Graph polynomial functions

We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions.

Example 1: 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].

Show Solution

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{align} f\left(0\right)&=-2{\left(0+3\right)}^{2}\left(0 - 5\right) \\ &=-2\cdot 9\cdot \left(-5\right) \\ &=90 \end{align}[/latex]

The y-intercept is (0, 90).

Figure 1

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.
• At (0, 90), the graph crosses the y-axis at the y-intercept.
• At (-3,0), the behavior is [latex]-2{\left(x+3\right)}^{2}\left(-3 - 5\right)=16(x+3)^{2}[/latex].  At (-3,0) the graph will resemble a parabola opening up.
• At (5,0), the behavior is [latex]-2{\left(5+3\right)}^{2}\left(x - 5\right)=-128(x-5)[/latex].  At (5,0) the graph will resemble a negative sloping line.

Figure 2

Figure 3

 

As [latex]x\to \infty [/latex] the function [latex]f\left(x\right)\to \mathrm{-\infty }[/latex], so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.

Using technology, we can create the graph for the polynomial function, shown in Figure 4, and verify that the resulting graph looks like our sketch in Figure 15.

Figure 4

Use 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. Figure 5 shows that there is a zero between a and b.

Figure 5. Using the Intermediate Value Theorem to show there exists a zero.

Example 2: 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].

Show Solution

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

We see that one zero occurs at [latex]x=2[/latex]. Also, since [latex]f\left(3\right)[/latex] is negative and [latex]f\left(4\right)[/latex] 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 [latex]x=1[/latex] and [latex]x=4[/latex].

Analysis of the Solution

We can also see in Figure 6 that there are two real zeros between [latex]x=1[/latex] and [latex]x=4[/latex].

Figure 6

Key Concepts

• To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most n – 1 turning points.
• Graphing a polynomial function helps to estimate local and global extremas.
• The Intermediate Value Theorem tells us that if [latex]f\left(a\right) \text{and} 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].

Section 4.2 Homework Exercises

For the following exercises, find the zeros, y-intercept, multiplicity, and end behavior.  Use these to graph each function.

1. [latex]f\left(x\right)=x^{4}-2x[/latex]

2. [latex]f\left(x\right)={\left(x+3\right)}^{2}\left(x - 2\right)[/latex]

3. [latex]g\left(x\right)=\left(x+4\right){\left(x - 1\right)}^{2}[/latex]

4. [latex]h\left(x\right)={\left(x - 1\right)}^{3}{\left(x+3\right)}^{2}[/latex]

5. [latex]k\left(x\right)={\left(x - 3\right)}^{3}{\left(x - 2\right)}^{2}[/latex]

6. [latex]m\left(x\right)=-2x\left(x - 1\right)\left(x+3\right)[/latex]

7. [latex]n\left(x\right)=-3x\left(x+2\right)\left(x - 4\right)[/latex]

8. [latex]f\left(x\right)=x^{4}+3x[/latex]

For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval.

9. [latex]f\left(x\right)={x}^{3}-100x+2[/latex], between [latex]x=0.01[/latex] and [latex]x=0.1[/latex]

10. [latex]f\left(x\right)={x}^{3}-9x[/latex], between [latex]x=-4[/latex] and [latex]x=-2[/latex].

11. [latex]f\left(x\right)={x}^{3}-9x[/latex], between [latex]x=2[/latex] and [latex]x=4[/latex].

12. [latex]f\left(x\right)={x}^{5}-2x[/latex], between [latex]x=1[/latex] and [latex]x=2[/latex].

13. [latex]f\left(x\right)=-{x}^{4}+4[/latex], between [latex]x=1[/latex] and [latex]x=3[/latex].

14. [latex]f\left(x\right)=-2{x}^{3}-x[/latex], between [latex]x=-1[/latex] and [latex]x=1[/latex].

15. [latex]f\left(x\right)={x}^{3}-100x+2[/latex], between [latex]x=0.01[/latex] and [latex]x=0.1[/latex]