Chapter 4.5, Problem 94AYU

Mixed Practice In Problems $93-104$, grapheach polynomial function.

$f\left(x\right)=2x^3+x^2+2x+1$

To determine

To graph: The polynomial function $f\left(x\right)=2x^3+x^2+2x+1$

Explanation of Solution

Given Information:

The polynomial function $f\left(x\right)=2x^3+x^2+2x+1$

Graph:

The steps of graphing the polynomial function as follows:

Step 1: First, find the end behavior of the graph of the function $f\left(x\right)=2x^3+x^2+2x+1$

Here, the degree of the polynomial function $f\left(x\right)$ is 3.

The graph of the function $f\left(x\right)=2x^3+x^2+2x+1$ behaves like $y=x^3$ for large values of $\left|x\right|$

Step 2: Find $x$ and $y-$ intercepts of the graph of the function.

For the $y-$ intercept, substitute $x=0$ in the function $f\left(x\right)=2x^3+x^2+2x+1$.

$f\left(0\right)=2{\left(0\right)}^3+{\left(0\right)}^2+2\left(0\right)+1=1$

Thus, the $y-$ intercept of the polynomial function is 1.

For $x-$ intercept, that is, zeros of the function, use the Rational Zeros Theorem

$f\left(x\right)=2x^3+x^2+2x+1=0$

Since there are no variations in the sign of the non-zero coefficients of $f\left(x\right)$, hence, by using the Descartes’ Rule of Signs, there are no positive real zeros.

$f\left(-x\right)=2{\left(-x\right)}^3+{\left(-x\right)}^2+2\left(-x\right)+1$

$\Rightarrow f\left(-x\right)=-2x^3+x^2-2x+1$

There are 3 variations in the signs of the non-zero coefficients of $f\left(-x\right)$, hence, by using the Descartes’ Rule of Signs, there are 3 negative real zeros or 1 negative real zero.

By using the Rational Zero Theorem, all possible rational zeroes are of the form $x=\frac{p}{q}$, where $p$ means all factors of the constant term in the polynomial and $q$ means all factors of the leading term of the polynomial.

In the polynomial $f\left(x\right)=2x^3+x^2+2x+1$, the constant term is 1 and the leading term is 2.

Factors of the constant term $p=1$ are $\pm 1$

Factors of the leading term $q=2$ are $\pm 1,\pm 2$

Therefore, all the possible rational zeros are:

$\frac{p}{q}:$ $\pm \frac{1}{1},\pm \frac{1}{2}$.

Simplifies to

$\frac{p}{q}:\pm 1,\pm \frac{1}{2}$.

As there are no positive real zeros, all the potential rational zeros of the polynomial function $f\left(x\right)=2x^3+x^2+2x+1$ are $x=-1,-\frac{1}{2}$

Now, test $x=-1$ using the synthetic division.

$\begin{array}{l}-1\overline{)2121}\\ \underset{\_}{-21-3}\\ 2-13-2\end{array}$

Here, since the remainder is $-2$, $x=-1$ is not a zero of $f$

Now, try $x=-\frac{1}{2}$ using the synthetic division.

$\begin{array}{l}-\frac{1}{2}\overline{)2121}\\ \underset{\_}{-10-1}\\ 2020\end{array}$

Here, since the remainder is 0, $x=-\frac{1}{2}$ is a zero of $f$ and $\left(x+\frac{1}{2}\right)$ is a factor of $f$

To write the factors of $f$, use the entries in the bottom row of the above synthetic division.

$f\left(x\right)=2x^3+x^2+2x+1=\left(x+\frac{1}{2}\right)\left(2x^2+0x+2\right)$

$\Rightarrow f\left(x\right)=2x^3+x^2+2x+1=\left(x+\frac{1}{2}\right)\left(2x^2+2\right)$

Here, $f\left(x\right)=2x^2+2$ is called a depressed equation.

Any solution to this depressed equation is also a zero of $f$, so now find the zeros of this depressed equation.

The depressed equation $f\left(x\right)=2x^2+2=0$ can be factored as follows:

$2x^2+2=2\left(x^2+1\right)=0$

$\Rightarrow 2\left(x^2+1\right)=0$

$\Rightarrow \left(x^2+1\right)=0$

$\Rightarrow x^2=-1$

This does not give a real solution.

Thus, the real zeros of polynomial function $f\left(x\right)=2x^3+x^2+2x+1$ is $x=-\frac{1}{2}$

Step 3: The maximum number of turning points on the graph of the function $f\left(x\right)=2x^3+x^2+2x+1$ are $3-1=2$, since the polynomial function $f\left(x\right)=2x^3+x^2+2x+1$ has degree 3.

The polynomial function $f\left(x\right)=2x^3+x^2+2x+1$ has degree 3 and one real zero at $x=-\frac{1}{2}$, hence, the graph crosses the x-axis at $x=-\frac{1}{2}$

Step 4: Now, use the information obtained in step 1 to 3 to draw the graph of the function $f\left(x\right)$ as follows:

Now, find additional points on the graph.

For $x=1$, the value of $f\left(x\right)$, at $x=1$, is $f\left(1\right)=2{\left(1\right)}^3+{\left(1\right)}^2+2\left(1\right)+1=6$.

For $x=-1$, the value of $f\left(x\right)$, at $x=-1$, is $f\left(-1\right)=2{\left(-1\right)}^3+{\left(-1\right)}^2+2\left(-1\right)+1=-2$.

For $x=1.5$, the value of $f\left(x\right)$, at $x=1.5$, is $f\left(1.5\right)=2{\left(1.5\right)}^3+{\left(1.5\right)}^2+2\left(1.5\right)+1=13$.

For $x=-1.5$, the value of $f\left(x\right)$, at $x=-1.5$, is

$f\left(-1.5\right)=2{\left(-1.5\right)}^3+{\left(-1.5\right)}^2+2\left(-1.5\right)+1=-6.5$.

Now, plot all these coordinates $\left(1,6\right),\left(1.5,13\right),\left(-1,-2\right),\left(-1.5,-6.5\right)$ on the graph and join them.

So, the graph of the function is obtained as follows:

Interpretation:

The graph represents the polynomial function $f\left(x\right)=2x^3+x^2+2x+1$