# Graph Polynomial Functions

### Learning Objectives

- Draw the graph of a polynomial function using end behavior, turning points, intercepts, and the intermediate value theorem

### How To: Given a polynomial function, sketch the graph.

- Find the intercepts.
- Check for symmetry. If the function is an even function, its graph is symmetrical about the
*y*-axis, that is,*f*(–*x*) =*f*(*x*). If a function is an odd function, its graph is symmetrical about the origin, that is,*f*(–*x*) =*–**f*(*x*). - Use the multiplicities of the zeros to determine the behavior of the polynomial at the
*x*-intercepts. - Determine the end behavior by examining the leading term.
- Use the end behavior and the behavior at the intercepts to sketch a graph.
- Ensure that the number of turning points does not exceed one less than the degree of the polynomial.
- Optionally, use technology to check the graph.

### Example: Sketching the Graph of a Polynomial Function

Sketch a graph of [latex]f\left(x\right)=-2{\left(x+3\right)}^{2}\left(x - 5\right)[/latex].Answer:
This graph has two *x-*intercepts. At *x *= –3, the factor is squared, indicating a multiplicity of 2. The graph will bounce at this *x*-intercept. At *x *= 5, the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.
The *y*-intercept is found by evaluating *f*(0).

[latex]\begin{array}{l}\hfill \\ f\left(0\right)=-2{\left(0+3\right)}^{2}\left(0 - 5\right)\hfill \\ \text{ }=-2\cdot 9\cdot \left(-5\right)\hfill \\ \text{ }=90\hfill \end{array}[/latex]

The*y*-intercept is (0, 90). Additionally, we can see the leading term, if this polynomial were multiplied out, would be [latex]-2{x}^{3}[/latex], so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. To sketch this, we consider that:

- As [latex]x\to -\infty [/latex] the function [latex]f\left(x\right)\to \infty [/latex], so we know the graph starts in the second quadrant and is decreasing toward the
*x*-axis. - Since [latex]f\left(-x\right)=-2{\left(-x+3\right)}^{2}\left(-x - 5\right)[/latex]
is not equal to
*f*(*x*), the graph does not display symmetry. - At [latex]\left(-3,0\right)[/latex], the graph bounces off of the
*x*-axis, so the function must start increasing.At (0, 90), the graph crosses the*y*-axis at the*y*-intercept.

### Try It

Sketch a graph of [latex]f\left(x\right)=\frac{1}{4}x{\left(x - 1\right)}^{4}{\left(x+3\right)}^{3}[/latex]. Check yourself with Desmos when you are done.### Try it

Use Desmos to write an odd degree function with one zero at (-3,0) whose multiplicity is 3 and another zero at (2,0) with multiplicity 2. The end behavior of the graph is: as [latex]x\rightarrow-\infty, f(x) \rightarrow\infty[/latex] and as [latex]x\rightarrow \infty, f(x)\rightarrow -\infty[/latex]## The Intermediate Value Theorem

In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the*x*-axis, we can confirm that there is a zero between them. Consider a polynomial function

*f*whose graph is smooth and continuous. The

**Intermediate Value Theorem**states that for two numbers

*a*and

*b*in the domain of

*f*, if

*a*<

*b*and [latex]f\left(a\right)\ne f\left(b\right)[/latex], then the function

*f*takes on every value between [latex]f\left(a\right)[/latex] and [latex]f\left(b\right)[/latex]. We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function

*f*at [latex]x=a[/latex] lies above the

*x*-axis and another point at [latex]x=b[/latex] lies below the

*x*-axis, there must exist a third point between [latex]x=a[/latex] and [latex]x=b[/latex] where the graph crosses the

*x*-axis. Call this point [latex]\left(c,\text{ }f\left(c\right)\right)[/latex]. This means that we are assured there is a solution

*c*where [latex]f\left(c\right)=0[/latex]. In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the

*x*-axis. The figure below shows that there is a zero between

*a*and

*b*.

### A General Note: Intermediate Value Theorem

Let*f*be a polynomial function. The

**Intermediate Value Theorem**states that if [latex]f\left(a\right)[/latex] and [latex]f\left(b\right)[/latex] have opposite signs, then there exists at least one value

*c*between

*a*and

*b*for which [latex]f\left(c\right)=0[/latex].

### Example: Using the Intermediate Value Theorem

Show that the function [latex]f\left(x\right)={x}^{3}-5{x}^{2}+3x+6[/latex] has at least two real zeros between [latex]x=1[/latex] and [latex]x=4[/latex].Answer: As a start, evaluate [latex]f\left(x\right)[/latex] at the integer values [latex]x=1,2,3,\text{ and }4[/latex].

x |
1 | 2 | 3 | 4 |

f (x) |
5 | 0 | –3 | 2 |

#### Analysis of the Solution

We can also see that there are two real zeros between [latex]x=1[/latex] and [latex]x=4[/latex].### Try It

Show that the function [latex]f\left(x\right)=7{x}^{5}-9{x}^{4}-{x}^{2}[/latex] has at least one real zero between [latex]x=1[/latex] and [latex]x=2[/latex].Answer: Because *f* is a polynomial function and since [latex]f\left(1\right)[/latex] is negative and [latex]f\left(2\right)[/latex] is positive, there is at least one real zero between [latex]x=1[/latex] and [latex]x=2[/latex].

## Licenses & Attributions

### CC licensed content, Original

- Interactive: Write a Polynomial Function.
**Provided by:**Lumen Learning (With Desmos)**License:**CC BY: Attribution. - Revision and Adaptation.
**Provided by:**Lumen Learning**License:**CC BY: Attribution.

### CC licensed content, Shared previously

- College Algebra.
**Provided by:**OpenStax**Authored by:**Abramson, Jay et al..**Located at:**https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites.**License:**CC BY: Attribution.**License terms:**Download for free at http://cnx.org/contents/[email protected].