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Algebra 2 Analyzing Graphs of Polynomial Functions

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Section 6: Polynomial Functions: Lecture 5 | 33:29 min

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1 answer

Last reply by: Hong Yang
Sun Nov 3, 2019 8:22 AM

Post by Hong Yang on November 3, 2019

fokmecok

2 answers

Last reply by: Hong Yang
Sun Nov 3, 2019 8:20 AM

Post by Daniel Cuellar on October 30, 2012

by looking at a polynomial of the degree of n, can we have "at most" n-1 relative max & mins? or is it that we WILL have n-1 relative max & mins. Thanks

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Analyzing Graphs of Polynomial Functions

If f(a) < 0 and f(b) > 0, then f has a zero between a and b.
The graph of a polynomial function of degree n has at most n – 1 local maximums and local minimums.

Analyzing Graphs of Polynomial Functions

Graph f(x) = x³ − 4x

Find the zeros
0 = x(x² − 4) = x(x − 2)(x + 2)
Using the Zero Product Property you'll have three equations
x = 0, x − 2 = 0, and x + 2 = 0
Solve and plot the roots/zeros
x = 0, x = 2, and x = − 2
Plot Some Points
x -3 -1 1 3
y
x -3 -1 1 3
y -15 3 -3 15
Sketch graph using end behavior by using the table below

End Behavior Chart	Even Degree	Odd Degree
+ leading coefficient	1. As x approaches +∞, y approaches +∞	1. As x approaches +∞, y approaches +∞
	2. As x approaches -∞, y approaches +∞	2. As x approaches -∞, y approaches -∞
- leading coefficient	1. As x approaches +∞, y approaches -∞	1. As x approaches +∞, y approaches -∞
	2. As x approaches -∞, y approaches -∞	2. As x approaches -∞, y approaches +∞

Since the graph is of Odd Degree and Leading Coefficient is positive, the graph will have the following properties:
As x approaches + ∞ , y approaches + ∞
As x approaches − ∞ , y approaches − ∞

Graph f(x) = x³ − 3x² − 4x

Find the zeros
0 = x(x² − 3x − 4) = x(x − 4)(x + 1)
Using the Zero Product Property you'll have three equations
x = 0, x − 4 = 0, and x + 1 = 0
Solve and plot the roots/zeros
x = 0, x = 4, and x = − 1
Plot Some Points
x -2 2 3 5
y
x -2 2 3 5
y -12 -12 -12 30
Sketch graph using end behavior by using the table below

End Behavior Chart	Even Degree	Odd Degree
+ leading coefficient	1. As x approaches +∞, y approaches +∞	1. As x approaches +∞, y approaches +∞
	2. As x approaches -∞, y approaches +∞	2. As x approaches -∞, y approaches -∞
- leading coefficient	1. As x approaches +∞, y approaches -∞	1. As x approaches +∞, y approaches -∞
	2. As x approaches -∞, y approaches -∞	2. As x approaches -∞, y approaches +∞

Since the graph is of Odd Degree and Leading Coesfficient is positive, the graph will have the following properties:
As x approaches + ∞ , y approaches + ∞
As x approaches − ∞ , y approaches − ∞

Graph f(x) = − x³ − x² + 6x

Find the zeros
0 = − x(x² + x − 6) = x(x − 2)(x + 3)
Using the Zero Product Property you'll have three equations
x = 0, x − 2 = 0, and x + 3 = 0
Solve and plot the roots/zeros
x = 0, x = 2, and x = − 3
Plot Some Points
x -4 -2 1 3
y
x -4 -2 1 3
y 24 -8 4 -18
Sketch graph using end behavior by using the table below

End Behavior Chart	Even Degree	Odd Degree
+ leading coefficient	1. As x approaches +∞, y approaches +∞	1. As x approaches +∞, y approaches +∞
	2. As x approaches -∞, y approaches +∞	2. As x approaches -∞, y approaches -∞
- leading coefficient	1. As x approaches +∞, y approaches -∞	1. As x approaches +∞, y approaches -∞
	2. As x approaches -∞, y approaches -∞	2. As x approaches -∞, y approaches +∞

Since the graph is of Odd Degree and Leading Coesfficient is Negative, the graph will have the following properties:
As x approaches + ∞ , y approaches − ∞
As x approaches − ∞ , y approaches + ∞

Graph f(x) = x⁴ − 5x² + 4

Find the zeros
0 = (x²)² − 5(x)² + 4
We know how to factor x² − 5x + 4 = 0
x² − 5x + 4 = 0
(x − 4)(x − 1) = 0 Since what we really want is to factor out is the polynomial to the 4th power,
all we have to do is replace the x with x²
(x² − 4)(x² − 1) = 0
Here we have difference of squares, so we'll use that pattern to factor out completely.
(x − 2)(x + 2)(x − 1)(x + 1) = 0
Using the Zero Product Property you'll have 4 equations
x − 2 = 0, x + 2 = 0, x − 1 = 0, and x + 1 = 0
Solve and plot the roots/zeros
x = 2, x = − 2, x = 1, and x = − 1
Plot Some Points
x -3 −[3/2] 0 [3/2] 3
y
x -3 −[3/2] 0 [3/2] 3
y 40 -2.19 4 -2.19 40
Sketch graph using end behavior by using the table below

End Behavior Chart	Even Degree	Odd Degree
+ leading coefficient	1. As x approaches +∞, y approaches +∞	1. As x approaches +∞, y approaches +∞
	2. As x approaches -∞, y approaches +∞	2. As x approaches -∞, y approaches -∞
- leading coefficient	1. As x approaches +∞, y approaches -∞	1. As x approaches +∞, y approaches -∞
	2. As x approaches -∞, y approaches -∞	2. As x approaches -∞, y approaches +∞

Since the graph is of Even Degree and Leading Coesfficient is Positive, the graph will have the following properties:
As x approaches + ∞ , y approaches + ∞
As x approaches − ∞ , y approaches + ∞

Graph f(x) = − x⁴ + 3x² − 2

Find the zeros
0 = − ((x²)² − 3(x)² + 2)
We know how to factor x² − 3x + 2 = 0
x² − 3x + 2 = 0
(x − 2)(x − 1) = 0 Since what we really want is to factor out is the polynomial to the 4th power,
all we have to do is replace the x with x²
− (x² − 2)(x² − 1) = 0
Here we have difference of squares, so we'll use that pattern to factor out completely.
− (x² − 2)(x − 1)(x + 1) = 0
Using the Zero Product Property you'll have 4 equations. For the first equation, solve by taking the square root of both sides.
x² − 2 = 0, x − 1 = 0, and x + 1 = 0
Solve and plot the roots/zeros
x = + √2 = 1.4, x = − √2 = − 1.4, x = 1, and x = − 1
Plot Some Points
x -2 −[5/4] 0 [5/4] 2
y
x -2 −[5/4] 0 [5/4] 2
y -6 [1/4] -2 [1/4] -6
Sketch graph using end behavior by using the table below

End Behavior Chart	Even Degree	Odd Degree
+ leading coefficient	1. As x approaches +∞, y approaches +∞	1. As x approaches +∞, y approaches +∞
	2. As x approaches -∞, y approaches +∞	2. As x approaches -∞, y approaches -∞
- leading coefficient	1. As x approaches +∞, y approaches -∞	1. As x approaches +∞, y approaches -∞
	2. As x approaches -∞, y approaches -∞	2. As x approaches -∞, y approaches +∞

Since the graph is of Even Degree and Leading Coesfficient is Negative, the graph will have the following properties:
As x approaches + ∞ , y approaches − ∞
As x approaches − ∞ , y approaches − ∞

Graph f(x) = x⁶ − 5x⁴ + 4x²

Find the zeros
0 = x²(x⁴ − 5x² + 4) = x²((x²)² − 5(x)² + 4)
We know how to factor x² − 5x + 4 = 0
x² − 5x + 4 = 0
(x − 4)(x − 1) = 0 Since what we really want is to factor out is the polynomial to the 4th power,
all we have to do is replace the x with x²
x²(x² − 4)(x² − 1) = 0
Here we have difference of squares, so we'll use that pattern to factor out completely.
x²(x − 2)(x + 2)(x − 1)(x + 1) = 0
Using the Zero Product Property you'll have 5 equations.
x² = 0, x − 2 = 0, x + 2 = 0, x − 1 = 0, and x + 1 = 0
Solve and plot the roots/zeros
x = 0, x = 2, x = − 2, x = 1, and x = − 1
Plot Some Points
x -2.5 -1.5 -0.5 0.5 1.5 2.5
y
x -2.5 -1.5 -0.5 0.5 1.5 2.5
y 73.83 -4.92 0.7 0.7 -4.92 73.83
Sketch graph using end behavior by using the table below

End Behavior Chart	Even Degree	Odd Degree
+ leading coefficient	1. As x approaches +∞, y approaches +∞	1. As x approaches +∞, y approaches +∞
	2. As x approaches -∞, y approaches +∞	2. As x approaches -∞, y approaches -∞
- leading coefficient	1. As x approaches +∞, y approaches -∞	1. As x approaches +∞, y approaches -∞
	2. As x approaches -∞, y approaches -∞	2. As x approaches -∞, y approaches +∞

Since the graph is of Even Degree and Leading Coesfficient is Positive, the graph will have the following properties:
As x approaches + ∞ , y approaches + ∞
As x approaches − ∞, y approaches + ∞

Graph f(x) = x³ − 2x² − 2x + 4

Find the zeros, factor by grouping
0 = (x³ − 2x²) + ( − 2x + 4)
0 = x²(x − 2) − 2(x − 2)
0 = (x² − 2)(x − 2)
Using the Zero Product Property you'll have 3 equations.
x² − 2 = 0 and x − 2 = 0
Solve and plot the roots/zeros. You must use the square root to solve the first equation.
x = + √2 = 1.41, x = − √2 = − 1.41, and x = 2
Plot Some Points
x -2 -1 0 1 3
y
x -2 -1 0 1 3
y -8 3 4 1 7
Sketch graph using end behavior by using the table below

End Behavior Chart	Even Degree	Odd Degree
+ leading coefficient	1. As x approaches +∞, y approaches +∞	1. As x approaches +∞, y approaches +∞
	2. As x approaches -∞, y approaches +∞	2. As x approaches -∞, y approaches -∞
- leading coefficient	1. As x approaches +∞, y approaches -∞	1. As x approaches +∞, y approaches -∞
	2. As x approaches -∞, y approaches -∞	2. As x approaches -∞, y approaches +∞

Since the graph is of Odd Degree and Leading Coesfficient is Positive, the graph will have the following properties:
As x approaches + ∞ , y approaches + ∞
As x approaches − ∞ , y approaches − ∞

Sketch the graph of f(x) = − x³ + x² + 7x − 7 by finding the zeros and using the end - behavior only

Find the zeros, factor by grouping
0 = ( − x³ + x²) + (7x − 7)
0 = − x²(x − 1) + 7(x − 1)
0 = (7 − x²)(x − 1) = − (x² − 7)(x − 1)
Using the Zero Product Property you'll have 3 equations.
x² − 7 = 0 and x − 1 = 0
Solve and plot the roots/zeros. You must use the square root to solve the first equation.
x = + √7 = 2.65, x = − √7 = − 2.65, and x = 1
Analyze the end behavior and use it to sketch the graph.

End Behavior Chart	Even Degree	Odd Degree
+ leading coefficient	1. As x approaches +∞, y approaches +∞	1. As x approaches +∞, y approaches +∞
	2. As x approaches -∞, y approaches +∞	2. As x approaches -∞, y approaches -∞
- leading coefficient	1. As x approaches +∞, y approaches -∞	1. As x approaches +∞, y approaches -∞
	2. As x approaches -∞, y approaches -∞	2. As x approaches -∞, y approaches +∞

Since the graph is of Odd Degree and Leading Coefficient is Negative, the graph will have the following properties:
As x approaches + ∞, y approaches − ∞
As x approaches − ∞, y approaches + ∞
With your pencil, start somewhere high on Quadrant II and draw a smooth curve going down passing through
the zero ( − 2.65,0) going down and then start going up going through the zero (1,0) and then start going down and pass through the
last zero at (2.65,0) until you are in Quadrant IV. Your lines should be smooth and the graph must pass through those three points.
The rest is just a sketch.
Compare your graph with the actual graph. How close/far were you?

Sketch the graph of f(x) = x³ − 6x² − 4x + 24 by finding the zeros, f(0), and end - behavior only.

Find the zeros, factor by grouping
0 = (x³ − 6x²) + ( − 4x + 24)
0 = x²(x − 6) − 4(x − 6)
0 = (x² − 4)(x − 6)
Using the difference of Squares, factor completely
0 = (x − 2)(x + 2)(x − 6)
Using the Zero Product Property you'll have 3 equations.
x − 2 = 0, x + 2 = 0, adn x − 6 = 0
Solve and plot the roots/zeros.
x = 2, x = − 2, and x = 6
Find f(0) and plot it
f(0) = (0)³ − 6(0)² − 4(0) + 24 = 24
Analyze the end behavior and use it to sketch the graph.

End Behavior Chart	Even Degree	Odd Degree
+ leading coefficient	1. As x approaches +∞, y approaches +∞	1. As x approaches +∞, y approaches +∞
	2. As x approaches -∞, y approaches +∞	2. As x approaches -∞, y approaches -∞
- leading coefficient	1. As x approaches +∞, y approaches -∞	1. As x approaches +∞, y approaches -∞
	2. As x approaches -∞, y approaches -∞	2. As x approaches -∞, y approaches +∞

Since the graph is of Odd Degree and Leading Coefficient is Positive, the graph will have the following properties:
As x approaches + ∞ , y approaches + ∞
As x approaches − ∞, y approaches − ∞
With your pencil, start somewhere low on Quadrant III and draw a smooth curve going up passing through
the zero ( − 2,0) going up and then start going down as you pass (0,24), then continue going down as you pass through (2,0) and continue going down about
24 down then start going up as you cross (6,0) and continue going up until you're in Quadrant I. . Your lines should be smooth and the graph must
pass through those 4 points. The rest is just a sketch.
Compare your graph with the actual graph. How close/far were you?

Sketch the graph of f(x) = − x⁴ + 5x² − 4 by finding the zeros, f(0), and end - behavior only.

Find the zeros, imagine you're factoring x² and then replace x with x² as in previous examples
0 = − (x⁴ − 5x² + 4)
0 = − (x − 4)(x − 1)
0 = − (x² − 4)(x² − 1)
Using the difference of Squares, factor completely
0 = − (x − 2)(x + 2)(x − 1)(x + 1)
Using the Zero Product Property you'll have 4 equations.
x − 2 = 0, x + 2 = 0, x − 1 = 0, and x + 1 = 0
Solve and plot the roots/zeros.
x = 2, x = − 2, x = 1, and x = − 1
Find f(0) and plot it
f(0) = − (0)⁴ + 5(0)² − 4 = − 4
Analyze the end behavior and use it to sketch the graph.

End Behavior Chart	Even Degree	Odd Degree
+ leading coefficient	1. As x approaches +∞, y approaches +∞	1. As x approaches +∞, y approaches +∞
	2. As x approaches -∞, y approaches +∞	2. As x approaches -∞, y approaches -∞
- leading coefficient	1. As x approaches +∞, y approaches -∞	1. As x approaches +∞, y approaches -∞
	2. As x approaches -∞, y approaches -∞	2. As x approaches -∞, y approaches +∞

Since the graph is of Even Degree and Leading Coefficient is Negative, the graph will have the following properties:
As x approaches + ∞ , y approaches − ∞
As x approaches − ∞ , y approaches − ∞
With your pencil, start on Quadrant III and go up and pass through ( − 2,0) and continue going up a little bit then start
coming down and go through point ( − 1,0). Continue going down and then start going up as you cross (0, − 4) heading towards (1,0) going up.
Continue going up and then start going down as you cross the point (2,0) and continue going down. You should end on Quadrant IV.
Compare your graph with the actual graph. How close/far were you?

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Answer

Analyzing Graphs of Polynomial Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Graphing Polynomial Functions

Example: Table and End Behavior

Location Principle

Maximum and Minimum Points

Example 1: Graph Polynomial Function

Example 2: Graph Polynomial Function

Example 3: Graph Polynomial Function

Example 4: Graph Polynomial Function

Intro 0:00
Graphing Polynomial Functions 0:11

Example: Table and End Behavior

Location Principle 4:43

Zero Between Two Points
Example: Location Principle

Maximum and Minimum Points 8:40

Relative Maximum and Relative Minimum
Example: Number of Relative Max/Min

Example 1: Graph Polynomial Function 11:57
Example 2: Graph Polynomial Function 16:19
Example 3: Graph Polynomial Function 23:27
Example 4: Graph Polynomial Function 28:35

Algebra 2

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

Transcription: Analyzing Graphs of Polynomial Functions

Welcome to Educator.com.0000

In a previous lecture, we introduced the concept of the graphs of polynomial functions.0002

Now, we are going to go further and actually talk about how to develop those graphs.0006

OK, in order to obtain the graph of a polynomial function, you make a table of values,0012

connect these points with a curve, and then use information about the end behavior of the function.0017

Earlier on, when you had to graph linear and quadratic functions, you used the technique of making a table.0024

So, it is the same idea here, but with the additional concept of thinking about end behavior.0032

For example, if you were to graph a function such as f(x) = 3x³ - 2x² + x,0039

you would approach this by finding some x and f(x) values and graphing those out.0054

Then, look at the areas of the function that tell you about end behavior.0063

I am not going to go ahead and make the table on this one.0070

But let's say that you were to graph it out, and it came out something like this.0072

OK, using points, and also finding the zeroes, can help you graph.0080

In addition to just finding various points, sometimes you can approach making a graph of a polynomial function by also finding the zeroes0088

(and recall that the zeroes are where the graph crosses the x-axis, so they are the x-intercepts), and then finding some other points.0097

And then, you think about what happens to the graph at the ends.0109

Well, recall that, if you have an odd-degree polynomial, such as this one, the ends are going to go in different directions.0112

So, one will increase; the other will decrease; and then you have to look at the leading coefficient.0124

The leading coefficient: if it is positive (if it is greater than 0), as x increases, y will also increase.0131

So, this is odd degree with a positive leading coefficient: x and y are both increasing, and as x gets very, very small,0140

y is going to get very, very small, as well; so this is odd degree, where a of n, the leading coefficient, is greater than 0.0152

If you have an odd degree where the leading coefficient is negative, then the graph is going to pretty much be the opposite,0168

where you are going to have y decreasing (f(x) decreasing) as x gets very large.0177

And when x gets very small, f(x) is going to increase; this is odd degree with a negative value for the leading coefficient.0184

Let's look at even degrees right here.0199

If you are talking about a function with an even degree, such as the largest exponent of the fourth power or the sixth power,0207

so it's an even degree, both ends go in the same direction--they both face up, or they both face down.0222

And the leading coefficient tells you which direction that is.0228

With a leading coefficient that is positive, you are going to get a graph where both ends are facing up.0231

So, as x gets very large, or as x gets very small, y is going to get very large; this is even with a leading coefficient that is positive.0238

Now, if you have an even degree where the leading coefficient is negative, then both ends will face down.0249

So, this could be an example of even with a negative leading coefficient.0260

This is review from a previous lecture; but recall that, in addition to finding what is going on0265

here in the middle of the graph, and finding your zeroes--finding various points--0270

you also need to know about end behavior in order to develop a graph of a polynomial function.0275

OK, I mentioned talking about finding the zeroes.0283

Well, sometimes it can be difficult or time-consuming to find the exact location of a zero.0286

But if you are just trying to sketch out a graph, you can use the location principle0292

to estimate where a zero will be--where a graph is going to cross the x-axis.0296

So, let's look at what this is saying: this is saying that, if you have a polynomial function f(x),0300

suppose that there are numbers a and b, such that f(a) < 0 and f(b) > 0.0306

Then, there is a zero, f(x), between these two points, a and b.0314

So, let's make this concrete: let's say that this is my point a--that I am going to let a equal 1.0318

And this is saying that if f(a) is less than 0 (I find f(a)--let's say it is somewhere down here: f(a) = -3), a is at 1, and f(a) is -3;0328

at (1,-3), that is (a,f(a)) on the graph; there is a negative value to the function.0349

OK, now let's say I find another value b, and let b equal 3.0362

So, here I had a equal 1, and let b equal 3; f(a) = -3; let's let f(b) equal 4.0367

Now, what this is telling me is that, somewhere between this point and this point,0382

we went from a negative value of the function to a positive value of the function.0394

That tells me that there has to be a zero somewhere...actually, it could even be over here; but somewhere between 1 and 3, this is going to cross.0398

So, I don't know that it is exactly here; but I know that somewhere between 1 and 3, there is a zero,0410

at an x-value somewhere between x = 1 and x = 3.0420

And I know that because the function changed from a negative value to a positive value.0432

Now, there actually could be more than one zero between those points.0442

For example, I could have a graph right here that goes like this.0446

And if I just graphed this point out here, I would see that the function is negative at this point, at -4: at f(-4), I have a negative value.0452

And then I say, "OK, let's say over here, at f(-2), I have a positive value."0464

I know that there is a zero in here; but as it turns out, there is actually more than one zero.0474

I know that there is at least one zero here, but there also could be more than one.0476

So, when you go ahead and make your table of values, x and f(x), if you see that you have,0480

say, positive, positive, and then it goes to negative, I know that at an x-value somewhere between here and here, there is a zero.0487

And then, I might have negative and negative, and then switch to positive again.0498

And I know that, in here somewhere, there is a zero.0501

So again, the idea is that, if you see the value of the function switch from positive to negative,0508

or negative to positive, you know that the graph has to pass through the x-axis, and that there is a zero in that location.0512

OK, maximum and minimum points: recall that, when we worked with quadratic functions, we talked about maximums and minimums.0521

And here, today, with polynomials, we are talking about relative maximums.0528

When we talked about quadratics that were downward-facing parabolas, we said, "OK, the vertex is a maximum."0533

We didn't say relative; we just said maximum, because this is the highest value that the function could achieve.0540

If we were dealing with a minimum value, we didn't say relative; we just said, "OK, this is the minimum,"0546

because this is the smallest value that the function could achieve.0552

However, when we talk about more complicated, or higher-degree, polynomials, we talk about relative maximums and minimums.0556

And what we mean is relative to the x-values around those.0564

For example, if I have a graph that looks like this, I could say that this right here--this point (we will call that a,0569

and then, right here, the y-value is going to be f(a)--f(a) would be right there,0581

so this point is going to be (a,f(a)))--this is a relative maximum.0589

And the reason it is a relative maximum is: if I look at any value of x around in this region, the function value is lower than it is right here.0597

So, it is the maximum, relative to the values of x near this point.0607

A relative minimum is the same idea: if I look at this point here, this gives me a relative minimum.0613

And that means that this value of the function (I will call this f(b), so this would be b, and then the y-value would be f(b),0621

so this is at (b,f(b)))--this is the smallest value for the function in that region, for these x's.0636

But you see here: the graph actually goes lower way over here.0645

So, it is not an absolute minimum; it is just relative to that region of the graph.0648

And there might be a maximum that is higher up at some area in the graph; but this is a relative maximum for that region.0652

The other important point is that the graph of a polynomial function of a certain degree,0660

degree n, can have at most n - 1 points that are relative maximums or relative minimums.0665

So, if I have a function--let's say f(x) = 4x⁵ + 3x³ - 2x + 6,0672

then here, the degree, n is 5 in this case, so the maximum number of relative maximums and relative minimums is 5 - 1, which equals 4.0682

So, in a graph like this, degree 5, I can have at most 4 of these maximums and minimums.0707

OK, putting together what we have discussed about finding points, the zeroes, using end behavior,0718

and relative maximums and minimums, we can develop a graph of some polynomials.0724

OK, looking at this first one, this is x³ - 2x; and I am just going to start this one by plotting out some points.0729

OK, so let's let x equal -2; this would equal -8; and then, -2 times -2 is 4; -8 + 4 is going to give me -4.0739

Here I have -1 for my x-value, and that is going to give me a -1 here, and then -1 times -2 is going to give me 2, so this will be 1.0756

Now, notice something: my value for the function switched from negative to positive.0767

That means that somewhere between -2 and -1, there is a zero.0774

Using the location principle, I know that, somewhere between x = -2 and x = -1, this graph of this function crosses the x-axis.0781

When x is 0, f(x) is 0; well, if f(x) is 0, then this point lies on the x-axis, and I actually found a zero--this is a zero.0793

Now, when x is 1, that gives me 1, minus 2; that is -1.0808

When x is 2, 2³ is 8, minus 2 times 2 (that is minus 4)--that is going to give me 4.0815

Oh, and notice also, here, another zero.0823

There is a zero between these values (between the corresponding x-values); somewhere between 1 and 2, there is a zero.0835

I have a zero between -2 and -1; I know I have a zero right here; and then, I have another zero between 1 and 2.0841

So, I am going to go ahead and plot these out; when x is -2, the function is -4.0848

When x is -1, the function is 1; when x is 0, the value of the function is 0;0854

when x is 1, we have a -1 right here; and when x is 2, the function value is 2.0862

I can see, between here and here: this had to cross the x-axis in between here and here.0870

Now, I connect these with a smooth line; I found my points; I know that there is a zero0875

somewhere in here; that there is one here; and that there is one here.0889

Now, I am going to think about end behavior; and the degree equals 3, and that is odd.0894

The leading coefficient, a of n, is 1, so that is positive.0902

And what that tells me is: it confirms that my graph is correct, because if it is an odd degree, the two ends are going to face in different directions.0910

And since it is a positive leading coefficient, that means that, as x gets very large, y gets very large.0920

As x gets very small, y gets very small.0927

The other thing I know is that the number of relative maximums and relative minimums is going to be one less than the degree of the polynomial.0930

And since the degree is 3, I can have, at most, 2 relative maximums and relative minimums.0939

And I see right here: I do have a relative maximum--this is the largest function value for the x-values in this region.0952

And right here, I have a relative minimum; so I found both relative maximums and minimums.0962

So, this is a pretty good graph of this polynomial, just based on using knowledge that I had and finding a few points on the graph.0969

In the second example, again, we have a polynomial of degree 3; but notice that it has a negative leading coefficient.0980

That is going to give the graph a different shape.0988

In the previous example, I just went ahead and graphed some points; I am going to take a slightly different approach here.0991

And I am going to find the zeroes; last time, I found one of the zeroes by luck at point (0,0),0997

and I also used the location principle to find the approximate location of two more zeroes.1003

Here, I am just going to go ahead and find them directly.1008

So, what I am going to do is look at the corresponding equation, -x³ + x² + 6x.1011

And I am going to set that equal to 0 to find values of x for which the function's value is 0; that will give me the zeroes.1019

So, I am going to approach this by factoring; and I see that I have a common factor of x in all three terms.1028

And I am actually going to factor out -x, so what I have left behind is simpler to factor.1035

If I pull a -x here, I am going to have an x² left.1040

If I pull a -x from x², that will leave me -x, because -x times -x is positive x².1043

Here, I am also going to pull out a negative x, leaving behind -6, and checking that -x and -6 is + 6x.1052

OK, now I can look here and see that I can factor this further.1062

So, I know that this is going to have the general form (x + something) (x - something).1070

And I know that, because I have a negative sign here; so a positive and a negative is going to give me a negative.1077

Factors of 6 are 1 and 6, and 2 and 3; and I need those factors when one is positive and one is negative.1084

I am going to make one positive and one negative; I want them to add up to the middle term of -1.1096

Well, these two are too far apart; now, if I take -3 + 2, that is going to equal -1.1102

So, I know that this is the right set of factors, and that I want 3 to be negative and 2 to be positive.1110

If I use FOIL to check this, I get x² - 3x + 2x (is going to give me -x) + 2 times -3 (is going to give me -6).1116

So, this is factored out as far as it can go.1133

And then, recall the zero product property that says that, if, say, a times b equals 0, then either a equals 0 or b equals 0, or they both equal 0.1135

So, any of these factors could equal 0, and that would give me a total value of 0.1147

And don't forget about this -x; that is part of it; x + 2 could equal 0, or x - 3 could equal 0, to make this equation true.1154

Dividing both sides by -1 just gives me x = 0; here, subtracting 2 from both sides gives me x = -2.1166

And adding 3 to both sides, I get x = 3.1176

These are the zeroes; these are the points at which the graph crosses the x-intercept.1179

So, x equals 0; x equals -2; and x equals 3.1185

OK, now, to flesh out this graph some more, I am going to find some other points.1192

I found three points--I found my zeroes; now I am just going to find some other points on the graph in this region.1195

When x is -1, if you work this out, it comes out to f(x) is -4.1203

When x is 1, this will give you -1 + 1 + 6, so that would give you 6.1212

When x is 2, working this out, it will give you f(x) is 8.1221

All right, so when x is -1, f(x) is 4; when x is 1, f(x) is 6 (about up here).1229

And notice here that this tells me that there is a zero in here--that between x is -1 and x is 1, there is a zero right in here.1244

And I defined that zero between x is -1 and x is 1; there is a zero that lies in here.1254

Let's see, when x is 2, f(x) is way up here; it is going to be way up here somewhere--it is at 8, so we will just put it right there.1264

All right--oh, and when x is...actually, I wrote that incorrectly; when x is -1, this is going to be -4.1282

And that is why, when you get the switch from negative to positive, there is a zero right in here (in the switch from here to here).1294

So, I have some points graphed, and I have my zeroes; so I am going to go ahead and connect these points.1305

I am also going to use end behavior; and since this is an odd degree, one end is going to go up, and the other is going to go down.1324

And since it is a leading coefficient that is negative, then I am going to have--as x becomes very large, y would be very small;1336

and as x becomes very small, over here, y becomes very large.1349

So, end behavior helped me figure out these portions of the graph.1356

Also, I know that, because my degree is 3, the greatest number of relative maximums and minimums I can have is 2.1360

2 relative maximums and minimums at most--that is the greatest number I can have.1369

And I do have a relative minimum here and a relative maximum there.1376

So, graphing is based on finding the zeroes, graphing a few points, and then thinking about end behavior.1384

And notice, also, that here, when my graph switched from a value of -4 (a negative value for the function)1391

to a positive value, up here at 6, I knew that there had to be a zero in between those; and I did find that zero.1399

OK, I am going to approach this, also, by first finding the zeroes.1406

Let's work with the corresponding equation and set this equal to 0.1411

Now, looking at this, there are no common factors to pull out; and I have four terms, so I am going to factor by grouping.1417

Grouping means that we are going to put x³ - x² together, and add that to -3x + 3.1424

Now, looking at this, I have a common factor of x²; that leaves behind an x and a -1.1434

Here, I have a common factor of -3; I pull that out--that leaves an x behind here, and a -1, because -3 and -1 would give me the 3 back.1442

Now, I see that I have a common binomial factor of x - 1; I am going to pull that out in front.1455

When that is pulled out, that leaves behind x² - 3.1462

OK, using the zero product property, I know that x - 1 = 0 or x² - 3 = 0.1467

If either of these expressions is 0, then the equation will hold true; this left side will be equal to 0.1476

So, this is easy: just add 1 to both sides--that gives me x = 1; and then, over here,1485

I am going to add 3 to both sides, and that is going to give me x² = 3.1490

I am going to take the square root of both sides, and that is going to give me x = ±√3.1496

OK, so I have zeroes at x = 1, x = √3, and x = -√3.1504

Well, the square root of 3 can be estimated at about 1.7; so let's rewrite this as 1.7 and -1.7, so that we can graph it.1517

OK, I have a zero here at 1; I have a zero here at 1.7 (which is about there) and at -1.7 (which is right about there).1528

All right, again, I need to just find a few more points to get a complete graph.1539

I am going to let x equal -2; and if I plug that into here, I will get out a value of a function that is -3.1547

I am going to let x equal -1, and if you do the calculation on that and substitute -1 in here, you are going to find that y is 4.1559

Plotting that out: when x is -2, f(x), or y, is -3, right here.1570

When x is -1, f(x) is 4; and you see the location principle at work--that, since we switched from negative to positive,1580

that means that there is a zero somewhere in here, somewhere between x is -2 and x is -1.1588

And we already know that there is a zero between those two, and it is at -1.7.1597

OK, finding a few more points: let's let x equal 0; f(x) is going to be 3.1605

When x is 2, plugging that in here, we get a value for the function of 1.1612

So, when x is 0, f(x) is 3; when x is 2, f(x) is 1.1621

So now, we have enough points to get some sense of what is happening.1633

And I am going to start here and connect these.1639

Now, using end behavior, I have an odd degree (this is degree 3), which means the two ends are going to go in opposite directions.1653

And I have a positive leading coefficient, meaning that, over here on the right, as x gets very large, y is going to get large.1663

Over here on the left, as x gets very small, f(x), or y, is going to get very small.1670

I am also going to look and see that my degree is 3; so I have, at most, 2 relative maximums and relative minimums.1677

And I have a relative maximum here, and I have a relative minimum here; so I have them both.1690

So again, I am handling this by finding the zeroes to get a few points on the x-axis,1699

finding some more points to find the shape of the rest of this area of the graph,1704

and using end behavior to predict how the graph will look at large and small values.1709

So, let's take this and set it equal to 0.1715

All right, as far as factoring this, this is a trinomial; and we are used to working with these when they are quadratic equations, and the first term is x².1722

But it is really not that much different with x⁴, because x² times x² is x⁴.1730

So, I am looking here, and I see that I have a positive term here, and a negative here.1740

And the only way to get that is if I am multiplying a negative times a negative; that gives me a positive.1745

And yet, when I sum up the outer and inner terms after multiplying, I will get a negative term here.1751

Factors of 8 are 1 and 8, and 2 and 4; and I need them to add up to 6.1757

So, if I added -2 and -4, I am going to get -6; so I know that these are the factors that I want to use.1764

OK, I can't do anything more with this, so I am just going to leave this as it is.1774

However, this is the difference of two squares, so I can factor that out a bit more to this.1777

OK, so using the zero product property: x² - 2 = 0 or x - 2 = 0 or x + 2 = 0.1785

So, starting with these easier ones: x = 2, x = -2; I found two zeroes.1794

Now, I am going to look over here: x² = 2--taking the square root of both sides gives me x = ±√2.1803

All right, so I have four zeroes; I have zeroes at x = 2, -2, and then √2 is approximately 1.4...1813

so let's rewrite that as 1.4 and -1.4 to help me with the graph.1824

So, plotting those out: I have a zero here, at -2, and then another one at 2.1830

I have a zero here at 1.4 and -1.4.1836

I need a few more points to complete my graph; so right up here, let's find x and f(x).1843

When x is -1, if you work this out, you get 1; you get -6, because this x² would just be 1, plus 8; so it is going to give you 9 - 6, or 3.1852

When x is 0, that cancels; that cancels; that becomes 0; that leaves me with 8 right here.1866

And then finally, when x is 1, this will be 1, minus 6, plus 8; so again, I will have 3.1873

All right, so when x is -1, f(x) is 3; when x is 0, f(x) is way up here, about here, at 8 (8 would be right about here); and then, when x is 1, f(x) is 3.1882

I am going to connect these points; and I am also going to use my knowledge of end behavior.1905

And since this is an even degree, both ends are either going to point up or down.1909

And since this is a positive leading coefficient, both ends will actually be up.1915

So, when x is very small, y is going to be large; and then, I know that after this point, I am going to go up,1921

because when x is very large, y is going to be very large.1938

After this point, I am going to continue on up, because when x is small, f(x) is going to be very large,1942

because it is an even degree with a positive leading coefficient.1948

Now, I am checking that I have degree 4; since this is degree 4, the greatest number of relative maximums and minimums I am going to have are 3.1951

And I have a relative minimum there; I have a relative maximum here; and I have a relative minimum here.1964

So, I found all three relative maximums and minimums.1973

So again, finding the zeroes, I found 1, 2, 3, 4 zeroes.1976

Then, I plotted a few other points to give me the shape of the graph.1983

I used my knowledge of end behavior to figure out what is going on out here and here.1987

And then, I just verified that I had the generally correct shape by seeing that I would expect, at most, three relative maximums and minimums.1991

And I found that I had all three.1999

That concludes this session of Educator.com on analyzing the graphs of polynomials.2003

And I will see you again next lesson!2008

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Algebra and Trigonometry: Structure and Method, Book 2

Author: McDougal Littell

ISBN: 395977258

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Dr. Carleen Eaton

Analyzing Graphs of Polynomial Functions

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro