x^4+2x^3-4x^2-2x+3$

  $\begin{array}{l}={(}^x(x+3)(x+1)\\ ={(}^{-}(-1+3)(x+1)\\ ={(}^{-}(2)(x+1)\\ =(4)(2)(x+1)\\ =8(x+1)\\ =8x+8\end{array}$

The line with slope 8.

Near $x=-3$

  $f(x)=x^4+2x^3-4x^2-2x+3$

  $\begin{array}{l}={(}^x(x+3)(x+1)\\ ={(}^{-}(x+3)(-3+1)\\ ={(}^{-}(x+3)(-2)\\ =(16)(-2)(x+3)\\ =-32(x+3)\\ =-32x-96\end{array}$

The line with slope -32.

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.