Topics
1A. Polynomial and Rational Functions
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United States of America
AP Precalculus

1.5B Polynomial functions and complex zeros

Lesson
Worksheet
Practice
Lesson

Introduction

Learning objective

  • 1.5.B Determine if a polynomial function is even or odd.

Even and odd functions

There are two special types of polynomial functions based on the degree of each term of the polynomial.

Even function

A function is even if f\left(-x\right)=f\left(x\right)

Example:

f\left(x\right) = x^4-5x^2+4

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The graph of an even function is symmetric about the y-axis.

The end behaviors as x approaches either positive or negative infinity will be the same, either both approaching positive infinity or both approaching negative infinity.

The degree of each term in an even function is even or 0.

When we substitute -x into the function and simplify, the result should be the same as the original function.

Odd function

A function is odd if f\left(-x\right)=-f\left(x\right)

Example:

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

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The graph of an odd function is symmetric about the origin. This means we can rotate the function 180\degree around the origin, and the graph would be in the same position it was originally.

The end behaviors as x approaches positive or negative infinity will be opposite one another.

The degree of each term in an odd function is odd.

When we substitute -x into the function and simplify, the signs of all the terms should be opposite of the terms in the original function.

If the degrees of some terms in a function are even and the degrees of other terms in a function are odd, then the function is neither even nor odd. This means its graph will not be symmetric to the y-axis nor the origin. When we substitute -x into the function and simplify, some signs will change, and others will not.

Examples

Example 1

Determine if the following functions are even, odd, or neither.

a

f\left(x\right)=-6x^5+15x^3-9x

Worked Solution
Create a strategy

Note that all the terms in the equation have an odd degree. To confirm algebraically that the function is odd, we need to show f\left(-x\right)=-f\left(x\right).

Apply the idea
\displaystyle f\left(x\right)\displaystyle =\displaystyle -6x^5+15x^3-9xOriginal function
\displaystyle f\left(-x\right)\displaystyle =\displaystyle -6\left(-x\right)^5+15\left(-x\right)^3-9\left(-x\right)Substitute -x
\displaystyle =\displaystyle -6\left(-x^5\right)+15\left(-x^3\right)-9\left(-x\right)Evaluate the exponents
\displaystyle =\displaystyle 6x^5-15x^3+9xEvaluate the multiplication
\displaystyle -f\left(x\right)\displaystyle =\displaystyle -\left(-6x^5+15x^3-9x\right)Multiply both sides of f\left(x\right) by -1
\displaystyle =\displaystyle 6x^5-15x^3+9xDistributive property

This shows f\left(-x\right)=-f\left(x\right), so f\left(x\right) is an odd function.

Reflect and check

We should also see that if \left(x,y\right) is a point on the curve, then \left(-x,-y\right) is also a point on the curve because this is an odd function. Testing this with x=2 and x=-2, we find:

\displaystyle f\left(x\right)\displaystyle =\displaystyle -6x^5+15x^3-9xOriginal function
\displaystyle f\left(2\right)\displaystyle =\displaystyle -6\left(2\right)^5+15\left(2\right)^3-9\left(2\right)Substitute x=2
\displaystyle =\displaystyle -6\left(32\right)+15\left(8\right)-9\left(2\right)Evaluate the exponents
\displaystyle =\displaystyle -192+120-18Evaluate the multiplication
\displaystyle =\displaystyle -90Evaluate the addition and subtraction

The point \left(2,-90\right) lies on the curve. Next, substitute x=-2:

\displaystyle f\left(-2\right)\displaystyle =\displaystyle -6\left(-2\right)^5+15\left(-2\right)^3-9\left(-2\right)Substitute x=-2
\displaystyle =\displaystyle -6\left(-32\right)+15\left(-8\right)-9\left(-2\right)Evaluate the exponents
\displaystyle =\displaystyle 192-120+18Evaluate the multiplication
\displaystyle =\displaystyle 90Evaluate the addition and subtraction

The point \left(-2,90\right) also lies on the curve. If we repeated this for multiple points, we would see that f\left(-x\right)=-f\left(x\right) for any point on the curve which confirms this is an odd function.

b

p\left(x\right)=\sqrt{6}x^4-\sqrt{11}x^2-\sqrt{5}

Worked Solution
Create a strategy

The constant in any function can also be written as c\cdot x^0 which means the degree of any constant is 0. This shows that all the terms in the equation have a degree that is even or 0. To confirm algebraically that the function is even, we need to show p\left(-x\right)=p\left(x\right).

Apply the idea
\displaystyle p\left(x\right)\displaystyle =\displaystyle \sqrt{6}x^4-\sqrt{11}x^2-\sqrt{5}Original function
\displaystyle p\left(-x\right)\displaystyle =\displaystyle \sqrt{6}\left(-x\right)^4-\sqrt{11}\left(-x\right)^2-\sqrt{5}Substitute -x
\displaystyle =\displaystyle \sqrt{6}\left(x^4\right)-\sqrt{11}\left(x^2\right)-\sqrt{5}Evaluate the exponents
\displaystyle =\displaystyle \sqrt{6}x^4-\sqrt{11}x^2-\sqrt{5}Evaluate the multiplication

This shows p\left(-x\right)=p\left(x\right), so p\left(x\right) is an even function.

Reflect and check

We can use technology to graph the function and see that it is symmetric to the y-axis.

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Worked Solution
Create a strategy

By looking at the graph, we can see that it is not symmetric about the y-axis because the function is not the same on both sides of the y-axis. As x\to -\infty, y\to \infty, but the other end of the graph goes to -\infty.

Next, we need to verify whether or not this function is odd. If it is not odd, then it must be neither even nor odd.

Apply the idea

If the function is odd, then reflecting the graph across the y-axis would show the same result as reflecting it across the x-axis. This is what is means to be symmetric to the origin. Let's first show the reflection across the y-axis, f\left(-x\right).

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Next, we will show the reflection across the x-axis, -f\left(x\right).

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Comparing the graphs, we see a relative maximum in the second quadrant of the first graph, but there is no relative maximum there in the second graph. Also, the point \left(-2,2\right) lies on the first graph but not the second graph.

Therefore, f\left(-x\right)\neq -f\left(x\right), so this function is neither even nor odd.

Reflect and check

The equation that was used for the graph in this problem is f\left(x\right)=-\dfrac{1}{2}x^3+2x^2-x. If we substitute -x into this equation, it will not be equal to f\left(x\right) nor -f\left(x\right).

\displaystyle f\left(-x\right)\displaystyle =\displaystyle -\dfrac{1}{2}\left(-x\right)^3+2\left(-x\right)^2-\left(-x\right)Substitute -x
\displaystyle =\displaystyle -\dfrac{1}{2}\left(-x^3\right)+2\left(x^2\right)-\left(-x\right)Evaluate the exponents
\displaystyle =\displaystyle \dfrac{1}{2}x^3+2x^2+xEvaluate the multiplication

In this case, some of the signs changed, but some of the signs did not change. This means f\left(-x\right)\neq f\left(x\right) and f\left(-x\right)\neq -f\left(x\right). This happens because the degrees of some terms are odd, and the degrees of other terms are even. Therefore, this function is neither even nor odd.

Idea summary

A function is even if f\left(-x\right)=f\left(x\right). In other words, if we substitute -x into the function and simplify, the result is the same as the original function. Even functions are symmetric about the y-axis, and the degree of each term is even or 0.

A function is odd if f\left(-x\right)=-f\left(x\right). In other words, if we substitute -x into the function and simplify, the signs of all the terms should be opposite of the terms in the original function. Odd functions are symmetric about the origin, and the degree of each term is odd.

Outcomes

1.5.B

Determine if a polynomial function is even or odd.

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