Question

Find the bounds to the zeros of each polynomial function. Use the bounds to obtain a complete graph of f.

$f\left(x\right)=2x^3-x^2-11x-6$

Short answer

Every zero of the polynomial function will lie between $-\frac{13}{2}$and $\frac{13}{2}$.

The graph of the function is

Step-by-step solution

Step 1. Given Information

We are given a polynomial function $f\left(x\right)=2x^3-x^2-11x-6$.

We need to find the bounds to the zeros of the function and using it we need to obtain its graph.

Step 2. Concept used

Let f denote a polynomial function whose leading coefficient is $1$.

localid="1646152790108" $f\left(x\right)=x^n+a_{n-1}{x}^{n-1}+...+a_{1}x+a_0$

A bound $M$ on the real zeros of f is the smaller of the two numbers

localid="1646152794514" $Max\left\{1,\left|a_0\right|+\left|a_1\right|+...+\left|a_{n-1}\right|\right\},1+Max\left\{\left|a_0\right|,\left|a_1\right|,...,\left|a_{n-1}\right|\right\}$

where Max $\left\{\right\}$means “choose the largest entry in $\left\{\right\}$.”

Step 3. Make the leading coefficient 1

The leading coefficient of the given polynomial is $2$. So divide the function side by $2$

$$\begin{gathered} f\left(x\right)=2x^3-x^2-11x-6 \\ f\left(x\right)=x^3-\frac{1}{2}{x}^2-\frac{11}{2}x-3 \end{gathered}$$

On comparing with standard form we get

$$\begin{gathered} a_2=-\frac{1}{2} \\ {a}_1=-\frac{11}{2} \\ {a}_0=-3 \end{gathered}$$

Step 4. Find the two numbers

Using the formula the two numbers are found as

$\begin{array}{rcl}Max\left\{1,\left|a_0\right|+\left|a_1\right|+\left|a_2\right|\right\}& =& Max\left\{1,\left|-3\right|+\left|-\frac{11}{2}\right|+\left|-\frac{1}{2}\right|\right\}\\ & =& Max\left\{1,3+\frac{11}{2}+\frac{1}{2}\right\}\\ & =& Max\left\{1,3+6\right\}\\ & =& Max\left\{1,9\right\}\\ & =& 9\end{array}$

and

$\begin{array}{rcl}1+Max\left\{\left|a_0\right|,\left|a_1\right|,...,\left|a_{n-1}\right|\right\}& =& 1+Max\left\{\left|-3\right|,\left|-\frac{11}{2}\right|,\left|-\frac{1}{2}\right|\right\}\\ & =& 1+Max\left\{3,\frac{11}{2},\frac{1}{2}\right\}\\ & =& 1+\frac{11}{2}\\ & =& \frac{13}{2}\end{array}$

Step 5. Find the bounds

Among the two numbers $9$and $\frac{13}{2}$, $\frac{13}{2}$is the smallest.

Therefore, $\frac{13}{2}$is the bound.

Every real zero of f lies between $-\frac{13}{2}$and $\frac{13}{2}$.

Step 6. Graph the function

Using the bound the graph of the function is given as