Chapter 2, Problem 38RE

To determine

To apply the leading coefficient test on a function.

Answer to Problem 38RE

The graph rises to the left and falls to the right.

Explanation of Solution

Given information:

The given function is:

  $\text{f}\left(\text{x}\right)=18+27\text{x}-2{\text{x}}^2-3{\text{x}}^3$

The degree of the polynomial is odd and the leading coefficient is negative, so the graph rises to the left and falls to the right.

b.

To determine

To find the zeroes of the polynomial.

Answer to Problem 38RE

The zeroes of the polynomial are $-1\text{ and }-3$

Explanation of Solution

Given information:

The given function is:

  $\text{f}\left(\text{x}\right)=18+27\text{x}-2{\text{x}}^2-3{\text{x}}^3\text{ or f}\left(\text{x}\right)=-3{\text{x}}^3-2{\text{x}}^2+27\text{x}+18$

Calculation :

The leading coefficient is $-3$ and the constant term is $18$

The possible zeroes of the function are $\frac{\text{factors of 18}}{\text{factors of }-\text{3}}=\frac{\pm 1,\pm 2,\pm 3,\pm 6,\pm 9\pm 18}{\pm 1,\pm 3}$

By synthetic division it is determined that $\text{3 and }-3$ is a zero of the function.

  $\begin{array}{l}3|-3-2\text{ 27 }18\\ \text{ | }-9-\text{3}3-\text{18 }\\ -3-11-60\\ \\ -3|-3-2\text{ 27 }18\\ \text{ | }9-21-\text{18 }\\ -3\text{ 7 }60\end{array}$

Hence, the zeroes of the polynomial are $\text{3 and }-3$

c.

To determine

To plot sufficient solution points of the polynomial.

Explanation of Solution

Given information:

The given function is:

  $\text{f}\left(\text{x}\right)=18+27\text{x}-2{\text{x}}^2-3{\text{x}}^3$

Graph:

The points of the polynomial are:

  

d.

To determine

To draw a continuous curve through the points.

Explanation of Solution

Given information:

The given function is:

  $\text{f}\left(\text{x}\right)=18+27\text{x}-2{\text{x}}^2-3{\text{x}}^3$

The zeroes of the polynomial are $\text{3 and }-3$

Graph:

The graph of the polynomial is: