Topics
3. Polynomial Functions
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Grades 9-12

3.01 Polynomial functions

Lesson
Practice

Introduction

In lesson  4.02 Function families  , we reviewed the function families that we studied in Math 2. This lesson will introduce us to the first new function we will learn in Math 3, polynomial functions. We have already worked with polynomials algebraically, but we will now look at some of the key features of polynomials and transform them in the same way as other functions.

Polynomial functions

A polynomial function is a function that involves variables raised to non-negative integer powers.

The standard form of a polynomial function is given by f\left(x\right)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_2x^2+a_1x+a_0 where n is a positive integer and a_n,a_{n-1},a_{n-2},\ldots,a_2,a_1,a_0 are constant coefficients.

Polynomial functions can be transformed in the same way as other functions:

\text{Vertical translation}f\left(x\right)+k\text{Shifts up when }k>0\text{Shifts down when }k<0
\text{Horizontal translation}f\left(x-h\right)\text{Shifts right when }h>0\text{Shifts left when }h<0
\text{Vertical stretch/compression}a\cdot f\left(x\right)\text{Stretch when }\left|a\right|>1\text{Compress when }\left|a\right|<1
\text{Horizontal stretch/compression}f\left(b\cdot x\right)\text{Stretch when }\left|b\right|<1\text{Compress when }\left|b\right|>1
\text{Reflection across }x\text{-axis}-f\left(x\right)
\text{Reflection across }y\text{-axis}f\left(-x\right)

Exploration

Use the applet to explore the graphs and answer the questions that follow:

  • Type the equation f\left(x\right) into the first line and press enter on your keyboard.

  • Select the circle next to f\left(x\right) to explore or hide its graph

  • Select the circle next to g\left(x\right) to explore or hide the graph of f\left(x\right) reflected about the x-axis

  • Select the circle next to h\left(x\right) to explore or hide the graph of f\left(x\right) reflected about the y-axis

  1. f\left(x\right)=2x^3-x: Compare the reflection of f\left(x\right) across the x-axis to the reflection of f\left(x\right) across the y-axis. What do you notice?

  2. f\left(x\right)=x^3-3x+1: Compare the reflection of f\left(x\right) across the x-axis to the reflection of f\left(x\right) across the y-axis. What do you notice? Why do you think this is different from the first cubic equation?

  3. f\left(x\right)=x^4-5x^2+4: Compare the graph of f\left(x\right) to its reflection across the y-axis. What do you notice?

  4. f\left(x\right)=x^4-4x^2+x: Compare the graph of f\left(x\right) to its reflection across the y-axis. What do you notice? Why do you think this is different from the first quartic equation?

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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.

The rate of change of a polynomial function is variable.

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In some polynomials, the function increases (or decreases) at a fast rate, then the rate of change slows around a point called an inflection point. In other words, the function continues increasing (or decreasing), but the rate is slower around the point of inflection.

This is one type of a point of inflection, sometimes referred to as a horizontal point of inflection, not all points of inflection will see the rate slow around the point.

Examples

Example 1

Given the polynomial function below:

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a

Find the average rate of change over the following intervals:

  • -3\leq x\leq-2
  • -2\leq x\leq 0
  • 0\leq x\leq 2
  • 2\leq x\leq 3
Worked Solution
Create a strategy

The average rate of change is found by \dfrac{f\left(b\right)-f\left(a\right)}{b-a} where a is one endpoint of the interval and b is the other endpoint of the interval.

Apply the idea

  • -3\leq x\leq-2

    \displaystyle \dfrac{f\left(b\right)-f\left(a\right)}{b-a}\displaystyle =\displaystyle \dfrac{f\left(-2\right)-f\left(-3\right)}{-2-\left(-3\right)}Substitute a=-3 and b=-2
    \displaystyle =\displaystyle \dfrac{-20-70}{-2-\left(-3\right)}Substitute f\left(-3\right)=70 and f\left(-2\right)=-20
    \displaystyle =\displaystyle -90Evaluate the subtraction and division

  • -2\leq t\leq 0

    \displaystyle \dfrac{f\left(b\right)-f\left(a\right)}{b-a}\displaystyle =\displaystyle \dfrac{f\left(0\right)-f\left(-2\right)}{0-\left(-2\right)}Substitute a=-2 and b=0
    \displaystyle =\displaystyle \dfrac{0-\left(-20\right)}{0-\left(-2\right)}Substitute f\left(-2\right)=-20 and f\left(0\right)=0
    \displaystyle =\displaystyle 10Evaluate the subtraction and division

  • 0\leq t\leq 2

    \displaystyle \dfrac{f\left(b\right)-f\left(a\right)}{b-a}\displaystyle =\displaystyle \dfrac{f\left(2\right)-f\left(0\right)}{2-0}Substitute a=0 and b=2
    \displaystyle =\displaystyle \dfrac{20-0}{2-0}Substitute f\left(0\right)=0 and f\left(2\right)=20
    \displaystyle =\displaystyle 10Evaluate the subtraction and division

  • 2\leq t\leq 3

    \displaystyle \dfrac{f\left(b\right)-f\left(a\right)}{b-a}\displaystyle =\displaystyle \dfrac{f\left(3\right)-f\left(2\right)}{3-2}Substitute a=2 and b=3
    \displaystyle =\displaystyle \dfrac{-70-20}{3-2}Substitute f\left(2\right)=20 and f\left(3\right)=-70
    \displaystyle =\displaystyle -90Evaluate the subtraction and division
Reflect and check

The average rate of change is the same for the intervals -3\leq x\leq 2 and 2\leq x\leq 3, and it is the same again for the intervals -2\leq x\leq 0 and 0\leq x\leq 2. However, the intervals around the point of inflection are much slower than the others.

b

Determine the end behavior of the function.

Worked Solution
Create a strategy

We want to determine what happens to the y-values when the x-values are very small and very large.

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Apply the idea

As x gets very small, y gets very large. So, as x\to -\infty, y\to \infty.

As x gets very large, y gets very small. So, as x\to \infty, y\to -\infty.

c

State the domain and range of the function.

Worked Solution
Create a strategy

The domain represents the x-values of the function, and the range represents the y-values of the function. Recall that the graph of a polynomial function is one smooth curve over a continuous domain.

Apply the idea

As a polynomial is defined for any real number input x, the domain is \left(-\infty ,\infty\right).This can be seen in the graph where the function is a smooth continuous curve and continues to be defined to the left, as the x-values get small, and the right, as the x-values get large.

Since the y-values continue to get increasingly small and increasingly large indefinitely, the range is also \left(-\infty ,\infty\right).

d

Determine if the function is even, odd, or neither.

Worked Solution
Create a strategy

The function is not symmetric to the y-axis, so we immediately know it is not an even function. If it is an odd function, then f\left(-x\right)=-f\left(x\right). We can determine this graphically by reflecting f\left(x\right) across the x-axis, then reflecting it across the y-axis, and determining if both graphs are the same.

Apply the idea

Reflecting f\left(x\right) across the x-axis:

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Reflecting f\left(x\right) across the y-axis:

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These graphs are the same. Therefore, f\left(-x\right)=-f\left(x\right) which shows the function is odd.

Reflect and check

If we look at individual points on the graph, we will also see f\left(-x\right)=-f\left(x\right). Consider the point \left(2,-20\right): \begin{aligned}f\left(2\right)&=20\\f\left(-x\right)&=-f\left(x\right)\\f\left(-2\right)&=20 \end{aligned} Looking at the graph, we see the point \left(-2,20\right) does lie on the graph.

We can also see that the points \left(3,70\right) and \left(-3,-70\right) lie on the graph.\begin{aligned}f\left(3\right)&=70\\f\left(-x\right)&=-f\left(x\right)\\f\left(-3\right)&=-70\end{aligned} This will apply to any point that lies on the graph.

Example 2

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.

Example 3

Consider the table of values for the function f \left( x \right) and the transformed function g \left( x \right) shown in the graph:

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Express g \left( x \right) in terms of f \left( x \right).

Worked Solution
Create a strategy

Determine the values of g\left(x\right) for the values of x shown in the table and think how the original function f\left(x\right) can be transformed to get these values.

Apply the idea

Determining the values of g\left(x\right) for the given values of x, we have the following table of values:

x-2-1012
g\left(x\right)-1046822

From the table of values above, we can see that the values of g\left(x\right) are twice that of f\left(x\right). Thus, we can express g\left(x\right) in terms of f\left(x\right) as g \left( x \right) = 2 f \left( x \right).

Reflect and check

Graphing both functions on the same coordinate plane, we can see that the original function has been vertically stretched by a factor of 2 because g\left(x\right) appears "taller" and thinner than f\left(x\right).

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Example 4

Consider the graph of y = x^{4}:

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Describe how to shift the graph of y = x^{4} and sketch the graph of y = \left(x + 2\right)^{4} - 1.

Worked Solution
Create a strategy

Check the form of the new function to determine its horizontal and vertical translation.

Horizontal Translation: If the function is in the form y=f\left(x-h\right), the graph of y=f\left(x\right) is moved to the left by h units when h is negative and to the right by h units when h is positive.

Vertical Translation: If the function is in the form y=f\left(x\right)+k, the graph of y=f\left(x\right) is moved upward by k units when k is positive and downward by k units when k is negative.

Apply the idea

Since h=-2 and k=-1 in y = \left(x + 2\right)^{4} - 1, we can obtain the graph of y = \left(x + 2\right)^{4} - 1 by moving the graph of y = x^{4} to the left by 2 units and down by 1 unit.

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Reflect and check

Notice that the transformed function is no longer an even function. If we expanded \left(x+2\right)^4-1, we would see that the expression contains both even and odd exponents. Therefore, the transformed 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

F.IF.B.4

For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.

F.IF.B.5

Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes.

F.IF.B.6

Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.

F.BF.B.3

Identify the effect on the graph of replacing f(x) by f(x) + k, kf(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing even and odd functions from their graphs and algebraic expressions for them.

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