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].