Chapter 3.2, Problem 31E

To determine

To graph: the polynomial function

Explanation of Solution

Given information: the function is $P\left(x\right)=-2x^3-x^2+x$ .

Graph:

  $\begin{array}{l}P\left(x\right)=x^4-3x^3+2x^2\\ p\left(x\right)=x^2\left(x^2-3x+2\right)\\ p\left(x\right)=x^2\left(x^2+2x-x+2\right)\\ p\left(x\right)=x^2\left[x\left(x-1\right)-2\left(x-1\right)\right]\\ p\left(x\right)=x^2\left(x-1\right)\left(x-2\right)\end{array}$

The $x$ and $y$ intercepts of the graph of $p\left(x\right)=x^2\left(x-1\right)\left(x-2\right)$ function is :

For $x$ intercept, solve equation $P\left(x\right)=0$ for $x$

  $\begin{array}{l}P\left(x\right)=0\\ x^2\left(x-1\right)\left(x-2\right)=0\\ x^2=0\text{ or }\left(x-1\right)=0\text{ or }\left(x-2\right)=0\\ x=0\text{ or }x=1\text{ or }x=2\end{array}$

Therefore if $x=0$ and estimate $P\left(0\right)$ then the $y$ intercept is

  $\begin{array}{l}P(0)=-0^2\left(0-1\right)\left(0-2\right)\\ P\left(0\right)=0\left(-1\right)\left(-2\right)\\ P(0)=0\end{array}$

Now multiply the leading terms of each factor of polynomials to get leading term of polynomial $p\left(x\right)=x^2\left(x-1\right)\left(x-2\right)$

Leading terms: $x^2\cdot x\cdot x=x^4$

Now, $P$ is of even degree and it leading coefficient is positiveso it has the following end behaviours

  $y\to +\infty \text{ as }x\to -\infty \text{ and }y\to +\infty \text{ as }x\to +\infty$

Create a table of value for several values of the variable $x$ to find more point on this curve.

$x$	$-2x^3-x^2+x$
−1	6
−1/2	0.187
3/2	−0.562
5/2	4.687

Finally draw a graph of polynomial function plot the point $\left(0,0\right),(1,0),\left(2,0\right)$ and connect them with smooth curve.

  

Interpretation:

From the above, it can be observed that the graph of $p\left(x\right)=x^2\left(x-1\right)\left(x-2\right)$ with coordinates $\left(0,0\right),(1,0),\left(2,0\right)$ .