# 5.3 Graphs of polynomial functions  (Page 6/13)

 Page 6 / 13

## Using the intermediate value theorem

Show that the function $\text{\hspace{0.17em}}f\left(x\right)={x}^{3}-5{x}^{2}+3x+6\text{\hspace{0.17em}}$ has at least two real zeros between $\text{\hspace{0.17em}}x=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=4.$

As a start, evaluate $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ at the integer values $\text{\hspace{0.17em}}x=1,2,3,$ and $4.\text{\hspace{0.17em}}$ See [link] .

 $x$ 1 2 3 4 $f\left(x\right)$ 5 0 –3 2

We see that one zero occurs at $\text{\hspace{0.17em}}x=2.\text{\hspace{0.17em}}$ Also, since $\text{\hspace{0.17em}}f\left(3\right)\text{\hspace{0.17em}}$ is negative and $\text{\hspace{0.17em}}f\left(4\right)\text{\hspace{0.17em}}$ is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4.

We have shown that there are at least two real zeros between $\text{\hspace{0.17em}}x=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=4.$

Show that the function $\text{\hspace{0.17em}}f\left(x\right)=7{x}^{5}-9{x}^{4}-{x}^{2}\text{\hspace{0.17em}}$ has at least one real zero between $\text{\hspace{0.17em}}x=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=2.$

Because $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is a polynomial function and since $\text{\hspace{0.17em}}f\left(1\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ is negative and $\text{\hspace{0.17em}}f\left(2\right)\text{\hspace{0.17em}}$ is positive, there is at least one real zero between $\text{\hspace{0.17em}}x=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=2.\text{\hspace{0.17em}}$

## Writing formulas for polynomial functions

Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function    written in factored form will have an x -intercept where each factor is equal to zero, we can form a function that will pass through a set of x -intercepts by introducing a corresponding set of factors.

## Factored form of polynomials

If a polynomial of lowest degree $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ has horizontal intercepts at $\text{\hspace{0.17em}}x={x}_{1},{x}_{2},\dots ,{x}_{n},\text{\hspace{0.17em}}$ then the polynomial can be written in the factored form: $\text{\hspace{0.17em}}f\left(x\right)=a{\left(x-{x}_{1}\right)}^{{p}_{1}}{\left(x-{x}_{2}\right)}^{{p}_{2}}\cdots {\left(x-{x}_{n}\right)}^{{p}_{n}}\text{\hspace{0.17em}}$ where the powers $\text{\hspace{0.17em}}{p}_{i}\text{\hspace{0.17em}}$ on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factor $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ can be determined given a value of the function other than the x -intercept.

Given a graph of a polynomial function, write a formula for the function.

1. Identify the x -intercepts of the graph to find the factors of the polynomial.
2. Examine the behavior of the graph at the x -intercepts to determine the multiplicity of each factor.
3. Find the polynomial of least degree containing all the factors found in the previous step.
4. Use any other point on the graph (the y -intercept may be easiest) to determine the stretch factor.

## Writing a formula for a polynomial function from the graph

Write a formula for the polynomial function shown in [link] .

This graph has three x -intercepts: $\text{\hspace{0.17em}}x=-3,2,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}5.\text{\hspace{0.17em}}$ The y -intercept is located at $\text{\hspace{0.17em}}\left(0,2\right).\text{\hspace{0.17em}}$ At $\text{\hspace{0.17em}}x=-3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=5,\text{\hspace{0.17em}}$ the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At $\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us

$f\left(x\right)=a\left(x+3\right){\left(x-2\right)}^{2}\left(x-5\right)$

To determine the stretch factor, we utilize another point on the graph. We will use the $\text{\hspace{0.17em}}y\text{-}$ intercept $\text{\hspace{0.17em}}\left(0,–2\right),\text{\hspace{0.17em}}$ to solve for $\text{\hspace{0.17em}}a.$

$\begin{array}{ccc}\hfill f\left(0\right)& =& a\left(0+3\right){\left(0-2\right)}^{2}\left(0-5\right)\hfill \\ \hfill -2& =& a\left(0+3\right){\left(0-2\right)}^{2}\left(0-5\right)\hfill \\ \hfill -2& =& -60a\hfill \\ \hfill a& =& \frac{1}{30}\hfill \end{array}$

The graphed polynomial appears to represent the function $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{30}\left(x+3\right){\left(x-2\right)}^{2}\left(x-5\right).$

Given the graph shown in [link] , write a formula for the function shown.

$f\left(x\right)=-\frac{1}{8}{\left(x-2\right)}^{3}{\left(x+1\right)}^{2}\left(x-4\right)$

## Using local and global extrema

With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph.

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