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Algebra 2 View details

4. Polynomial Functions and Their Graphs

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Algebra 2 /

Chapter 1

Polynomial Functions and Their Graphs

Polynomial functions and their graphs play a pivotal role in understanding the behavior of mathematical equations. The lesson delves deep into the nuances of graphing polynomial functions, emphasizing the significance of x-intercepts, end behavior, and turning points. By analyzing polynomial functions, one can determine the regions where the function increases or decreases, and where it reaches its maximum or minimum values. Furthermore, the symmetry of a graph can reveal if a function is even, odd, or neither. Through various examples and exercises, the lesson provides a holistic approach to mastering the art of graphing polynomial functions, ensuring a robust foundation in algebra.

Lesson Settings & Tools

	16 Theory slides
	12 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Image Credits

Slide of 16

Graphing functions is essential in different fields of research. This has led many people to develop programs and technologies that draw accurate graphs, such as graphing calculators. However, knowing how to draw a graph by hand is also essential. After all, that is how they did it not too many years ago!

Unfortunately, graphing an arbitrary function usually requires knowing advanced algebraic techniques. However, when the function is a polynomial, some key features of the graph can be derived directly from the function rule, leading to an accurate graph. As such, this lesson aims to teach how to graph polynomials.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Polynomial
Degree of a Polynomial
Standard Form of a Polynomial
Leading Coefficient
Function Notation
Evaluating a Function
Table of Values
Making a Table of Values

Challenge

Investigating the End Behavior

In order to improve her graphing polynomial skills, Emily decided to try graphing the fourth-degree polynomial function $f (x) = x^{4} - 4 x^{3} - 39 x^{2} + 46 x + 80 .$ To start, she made a table of values.

$x$	$- 6$	$- 5$	$- 4$	$- 3$	$- 2$	$- 1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$f (x)$	$560$	$0$	$- 216$	$- 220$	$- 120$	$0$	$80$	$84$	$0$	$- 160$	$- 360$	$- 540$	$- 616$

Next, she plotted all the points in a coordinate plane and connected them with a smooth curve. She noticed that the left part of the curve seemed to be going up while the right part seemed to be going down. She decided to extend the curve based on this intuition.

Curve connecting the points from the table. The left part of the curve is extended upwards and the right-part is extended downwards.

However, Emily was not so sure about her reasoning. Is Emily's graph correct? Is there a way she can be sure about the left and right parts of the graph?

Discussion

Polynomial Functions and End Behavior

Functions are usually named after the algebraic expression that defines them. For example, if the function rule is a real number, the function is called a constant function. Something similar happens when the function rule is a linear, quadratic, or exponential expression.

Function Rule	Name
$f (x) = - 7$	Constant Function
$g (x) = 2 x + 1$	Linear Function
$h (x) = x^{2} + 3 x - 1$	Quadratic Function
$t (x) = 2 e^{3 x}$	Exponential Function

The same holds true to those functions whose function rule is a polynomial.

Concept

Polynomial Function

A polynomial function is a function whose rule is a polynomial. In general, a polynomial function has the following form.

$f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$

Here, all exponents are whole numbers and all the coefficients are real numbers with

a_{n} \neq = 0 .

Unless explicitly stated, the domain of a polynomial function is all real numbers and its range depends on the end behavior.

Concept

End Behavior

The end behavior of a function is the value to which

f (x)

tends as

x

extends to the left or the right infinitely. If

f (x)

keeps increasing without bound, it is said to tend to positive infinity. The end behavior of this case is stated as up.

f (x) \to + \infty

Conversely, if

f (x)

keeps decreasing without bound, it is said to tend to negative infinity. In this case, the end behavior is stated as down.

f (x) \to - \infty

For example, consider the graph of a function

g (x) .

From the arrows on the graph, it can be seen that the left end of the graph extends downward, while the right end extends upward. The end behavior of

g

can then be expressed as follows.

As x \to - \infty, As x \to + \infty, g (x) \to - \infty g (x) \to + \infty

To state the end behavior of a function in words, begin by stating the left-end behavior, then state the right-end behavior. A dash can also be used to separate the words. For instance, the end behavior of the graph of

g (x)

can be written as down and up or as down-up.

End Behavior of Polynomial Functions

When the function is a polynomial function, the end behavior can be determined from the function rule.

P (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}

Particularly, the end behavior is given by the sign of the leading coefficient and the degree of the polynomial.

Leading Coeff. positive and degree even: Up-Up; Leading Coeff. positive and degree odd: Down-Up; Leading Coeff. negative and degree even: Down-Down; Leading Coeff. negative and degree even: Up-Down;

Note that when the degree of the polynomial is even, both ends have the same behavior, which depends on the sign of the leading coefficient. By contrast, when the degree is odd, both ends have opposite behaviors.

Pop Quiz

Determining the End Behavior

Determine the end behavior of the given polynomial function.

Either the function rule or the graph of a polynomial function.

Example

Image Editor Running Time

Emily loves to record every trip she takes on her blog, including amazing photos. However, since she does not have a professional camera, some of her photos need some retouching. To help with this, she plans to buy an image editing software. She is torn between two options whose running times to process an $x - by - x$ pixels image are given by the following polynomials.

	Running Time (nanoseconds)
Software $1$	$P (x) = x^{3} - 520 x^{2} + 10000 x + 20000000$
Software $2$	$Q (x) = 100 x^{2} + 1000000$

a Emily used online software to plot both polynomials in an effort to clear things up before deciding which software to purchase.

Graph of both polynomials between x=0 and x=325.

From the graph, Emily noticed that the right-end behavior of the graph of

y = P (x)

is down, while the right-end behavior of the graph of

y = Q (x)

is up. Therefore, she concluded that for images with realistic dimensions, Software

1

is faster and plans to buy it. Should Emily buy Software

1 ?

b What is the end behavior of the graph of

y = P (x) ?

c What is the end behavior of the graph of

y = Q (x) ?

Hint

a Evaluate both polynomials at different values and compare the results. In general, images have sides of at least

600

pixels. In order to study the end behavior of the curves, the graph should be plotted in a large enough domain.

b The end behavior of a polynomial graph can be determined by studying the sign of the leading coefficient and the degree of the polynomial rule. When the degree is odd, both ends of the graph have opposite behaviors.

c When the degree is even, both ends of the graph have the same behavior.

Solution

a Before analyzing Emily's graph, evaluate the polynomial functions at different values to compare algebraically the running times of both software.

Image Size (pixels)	Running Times (nanoseconds)	Conclusion
$100$ -by- $100$	$P (100) Q (100) = 16800000 = 2000000 ✓$	Software $2$ is faster
$240$ -by- $240$	$P (240) Q (240) = 6272000 ✓ = 6760000 ✓$	Similar Running Times
$300$ -by- $300$	$P (300) Q (300) = 3200000 ✓ = 10000000$	Software $1$ is faster
$600$ -by- $600$	$P (600) Q (600) = 54800000 = 37000000 ✓$	Software $2$ is faster
$1200$ -by- $1200$	$P (1200) Q (1200) = 1011200000 = 145000000 ✓$	Software $2$ is faster
$2000$ -by- $2000$	$P (2000) Q (2000) = 5960000000 = 401000000 ✓$	Software $2$ is faster

The table suggests the following conclusions.

For very small images, Software $1$ is faster than Software $2 .$
For images of about $240 - by - 240$ pixels, both software have similar running times.
For images with more realistic dimensions, $600 - by - 600$ pixels images or larger, Software $2$ is faster than Software $1 .$

Emily's graph also suggests the first two conclusions. However, that graph corresponds only to a small portion of the domain. Therefore, before making any decisions about end behavior, graph both polynomials in a bigger domain.

It can be seen that the right-end behavior of the graph of $y = P (x)$ is up, and not down as Emily concluded. Additionally, for images with $600 -$ pixel sides or bigger, the running time of Software $2$ is considerably shorter than the running time of Software $1 .$

Conclusion
In general, Software $2$ is faster.

Although the right-end behavior of both graphs is up, the graph of $y = P (x)$ increases way much faster than the graph of $y = Q (x) .$ Therefore, Emily should buy Software $2 .$

b Although the behavior of the right end of the graph of

y = P (x)

was studied in Part A, it can also be determined using the function rule.

P (x) = x^{3} - 520 x^{2} + 10000 x + 20000000

Start by writing the end behavior of a polynomial graph based on the leading coefficient and degree.

Leading Coefficient	Degree	End Behavior
Positive	Even	Up-Up
Positive	Odd	Down-Up
Negative	Odd	Up-Down
Negative	Even	Down-Down

The leading term of $P (x)$ is $x^{3},$ which means that the leading coefficient is $1$ and the degree is $3 .$ Since the leading coefficient is positive and the degree is odd, the end behavior of the graph of $y = P (x)$ is Down-Up.

Despite the $y -$ axis being measured in seconds rather than nanoseconds, the conclusion about the end behavior or the fastest software remains the same.

c As in Part B, the end behavior of the graph of

y = Q (x)

can be obtained from its function rule.

Q (x) = 100 x^{2} + 1000000

Here, the leading term is

100 x^{2},

so the leading coefficient is

100

and the degree is

2 .

Since the leading coefficient is positive and the degree is even, the end behavior of the graph of

y = Q (x)

is Up-Up.

Again, the $y -$ axis is measured in seconds, not in nanoseconds.

Discussion

Alternative Methods for Finding the $x -$ intercepts

When graphing polynomial functions, the

x -

intercepts is a key feature. Such information can sometimes be derived from the function rule, but it is not always self-evident. Consider the following polynomial.

f (x) = x^{4} - 3 x^{3} - 15 x^{2} + 19 x + 30

At first glance, it is not clear whether the graph of

y = f (x)

intercepts the

x -

axis, and if so, where. However, if such a point existed, then it would have the form

(k, 0),

which means that

f (k) = 0 .

A portion of a graph crossing the x-axis at the point (k,0)

Note that the fact that

f (k) = 0

has two implications — that

k

is a zero of the polynomial function and that

x = k

is a solution to the equation

f (x) = 0 .

f (k) = 0 ⇕ k^{4} - 3 k^{3} - 15 k^{2} + 19 k + 30 = 0

Consequently, finding the

x -

intercepts of the graph of

y = f (x)

is equivalent to finding the real zeros of the polynomial

f (x),

which in turn is equivalent to finding the real solutions to the equation

f (x) = 0 .

The following diagram explains what this equivalence means.

Let f(x) be a polynomial function. (a) k is a zero of f(x) if and only if k is a solution to the equation f(x)=0. (b) A real number k is a zero of f(x) if and only if k is an x-intercept of the graph of f(x). (c) k is a real solution to the equation f(x)=0 if and only if k is an x-intercept of the graph of f(x).

Remember, in an equivalence, both the conditional statement and its converse are true.

Example

Graphing a Polynomial Function

Consider the following polynomial function.

f (x) = x^{4} - 3 x^{3} - 15 x^{2} + 19 x + 30

a List all the values where the graph of

y = f (x)

intercepts the

x -

axis.

b List the missing numbers in the following table of values. Write them in the order they should be in the table from left to right.

$x$	$- 4$	$- 2$	$0$	$1$	$3$	$4$	$6$
$f (x)$		$- 28$			$- 48$		$252$

c What is the end behavior of the graph of

y = f (x) ?

d Which of the following is the graph

y = f (x) ?

Hint

a Finding the

x -

intercepts is the same as finding the solutions to the equation

f (x) = 0 .

Factor the polynomial and use the Zero Product Property.

b Evaluate the polynomial at the missing values.

c Study the sign of the leading coefficient and the degree of the polynomial. When the degree is even, both ends have the same behavior.

d Start by plotting the

x -

intercepts obtained in Part A. Next, plot the points from the table of values. Connect all the points using a smooth curve and extend the graph according to the end behavior.

Solution

a Determining the places where the graph of

y = f (x)

intercepts the

x -

axis is equivalent to finding the solutions to the equation

f (x) = 0 .

f (x) = 0 ⇓ x^{4} - 3 x^{3} - 15 x^{2} + 19 x + 30 = 0

To solve the polynomial equation, factor the polynomial. This can be done by grouping, but some rewrites are needed first.

- 3 x^{3} - 15 x^{2} 19 x = - 4 x^{3} + x^{3} = - 11 x^{2} - 4 x^{2} = 30 x - 11 x

Substitute these expressions into the polynomial equation.

x^{4} - 3 x^{3} - 15 x^{2} + 19 x + 30 = 0

▼

Factor

SubstituteExpressions

Substitute expressions

x^{4} - 4 x^{3} + x^{3} - 11 x^{2} - 4 x^{2} + 30 x - 11 x + 30 = 0

CommutativePropAdd

Commutative Property of Addition

(x^{4} - 4 x^{3} - 11 x^{2} + 30 x) + (x^{3} - 4 x^{2} - 11 x + 30) = 0

FactorOut

Factor out $x$

x (x^{3} - 4 x^{2} - 11 x + 30) + (x^{3} - 4 x^{2} - 11 x + 30) = 0

FactorOut

Factor out $(x^{3} - 4 x^{2} - 11 x + 30)$

(x^{3} - 4 x^{2} - 11 x + 30) (x + 1) = 0

The cubic polynomial within the parentheses can also be factored by groping. As before, some of the terms have to be rewritten first.

- 4 x^{2} - 11 x = - 2 x^{2} - 2 x^{2} = - 15 x + 4 x

Now, substitute the expressions into the cubic polynomial.

(x^{3} - 4 x^{2} - 11 x + 30) (x + 1) = 0

▼

Factor

SubstituteII

$- 4 x^{2} = - 2 x^{2} - 2 x^{2}$ , $- 11 x = - 15 x + 4 x$

(x^{3} - 2 x^{2} - 2 x^{2} - 15 x + 4 x + 30) (x + 1) = 0

CommutativePropAdd

Commutative Property of Addition

((x^{3} - 2 x^{2} - 15 x) + (- 2 x^{2} + 4 x + 30)) (x + 1) = 0

FactorOut

Factor out $x & - 2$

(x (x^{2} - 2 x - 15) - 2 (x^{2} - 2 x - 15)) (x + 1) = 0

FactorOut

Factor out $(x^{2} - 2 x - 15)$

(x^{2} - 2 x - 15) (x - 2) (x + 1) = 0

Once again, factor the quadratic polynomial within the parentheses by grouping. To do so, start by rewriting

- 2 x

- 5 x + 3 x .

(x^{2} - 2 x - 15) (x - 2) (x + 1) = 0

▼

Factor

Rewrite

Rewrite $- 2 x$ as $- 5 x + 3 x$

(x^{2} - 5 x + 3 x - 15) (x - 2) (x + 1) = 0

FactorOut

Factor out $x & 3$

(x (x - 5) + 3 (x - 5)) (x - 2) (x + 1) = 0

FactorOut

Factor out $(x - 5)$

(x - 5) (x + 3) (x - 2) (x + 1) = 0

The polynomial has been completely factored. Next, by the Zero Product Property, at least one of the four factors is equal to zero. This produces four linear equations that can be solved for

x .

(x - 5) (x + 3) (x - 2) (x + 1) = 0 ⇓ ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ x - 5 = 0 x + 3 = 0 x - 2 = 0 x + 1 = 0 \Rightarrow ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ x = 5 x = - 3 x = 2 x = - 1

Consequently, the solutions to the polynomial equation, and therefore the

x -

intercepts, are

x = - 3,

- 1,

2,

and

5 .

b To find the values missing from the table, evaluate

f (x)

at the corresponding

x

value. For example, begin by substituting

x = - 4 .

f (x) = x^{4} - 3 x^{3} - 15 x^{2} + 19 x + 30

▼

Substitute

- 4

for

x

and simplify

Substitute

$x = - 4$

f (- 4) = (- 4)^{4} - 3 (- 4)^{3} - 15 (- 4)^{2} + 19 (- 4) + 30

CalcPowProd

Calculate power and product

f (- 4) = 256 - 3 (- 64) - 15 (16) - 76 + 30

Multiply

f (- 4) = 256 + 192 - 240 - 76 + 30

AddSubTerms

Add and subtract terms

f (- 4) = 162

The remaining missing values can be found in a similar way.

$x$	$f (x) = x^{4} - 3 x^{3} - 15 x^{2} + 19 x + 30$	Result
$0$	$f (0) = 0^{4} - 3 (0)^{3} - 15 (0)^{2} + 19 (0) + 30$	$30$
$1$	$f (1) = 1^{4} - 3 (1)^{3} - 15 (1)^{2} + 19 (1) + 30$	$32$
$4$	$f (4) = 4^{4} - 3 (4)^{3} - 15 (4)^{2} + 19 (4) + 30$	$- 70$

Consequently, the complete table of values looks as follows.

$x$	$- 4$	$- 2$	$0$	$1$	$3$	$4$	$6$
$f (x)$	$162$	$- 28$	$30$	$32$	$- 48$	$- 70$	$252$

c To determine the end behavior of the graph of a polynomial, start by identifying the leading term.

f (x) = x^{4}_{1} - 3 x^{3} - 15 x^{2} + 19 x + 30

Since the degree is

4,

which is an even number, both ends of the graph have the same behavior. Additionally, since the leading coefficient is

1,

which is a positive number, the end behavior of the graph of

y = f (x)

is up-up.

d To determine which of the graphs is correct, plot the graph of

y = f (x)

by using the information obtained in the previous parts.

Begin by plotting the $x -$ intercepts from Part A.
Next, graph the points from the table of values in Part B.
Connect the points using a smooth curve.
Extend both ends of the graph upwards, since the end behavior of the graph of $y = f (x)$ was found to be up-up in Part C.

Comparing the graph obtained with the given ones, it can be concluded that the graph of

y = f (x)

is Graph

4 .

Discussion

Locating the $x -$ Intercepts of a Graph

Given a polynomial function, the end behavior of the graph can be determined directly from the function rule without doing anything to it. For example, consider the following function.

The end behavior of the graph of

y = f (x)

is down-up. In contrast, the zeros are not easy to identify simply by looking at the function rule. Usually, the polynomial must be factored first. However, not all polynomials have a suitable factorization. Therefore, it is worth considering whether there is a simple way to estimate the zeros of a function. The following principle answers this.

Rule

The Location Principle

Let $f$ be a polynomial function and $a$ and $b$ be two real numbers. If $f (a)$ and $f (b)$ have opposite signs, then there is at least one zero between $a$ and $b .$

A polynomial graph having one x-intercept. A point of the curve being above the x-axis is marked and another point of the curve below the x-axis is marked. The x-intercept is highlighted.

Notice that if $f (a)$ and $f (b)$ have opposite signs, the principle guarantees one zero between $a$ and $b,$ but there could be more. Conversely, the fact that $f (a)$ and $f (b)$ have the same sign does not imply that there are no zeros between $a$ and $b .$

Proof

Informal Justification

Let $a$ and $b$ be two real numbers with $a < b$ and let $f$ be a polynomial function such that $f (a) > 0$ and $f (b) < 0 .$ A sketch of the graph of $y = f (x)$ around $x = a$ and around $x = b$ could look as follows.

Since graphs of polynomial functions are continuous — that is, they have no gaps — the points

(a, f (a))

and

(b, f (b))

must be connected by a smooth curve that will cross the

x -

axis at least once, implying that there is at least one zero between

a

and

b .

For $f (x) = 8 x^{3} - 12 x^{2}$ $- 26 x$ $+ 15,$ it can be found that $f (0) = 15 > 0$ and $f (1) = - 15 < 0 .$ Notice that $f (0)$ and $f (1)$ have opposite signs. Consequently, $f$ has at least one zero between $x = 0$ and $x = 1 .$

A coordinate plane. Segment between x=0 and x=1 highlighted.

The interval where the zero is located can be reduced by considering closer testing values.

A coordinate plane. Segment between x=0.25 and x=0.75 highlighted.

All real zeros of a polynomial can be estimated in this fashion.

Pop Quiz

Locating Zeros

Use the Location Principle to indicate whether the given polynomial function is guaranteed to have a zero in the indicated interval.

Random Polynomial Functions and random testing values

Discussion

Maximum and Minimum Points of a Graph

Sometimes knowing that a graph passes through a point does not give enough information about how the curve is before and after the point. However, when the point is either a maximum or a minimum, there is a clear view of the behavior of the graph around it.

Concept

Relative Minimum and Maximum

The value $f (c)$ is a relative minimum, or local minimum, of a function if $f (c)$ is the least output of $f$ around $x = c .$ Likewise, the value $f (d)$ is a relative maximum, or local maximum, of a function if $f (d)$ is the greatest output of $f$ around $x = d .$

If the function is continuous, the function switches from increasing to decreasing at a relative maximum or from decreasing to increasing at a relative minimum.

Sometimes the phrase relative extrema is used to refer to both relative maximums and relative minimums. Note that a function can have one or more relative extrema, or none at all.

When a relative extrema is greater or lower than any other point in the graph, it has a particular name.

Concept

Absolute Minimum and Maximum

The absolute minimum, or global minimum, of a function is the least output in its whole domain.

Absolute minimum of a quartic function 0.3*(x+3)*(x+2)*(x+1)*(x-1) located at (0.326345,-2.07423)

The absolute maximum, or global maximum, of a function is defined in a similar way. It is the greatest output of the function in its whole domain.

Absolute minimum of a quartic function -0.3*(x+2.5)*(x+1)*(x)*(x-2) located at (1.3,2.38602)

The absolute maximum of a function is also a relative maximum, and the absolute minimum is also a relative minimum. If a function increases indefinitely, it does not have an absolute maximum. Likewise, if a function decreases indefinitely, it does not have an absolute minimum. The function might still have relative extrema.

Different functions with different extrema

Discussion

Turning Points of Polynomial Functions

Relative or absolute extrema are very useful when graphing functions, but the method for finding them is beyond the scope of this lesson. That said, when working with polynomial functions, there is a way to estimate the maximum number of relative extrema by using the turning points of the graph.

Concept

Turning Point

A turning point is a point where the graph of a function changes from increasing to decreasing or vice versa. In other words, the turning points correspond to the relative extrema of the graph of a function.

Graph of f(x)=x^5 + 0.5 x^4 - 23.5 x^3 + 7 x^2 + 60 x

The graph corresponds to a fifth-degree polynomial with four turning points. When dealing with polynomial functions, there is a close connection between the degree, the zeros, and the turning points.

The graph of a polynomial function of degree $n$ can have at most $n - 1$ turning points.
If a polynomial function has $n$ distinct real zeros, then its graph has exactly $n - 1$ turning points.

Pop Quiz

Determining the Maximum Number of Turning Points

Consider the given polynomial function. Determine either the maximum or the exact number of turning points that its graph may have.

Example

Graphing a Fifth-Degree Polynomial Function

After purchasing the image editing software, Emily was curious about the online software she used to plot the polynomials and wanted to try graphing a polynomial by hand. To challenge herself, she chose the polynomial function $f (x) = - 0.5 x^{5} - 3.5 x^{4}$ $+ 91 x^{3} + 374 x^{2}$ $- 3556 x - 7840 .$ So far, she has created the following table of values.

$x$	$- 15$	$- 10$	$- 7$	$- 3$	$0$	$6$	$8$	$11$
$f (x)$	$25025$	$- 10880$	$4165$	$3575$	$- 7840$	$- 4480$	$3520$	$- 12350$

She next plans to find the zeros of the polynomial and determine the end behavior of the graph. She also knows that $(- 11.83, - 16791.37),$ $(- 5.16, 7312.41),$ $(2.67, - 13181.92),$ and $(8.71, 4482.69)$ are turning points. Emily thinks it is enough information to make an accurate sketch of the graph. Follow Emily's plan and graph $y = f (x) .$

a Find all the zeros of the polynomial function.

b What is the end behavior of the graph of

f ?

c Sketch of the graph of

f .

Answer

a Zeros:

- 14, - 8, - 2, 7, 10

b End behavior: Up-Down

c Graph:

Hint

a Use the Location Principle.

b Take a look at the degree and the sign of the leading coefficient of the polynomial.

c Use the fact that the zeros are the

x -

intercepts of the graph. Plot the turning points. Connect all the points with a smooth curve and extend it according to the end behavior.

Solution

a One way of finding the zeros of the function is by factoring the given polynomial. However, it does not seem easy to factor. Alternatively, the Location Principle can help locate where the zeros could be. In the given table, highlight the positive values with one color and the negative values with a different color.

$x$	$- 15$	$- 10$	$- 7$	$- 3$	$0$	$6$	$8$	$11$
$f (x)$	$25025$	$- 10880$	$4165$	$3575$	$- 7840$	$- 4480$	$3520$	$- 12350$

The first sign change occurs between

- 15

and

- 10 .

Therefore, there is a zero between those two values. For simplicity, test the integer values within this interval. For example, start with

- 14 .

f (x) = - 0.5 x^{5} - 3.5 x^{4} + 91 x^{3} + 374 x^{2} - 3556 x - 7840

▼

Substitute

- 14

for

x

and evaluate

Substitute

$x = - 14$

f (- 14) = - 0.5 (- 14)^{5} - 3.5 (- 14)^{4} + 91 (- 14)^{3} + 374 (- 14)^{2} - 3556 (- 14) - 7840

CalcPowProd

Calculate power and product

f (- 14) = - 0.5 (- 537824) - 3.5 (38416) + 91 (- 2744) + 374 (196) + 49784 - 7840

Multiply

f (- 14) = 268912 - 134456 - 249704 + 73304 + 49784 - 7840

AddSubTerms

Add and subtract terms

f (- 14) = 0

Since

f (- 14) = 0,

it means that

x = - 14

is a zero of

f .

From the table, the second sign change happens between

- 10

and

- 7 .

As before, test the integer values between

- 10

and

- 7 .

$f (x) = - 0.5 x^{5} - 3.5 x^{4} + 91 x^{3} + 374 x^{2} - 3556 x - 7840$
$x = - 9$	$x = - 8$
$f (- 9) = - 0.5 (- 9)^{5} - 3.5 (- 9)^{4} + 91 (- 9)^{3} + 374 (- 9)^{2} - 3556 (- 9) - 7840 ⇓ f (- 9) = - 5320$	$f (- 8) = - 0.5 (- 8)^{5} - 3.5 (- 8)^{4} + 91 (- 8)^{3} + 374 (- 8)^{2} - 3556 (- 8) - 7840 ⇓ f (- 8) = 0$

The polynomial has a second zero at $x = - 8 .$ The remaining zeros can be found by applying a similar reasoning.

The third sign change occurs between $- 3$ and $0 .$
The fourth sign change occurs between $6$ and $8 .$
The fifth sign change occurs between $8$ and $11 .$

After testing some integers that belong to each of the mentioned intervals, the remaining zeros can be found at $x = - 2,$ $x = 7,$ and $x = 10 .$

Zeros
$- 14, - 8, - 2, 7, 10$

Since the polynomial has degree five, it has no more zeros.

b The end behavior of the graph of a polynomial function can be determined by the sign of the leading coefficient and the degree of the polynomial.

Leading Coefficient	Degree	End Behavior
Positive	Even	Up-Up
Positive	Odd	Down-Up
Negative	Odd	Up-Down
Negative	Even	Down-Down

For the given polynomial, the leading term is $- 0.5 x^{5} .$ The leading coefficient is negative and the degree is odd. Consequently, the end behavior of the graph of $f$ is Up-Down.

c Since the zeros of a polynomial are the

x -

intercepts, start by plotting these points on the coordinate plane. It also may help to make a sketch reflecting the end behavior of the graph.

Now, graph the points from the table.

Before connecting the points with a smooth curve, plot the turning points $(- 11.83, - 16791.37),$ $(- 5.16, 7312.41),$ $(2.67, - 13181.92),$ and $(8.71, 4482.69) .$ At these points, the graph changes from increasing to decreasing or vice versa.

According to the distribution of the points, from left to right, the first turning point is a relative minimum, the second one a relative maximum, the third another relative minimum, and the fourth and final one another relative maximum. There are no absolute extrema. With all this information, the graph can finally be drawn.

Discussion

Even and Odd Functions

Some functions have the property that their graphs look the same on both sides of the $y -$ axis, as if the $y -$ axis were a mirror. Such a property can be easily identified once the graph is drawn but, can it be deduced algebraically? The next concepts can answer this question.

Concept

Even Function

An even function is a function for which the value of $f (- x)$ is equal to the value of $f (x)$ for all the values in its domain. That is, opposite inputs have the same output.

$f (- x) = f (x)$

The graph of an even function is symmetric about the

y -

axis. The functions

y = x^{2}

and

y = 2 ∣ x ∣

are two examples of even functions.

Graphs of f(x)=x^2 and graph of g(x)=2|x|

Notice that if a function is even and the point

(x, y)

is on the graph, then the point

(- x, y)

is also on the graph.

Extra

Determining If a Function is Even

To determine algebraically whether a function is even, substitute

- x

into the function rule and simplify. If the resulting expression is equal to

f (x),

then the function is even; otherwise, it is not. For example, consider the following function.

f (x) = 3 x^{4} - 2 x^{2} + 1

Substitute

- x

for

x

and simplify.

f (x) = 3 x^{4} - 2 x^{2} + 1

Substitute

$x = - x$

f (- x) = 3 (- x)^{4} - 2 (- x)^{2} + 1

NegBaseToPosPow

$(- a)^{4} = a^{4}$

f (- x) = 3 x^{4} - 2 (- x)^{2} + 1

NegBaseToPosBase

$(- a)^{2} = a^{2}$

f (- x) = 3 x^{4} - 2 x^{2} + 1

Substitute

$3 x^{4} - 2 x^{2} + 1 = f (x)$

f (- x) = f (x)

Since

f (- x) = f (x),

the given function is even.

Some other functions seem to be reflected across the origin. Functions with this property are called odd functions.

Concept

Odd Function

An odd function is a function for which the value of $f (- x)$ is equal to the value of $- f (x)$ for all the values in its domain. It is like if the function allows moving the negative sign from the input to the output.

$f (- x) = - f (x)$

The graph of an odd function is symmetric about the origin, meaning that the graph looks the same after a

18 0^{\circ}

rotation about the origin. The functions

y = x

and

y = x^{3}

are two examples of odd functions.

Notice that if a function is odd and the point

(x, y)

is on the graph, then the point

(- x, - y)

is also on the graph.

Extra

Determining If a Function is Odd

To determine algebraically whether a function is odd, substitute

- x

into the function rule and simplify. If the resulting expression is equal to

- f (x),

then the function is odd; otherwise, it is not. For example, consider the following function.

f (x) = \frac{x}{x ^{2} - 1}

Substitute

- x

for

x

and simplify.

f (x) = \frac{x}{x ^{2} - 1}

Substitute

$x = - x$

f (- x) = \frac{- x}{( - x ) ^{2} - 1}

NegBaseToPosBase

$(- a)^{2} = a^{2}$

f (- x) = \frac{- x}{x ^{2} - 1}

MoveNegNumToFrac

Put minus sign in front of fraction

f (- x) = - \frac{x}{x ^{2} - 1}

Substitute

$\frac{x}{x ^{2} - 1} = f (x)$

f (- x) = - f (x)

Since

f (- x) = - f (x),

the given function is odd.

Be aware that these concepts are exclusive — in other words, if a function is even, it cannot be odd, and vice versa. However, unlike integers, a function may be neither even nor odd.

f(x) can be either even, odd, or neither

Notice that if a function is even or odd, it is enough to make the graph only for positive values because the graph for negative values will be its reflection either across the

y -

axis or across the origin.

Pop Quiz

Classifying Functions

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

Closure

Verifying End Behavior

At the beginning of the lesson, Emily graphed the polynomial function $f (x) = x^{4} - 4 x^{3} - 39 x^{2} + 46 x + 80$ using a table of values, but she was not so sure about the end behavior of the graph. She concluded that the end behavior was "Up-Down."

Since Emily learned that the end behavior of the graph of a polynomial can be deduced from the function rule, she can now verify whether her gut was correct or not.

f (x) = x^{4}_{1} - 4 x^{3} - 39 x^{2} + 46 x + 80 ⇓ Leading Coefficient : 1 Degree : 4

The leading coefficient is positive and the degree is even. Therefore, the end behavior is up-up. Consequently, Emily's graph was not correct. She had to extend the right part upward, and not downward.

Graph of f(x)=x^4 - 4x^3 - 39x^2 + 46x + 80.

Notice that if Emily had determined the end behavior of the graph from the function rule first, she probably would have made a table of values that included whole numbers greater than

6 .

This would have led her to graph

y = f (x)

correctly in her first attempt.

Level 1

Level 2

Level 3

Describe the end behavior of each of the following graphs.

Graph of f(x)=(x+3.5)*(x-0.75)*(x-4.5)/15

Graph of f(x)=(x+4)*(x+1.75)*(x-2)*(x-3)/15

Graph of f(x)=-(x+4.25)*(x+2.5)*(x)*(x-4)^2/100

Let's take a look at the ends of the given graph. We can see that the left-end arrow points downward, while the right-end arrow points upward.

Therefore, the left-end behavior of the graph is down and the right-end behavior is up. Consequently, the end behavior of the given graph is down-up.

As we did in Part A, let's consider the ends of the graph. This time both the left-end arrow and the right-end arrow point upward.

Therefore, both the left-end behavior and the right-end behavior of the graph are up. Consequently, the end behavior of the given graph is up-up.

Let's check the ends of the final graph like we did in the previous parts. This time we can see that the left-end arrow points upward while the right-end arrow points downward.

Therefore, the left-end behavior of the graph is up and the right-end behavior is down. Consequently, the end behavior of the given graph is up-down.

Consider the following polynomial function.

f (x) = x^{4} + 10 x^{3} - 38 x^{2} - 77 x + 164

According to the Location Principle, is

f

guaranteed to have a zero in the interval

[2, 3] ?

In which of the following intervals is

f

guaranteed to have at least one zero?

We can use the Location Principle to determine whether f is guaranteed to have a zero in the interval [ 2, 3]. This principle states that if the polynomial function has different signs at the endpoints of an interval, then the function has at least one zero on that interval. Let's start by evaluating f(x) at x= 2.

Similarly, let's evaluate f(x) at x= 3.

After evaluating the function f at the endpoints of the interval [ 2, 3] we got two negative outputs. f( 2) < 0 and f( 3) < 0 Since f( 2) and f( 3) have the same sign, we cannot use the Location Principle. Therefore, we cannot guaranteed that f has a zero in the interval [ 2, 3].

As in Part A, we can use the Location Principle to determine if f could have a zero in the given intervals. Let's start by listing the intervals. I_1 &= [-13,-12] & I_4 &= [0,1] I_2 &= [-12,-10] & I_5 &= [1,2] I_3 &= [-3,-2] & I_6 &= [3,5] Next, let's evaluate the function at each of the endpoints of the intervals. We can being by finding f(-13).

We can evaluate f(x) at the remaining endpoints by following the same procedures for each value. The following table summarizes the results.

Interval	Endpoint	Evaluate	Sign
I_1 = [-13,-12]	-13	f(-13)=1334	Positive
I_1 = [-13,-12]	-12	f(-12)=-928	Negative

I_2 = [-12,-10]	-12	f(-12)=-928	Negative
I_2 = [-12,-10]	-10	f(-10)=-2866	Negative

I_3 = [-3,-2]	-3	f(-3)=-136	Negative
I_3 = [-3,-2]	-2	f(-2)= 102	Positive

I_4 = [0,1]	0	f(0)=164	Positive
I_4 = [0,1]	1	f(1)= 60	Positive

I_5 = [1,2]	1	f(1)=60	Positive
I_5 = [1,2]	2	f(2)= -46	Negative

I_6 = [3,5]	3	f(3)=-58	Negative
I_6 = [3,5]	5	f(5)= 704	Positive

According to the Location Principle, f has at least one zero in [a,b] if f(a) and f(b) have different signs. Therefore, let's look for the intervals in which the outputs of the endpoints have opposite signs. From the table, we can identify four intervals meeting this condition. I_1 &= [-13,-12] & ✓ & & I_4 &= [0,1] * I_2 &= [-12,-10] & * & & I_5 &= [1,2] ✓ I_3 &= [-3,-2] & ✓ & & I_6 &= [3,5] ✓ From these results, we can guarantee that f has at least one zero in the intervals [-13,-12], [-3,-2], [1,2], and [3,5].

Consider the following graph of a polynomial function.

Graph of f(x)=(x^6/6 + (2*x^5)/5 - (25*x^4)/4 - (26*x^3)/3 + 60*x^2)/29.6 + 1

Which of the following statements about the graph of

f

are true? Select all that apply.

I . II . III . IV . It has five relative extrema . It has both absolute minimum and absolute maximum . It has an absolute minimum but not an absolute maximum . It has an absolute maximum but not an absolute minimum .

List the

x -

values at which the graph of

f

reaches a local maximum.

List the

x -

values at which the graph of

f

reaches a local minimum.

List all the relative minima of the graph of

f .

Let's start by recalling that the phrase relative extrema involves both relative maximums and relative minimums. Therefore, we want to count the number of relative maximums and minimums in the graph of f(x). At relative extrema, the graph changes from increasing to decreasing or vice versa.

From the graph, we get the following information.

The graph changes its behavior five times, so it has five relative extrema.
The graph of f(x) increases indefinitely, so it does not have an absolute maximum.
The graph of f(x) is always greater than or equal to -2.5, so it has an absolute minimum.

Comparing the information we gathered to the given statements, we can say that only I and III are true.

Item	Statement	True/False
I	It has five relative extrema.	True
II	It has both absolute minimum and absolute maximum.	False
III	It has an absolute minimum but not an absolute maximum.	True
IV	It has an absolute maximum but not an absolute minimum.	False

At a local maximum, the graph of a function changes from increasing to decreasing. In other words, the local maximums are the y-coordinates of the tops of the hills in the graph of a function.

As we can see, the graph has two hills and, therefore, it has two local maximums. Let's identify the x-coordinate of each point.

The x-values at which f reaches a local maximum are -3 and 2.

While we looked for hills in Part B, this time we are going to look for the valleys in the graph. This is because at a local minimum, the graph of a function changes from decreasing to increasing. The local minimums are the y-coordinates of the bottoms of the valleys in the graph of a function.

As we can see, the graph has three valleys, so it has three local minimums. Let's identify the x-coordinate of each point.

The x-values at which f reaches a local minimum are -5, 0, and 4.

From Part C, we know that the graph of f(x) reaches a local minimum at x=-5, 0, and 4. This time, we were asked to state the local minimums. This means that we will look at the y-coordinates of the points labeled in Part C.

As we can see, local minimums of the graph of f are -2.5, 1, and 2.

Consider the following graph of a polynomial function.

Graph of f(x)=87.75*x - 32.25*x^2 - 9.41667*x^3 + x^4 + 0.2*x^5

How many turning points does the graph have?

Which of the following points correspond to turning points of the graph of

f ?

Select all that apply.

A turning point is a point where the graph of a function changes from increasing to decreasing or vice versa. Let's begin by identifying where the function increases and where it decreases.

As we can see there are four points where the graph changes its behavior. Therefore, the graph has 4 turning points.

Let's label each of the given points. ll A(-6.5,118) & D(1,47) B(-6,87) & E(4,-307) C(-3,-200) & F(4.5,-337) From the graph, we can tell that the x-coordinates of the turning points are x_1=-6.5, x_2=-3, x_3=1, and x_4=4.5. With this information, we can discard those points whose x-coordinates are not included in this list. llll A(-6.5,118) & & D(1,47) & B(-6,87) & * & E(4,-307) & * C(-3,-200) & & F(4.5,-337) & Of the remaining points, we need to verify that their y-coordinates correspond to those of the turning points. Let's plot these points along with the graph.

As we can see, point C is not on the graph, meaning we can discard it. Notice that points A, D, and F are on the graph and correspond to turning points. This means that of the six given points, only these three are turning points. llll A(-6.5,118) & ✓ & D(1,47) & ✓ B(-6,87) & * & E(4,-307) & * C(-3,-200) & * & F(4.5,-337) & ✓ The fourth turning point (-3,-267) is not listed among the choices.

Determine whether the given graphs are even, odd, or neither.

Graph of h(x)=(x+4)^2*(x-4)^2*(x+2)*(x-2)/500

Let's begin by checking whether the graph is even. If a function is even, then the y-axis should act like a mirror. However, we can see that the point (0.5,3) is on the graph but the point (-0.5,3) — its reflection on the y-axis — is not on the graph.

Therefore, we know that the graph is not even. Now let's check if the graph is odd. A function is odd if its graph is symmetric about the origin. This implies that if the point (x,y) is on the graph, then (-x,-y) must also be on the graph. However, we can see that (1.5,2) is on the graph but (-1.5,-2) is not.

This means that the graph is not odd. With this information, we can conclude that the given graph is neither even nor odd.

As we did in Part A, let's begin by checking if the given graph is even. We can see that the y-axis does not act like a mirror because the point (0.5,1.5) is on the graph but its reflection on the y-axis (- 0.5,1.5) is not.

This is how we know that the graph is not even. To check if the graph is odd, let's pick some arbitrary points on the graph and reflect them about the origin. We have to verify that the reflections are on the graph.

We can see that the reflections about the origin are also in the graph. Furthermore, this happens to every point on the graph we pick. As such, the given graph is odd. We can also confirm this claim by rotating the graph 180^(∘) about the origin.

At a glance, we can see that the given graph seems to be symmetric about the y-axis. If this is the case, the graph is even. Let's verify this assumption by picking some points on the graph and reflecting them in the y-axis.

As we can see, all the reflections are also in the graph. This happens to every point on the graph we pick, so we can say that the given graph is even. Since a graph cannot be even and odd at the same time, we do not need to verify if it is odd. In fact, it is not odd!

Polynomial Functions and Their Graphs

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