Chapter 11.8, Problem 26PPS

(a)

To determine

To describe: the shape of the graph.

Answer to Problem 26PPS

The shape of the function is a parabola.

Explanation of Solution

Given:

  $f\left(x\right)=\frac{x^3+x^2-2x}{x+2}$

Calculation:

The function is as follows.

  $f\left(x\right)=\frac{x^3+x^2-2x}{x+2}$

In the graphing calculator, press $Y=$ and enter the function.

The display is shown below.

  $\begin{array}{l}Plot1 plot2 plot3\\ \backslash Y_{1}B{\left(X\wedge 3+X^2-2X\right)}^{\prime }\\ \left(X+2\right)\\ \backslash Y_2=\\ \backslash Y_3=\\ \backslash Y_4=\\ \backslash Y_5=\\ \backslash Y_6=\end{array}$

Press GRAPH

The display is shown below.

  

The shape of the function is a parabola.

Conclusion:

Therefore, the shape of the function is a parabola.

(b)

To determine

To simplify: the given function.

Answer to Problem 26PPS

Therefore, the simplified factor is $x\left(x-1\right)$ .

Explanation of Solution

Given:

  $f\left(x\right)=\frac{x^3+x^2-2x}{x+2}$

The objective is to use factoring and simplify the function.

The function is factorized below.

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

Therefore, the simplified factor is

  $x\left(x-1\right)$ .

Conclusion:

Therefore, the simplified factor is $x\left(x-1\right)$ .

(c)

To determine

To find: the zeros of the function.

Answer to Problem 26PPS

Therefore, the zeros of the function is $0,1$ .

Explanation of Solution

Given:

  $f\left(x\right)=\frac{x^3+x^2-2x}{x+2}$

The objective is to find the zeros of the function.

Factor the given equation as

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

Set $f=0$ and solve for x

The zeros of the function is calculated below.

  $\begin{array}{l}x\left(x-1\right)=0\\ x=0,1\end{array}$

Therefore, the zeros of the function is $0,1$ .

Conclusion:

Therefore, the zeros of the function is $0,1$ .