Chapter 11.8, Problem 27PPS

(a)

To determine

To describe: the shape of the graph.

Answer to Problem 27PPS

Therefore, the shape of the function is a parabola.

Explanation of Solution

Given:

  $f\left(x\right)=\frac{x^3+6x^2+12x}{x}$

Calculation:

The function is as follows.

  $f\left(x\right)=\frac{x^3+6x^2+12x}{x}$

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

The display is shown below.

  $\begin{array}{l}Plot1 plot2 plot3\\ \backslash Y_{1}B{\left(X\wedge 3+6X^2+12X\right)}^{\prime }\\ \left(X\right)\\ \backslash Y_2=\\ \backslash Y_3=\\ \backslash Y_4=\\ \backslash Y_5=\\ \backslash Y_6=\end{array}$

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 27PPS

Therefore, the simplified factor is $x^3+6x^2+12$ .

Explanation of Solution

Given:

  $f\left(x\right)=\frac{x^3+6x^2+12x}{x}$

The objective is to use factoring and simplify the function.

The function is factorized below.

  $\begin{array}{l}f\left(x\right)=\frac{x^3+6x^2+12x}{x}\\ =\frac{x\left(x^3+6x^2+12x\right)}{x}\\ =x^2+6x+12\end{array}$

Therefore, the simplified factor is $x^3+6x^2+12$

Conclusion:

Therefore, the simplified factor is $x^3+6x^2+12$ .

(c)

To determine

To find: the zeros of the function.

Answer to Problem 27PPS

Therefore, the zeros of the function no real values.

Explanation of Solution

Given:

  $f\left(x\right)=\frac{x^3+6x^2+12x}{x}$

The objective is to find the zeros of the function.

Set $f=0$ and solve for x

  $x^3+6x^2+12x=0$

Use quadratic formula

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-6\pm \sqrt{6^2-4(1)(12)}}{2(1)}\\ =\frac{-6\pm \sqrt{-12}}{2}\end{array}$

The given function has no real solutions

Therefore, there are no zeros

Conclusion:

Therefore, the zeros of the function no real values.