Chapter 3.2, Problem 33E

To determine

To graph: the polynomial function

Explanation of Solution

Given information: the function is $P\left(x\right)=x^3+x^2-x-1$ .

Graph:

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

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

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

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

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

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

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

Leading terms: $x^2\cdot x=x^3$

Now, $P$ is of odd 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.

$x$	$x^3+x^2-x-1$
−2	−3
2	9

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

  

Interpretation:

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