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Dedication

To our fabulous Pre-Calculus teacher Mrs. DeSimone whose help is always

inspiring!

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Table of Contents

Introduction .......................................................................................................................................... 3

Chapter 1 ................................................................................................................................................. 4

Graph Families ......................................................................................................................................... 4

Vocabulary .............................................................................................................................................. 5

Chapter 2 ................................................................................................................................................ 6

Concepts and Applications in Graphing ................................................................................................ 7

Chapter 3 ............................................................................................................................................... 15

Examples ............................................................................................................................................... 15

Example 1 - Quadratic............................................................................................................................ 15

Example 2 – Cubic that can be factored ............................................................................................... 17

Example 3 – Cubic that cannot be factored........................................................................................... 20

Chapter 4 ............................................................................................................................................... 21

Practice and Test ................................................................................................................................ 21

Test Answers ......................................................................................................................................... 22

About the Authors ........................................................................................................................... 24

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Introduction Dear Students

Analyzing graphs can be a very assiduous task for young learners. That is why we decided to

instruct all math students with this new comprehensive manual. When you have finished this

manual, you should be able to:

1. Identify a graph into types

2. Show understanding of graph behavior

3. Analyze the concepts behind graphs

To better your results with this manual you should practice the problems given at the end of

each section by following the instructions. It is greatly suggested that you complete the test at

the end of the manual to check your understanding.

Dear Teachers and Parents:

Choosing this manual to help students understand and analyze graphs is the right step to take.

This manual has clear instructions that a student who has taken Algebra and Geometry can

easily follow. You can assist the student if necessary with reinforcement of basic knowledge.

To see the electronic version of this manual go to this webpage, designed and build on graph

analyzing.

http://graph-analysis.tk

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Chapter 1

A graph is the visual representation of certain data’s behavior. While in Algebra 1 you focused

on linear relationships (i.e. y=2x+1), in more advanced classes there you will need to be able to

understand the behavior of other types of graphs. Below there is a chart representing graph

families.

Classification:

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This manual is focused in helping you understand quadratic and cubic equation and analysis of

their graphs. Before we start working on any concepts, let’s take a look at some important

vocabulary terms. It is strongly suggested that you know their meaning in order to be able to

understand later activities. Explanation will be given in the next chapter on how to perform

certain tasks and why they are important.

Vocabulary

End Behavior – the behavior of f(x) as x becomes very large

Intercepts – Points on the graph that touch the axis

X Intercept (Root) – Where the graph touches the X-Axis

Y intercept – Where the graph touches the Y-Axis

Critical Points – Points where the graph behavior changes. A curve may have the following

three types of critical points

Minimum – the curve changes from an increasing curve to a decreasing

Maximum – the curve changes from an increasing curve to a decreasing

Point of Inflection – point where the graph changes the direction of the curvature

Derivative - a derivative means how much a quantity is changing at some point. It is the slope of the

line tangent to the graph of a function at a point. Performing the following tests with help in graph

analysis:

First Derivative Test - determines the critical point of a function

Second Derivative Test- determines whether a critical point is a minimum, maximum, or point

of inflection.

Factoring – writing an algebraic expression as a multiplication problem

Factoring a Trinomial – writing a three-term-expression into two factors

Quadratic Formula – � � �����2�4�2 , used to find x in a quadratic equation.

Synthetic Division – method used to simplify a polynomial

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Chapter 2 Concepts and Applications in Graphing

There are four parts to completely analyzing a graph

1) Understand the general behavior of it.

2) Identify important points of the graph

3) Graph

1) Understanding the general behavior means to know the basic structure of the graph.

Important pieces of information about a graph are its end behavior and the intercepts.

End Behavior

End behavior is how the graph looks when it zoomed in a graphic calculator. The

following statements are going to be true for any equation:

Quadratic: if in the vertex form, y=a(x±h)2±k, a>0, the graph ends are going to show in

the 1st

and 2nd

quadrant, like the parent graph. If a<0, the end behavior is in 3rd

and 4th

quadrant.

Cubic: If in y=a(x±h)3±k, a>0, the end behavior is in the 1

st and 3

rd quadrant like the

parent graph of a cubic. If a<0, the end behavior is in the 2nd

and 4th

quadrant.

By analyzing end behavior you are going to have a simple idea on how your graph is

going to look.

Example: f(x)= 2x3+2x-2

The end behavior is in quadrant 1 and 3, because a=2 and 2>0.

f(x)=-2x3+2x-2

The end behavior is in quadrant 2 and 4 because a=-2 and -2<0

Practice:

What is the end behavior?

a) f(x)=2x3+4

b) q(x)=-.5x2-6x-1

c) p(x)=-x3+7x

2

d) g(x)=9x2+2x-4

Intercepts

Finding where the graph touches the axis is very important for an accurate sketch. To

find the point where any graph touches the Y-axis, you can solve the equation for x=0.

Example: f(x)=2x3+2x-2

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If x=0: f(0)=2(0)3+2(0)-2

=-2.

The Y-intercept of the graph whose equation is f(x)=2X3+2x-2, is y=-2.

X-Intercepts are also called the roots of a graph. To find a root therefore means to find

the x that when substituted in the equation make y=0. Factoring is the easiest method

to find the roots, most of the time.

Factoring

Factoring is the easiest way to find the roots of an equation, but it cannot be used

all the time.

For trinomials of the form: x2

± bx ± c

1) Take out the greatest common factor, if there is one, which will make things

easier later on.

x3-7x

2+6x = x(x

2-7x+6) Notice! The x that is factored out needs to be

brought along all the steps of the problem

2) Find the factors of the product of the constant and the leading coefficient in your

equation.

Example: x2-7x+6

Constant is 6, Leading coefficient is 1. 1*6=6 Factors of 6 are ±{ 1,2,3,6}

3) Find two factors that can be added or subtracted together to equal the coefficient of

the number of the degree one monomial.

Example: 6+1=-7

4) Make them an equation with parenthesis to show factors.

Example: (x+6)(x+1)

5) Use the foil method to double check your equation.

Example: (x+6)(x+1) = x2-7x+6

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Factoring Perfect Square trinomials: x2±bx±c

A perfect square trinomial is the square of a binomial, therefore to factor a perfect

square trinomial means to find the binomial that is being squared. These formulae can

help you go between the two ways:

(a+b)2 = a

2 + 2ab + b

2 (a-b)

2 = a

2 - 2ab + b

2

If you are given this equation and you need to factor it first check to see if the

expression is a perfect square trinomial. Then take a square root of first term and the

square root of last term.

If the sign of the middle term is positive add the two square roots together. If the sign of

the middle term is negative, subtract the square root of second term from the square

root of the first term. Then square the parenthesis.

Example: x2 + 6x + 9 (This expression is a perfect square trinomial, because the

first and the last term are perfect squares, and the middle term appears to be twice

the product of their square roots. These two conditions must be met in order to

proceed)

x2 + 6x + 9

√x2=x; √9=3

(x+3)2

Factoring: Difference between Two Squares

The two terms in the expression have to be perfect squares (a number whose square

root is an integer. Ex 1, 4, 9, 16, 25, 36, 49, 64, 81, 100)

To write the expression as factors follow this pattern: a2-b

2=(a+b)(a-b). So:

x2-9 can be written as x

2-3

2 = (x+3)(x-3)

Practice:

Factor the following expressions:

a) 2x2 + 5x + 3

b) 2x² + x – 15

c) x2

– 25

d) 3x3 - 5x

2 – 2x

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The processes of Synthetic Division, Factoring, and Quadratic Formula are used to help

find the roots of the equation.

Synthetic division can be used with any degree polynomial.

Synthetic Division is a way to test numbers to determine if they are possible roots for a

polynomial. You know that they are roots when the final column (usually the one that

represents the constant) solves out to be zero.

Steps are as follows:

1) Find the amount of roots of your polynomial by looking at the degree of the function.

This is called the Theorem of Algebra and its Corollary: A polynomial has as many roots

as its highest degree.

Example: x3-4x+6 has three roots because the highest degree term is x

3

2) Use the Rational Root Theorem to provide yourself with possible roots which you will

test first in your synthetic division chart. To do this take factors of the constant term

(called p) and divide them by the factors of the leading coefficient (called q) to form a

fraction.

Example: If your polynomial is x3-4x+6

~your constant (p) is 6 and its factors are ±{1,2,3,6}

~your leading coefficient (q) is 1, and its factors are ±1

* your possible (not necessarily definite!) roots are ±{1/1, 2/1, 3/1, 6/1} or

1, -1, 2, -2, 3, -3, 6, -6

- Take the highest (6) and lowest (-6) numbers, these show the

range of the possible roots

3) Use the Descartes Rule of Signs to determine what type of roots you have. To do this

you must count the amount of sign changes in your polynomial for both p(x) (to

determine the maximum number of positive roots) and p(-x) (the maximum number of

negative roots).

Example: Using the same polynomial as above

P(x)= x3-4x+6 ~ There are 2 sign changes (maximum of 2 positive roots)

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P(-x)= -x3+4x+6 ~ There is 1 change in signs (maximum of 1 negative roots)

Then construct a chart while integrating the idea of imaginary numbers

Play with different combinations of roots that add up to the total identified by the

Theorem of Algebra and its corollary.

Positive Negative Imaginary Total

2 1 0 3

0 1 2 3

Notice! How 2 roots from the positive column are “converted” to imaginary. You can

get imaginary roots by taking pairs from other columns and designating them into the

imaginary column.

4) Construct the beginning of your chart by taking the coefficient and constant and putting

them at the top of the columns. Also, put the possible roots that you found in step 2 in

the rows or down the r (roots) column from the smallest down to the largest.

Notice! The coefficients you are writing on the top of the column MUST go from the

highest degree to the lowest. If there is not a placeholder for any of the degrees put a 0

on that column. Also, it is always helpful to include 0 in your chart.

Lower Boundary= Upper Boundary=

r 1

x3 0

x2 -4

x1 6

x0

-6 1 -6 32 -192

-3 1 -3 5 -9

-2 1 -2 0 6

-1 1 -1 -3 9

0 1 0 -4 6

1 1 1 -3 3

2 1 2 0 6

3 1

6 1

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5) Multiply your next coefficient by r and then add it to the coefficient in your next

column. Then put the final product in this column. Continue doing this process until you

have filled out the entire row.

Look for trends in the chart, without completing the whole chart. The following can

narrow down your choices.

Lower Bound- Alternating signs across a row. This means there is no root lesser than

this r value.

Upper Bound- All positive values across a row. This means that there is no root greater

than this r value.

Try to locate the roots with these methods. Again, you know you have a root if the

remainder column is 0. Also, a change of the sign in the last column means that the root

is located between the two. You do not need to locate the specific decimal!

In the range of the roots explained earlier, you have found 1 negative root.

Because you did not find a positive real root, it means that there are 2 imaginary

for a total of 3.

Another Example: 2x3-3x

2+4x-1 has three roots because the highest degree term is

2x3 ~your constant (p) is 1 and its factors are ±1

~your leading coefficient (q) is 2, and its factors are ±{1,2}

* your possible (not necessarily definite) roots are ±{1/1, 1/2} or

1, -1, � , - �

P(x)= 2x3-3x

2+4x-1 ~ There are 3 sign changes (maximum of 3 positive roots)

P(-x)= -2x3-3x

2-4x-1 ~ There is no change in signs (maximum of 0 negative roots)

Positive Negative Imaginary Total

3 0 0 3

1 0 2 3

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r 2x3

-3x2

4x1

-1x0

-1 2 -5 9 -10

-½ 2 -4 6 -4

0 2 -3 4 -1

½ 2 -2 3 ½

1 2 -1 3 2

You still haven’t found the roots and you still haven’t found the upper boundary. This is

perfectly normal. Before you give up, put in numbers greater than 1 to try as roots till

you find the lower boundary.

r 2x3

-3x2

4x1

-1x0

-1 2 -5 9 -10

-½ 2 -4 6 -4

0 2 -3 4 -1

½ 2 -2 3 ½

1 2 -1 3 2

2 2 1 6 11

You just found an upper boundary. Now look of sign changes in the remainder column,

because it indicates a root is in between the two roots you have just tested, but it is not

an integer.

The sign change appears between 0< r < ½

Most likely you will not be asked to test roots that are not integers. This example was

given to show it can be done, but it can get messy in calculations. It was also a way to

show what you can do if you cannot locate any roots right away.

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Quadratic Formula is a simple formula that can be used to find the roots of a degree 2

polynomial.

The steps are as followed:

1) Find the A, B, and C of your equation using the Ax+Bx+C=0 as a reference.

2) Find the discriminant of your equation by using the formula b2-4ac. The discriminant

helps you find what type of roots you will get.

Discriminant Nature of Roots

b2-4ac > 0 two distinct real roots

b2-4ac < 0 exactly one real root

b2-4ac = 0 two distinct imaginary roots

3) Substitute you’re A,B, and C values into the equation –b+/- √(discriminant)

2a

*This will give you two values and these are your roots.

1. Identifying important points of the graph means to locate the critical points.

To find where these points are use the first derivative test, f’(x)=n*xn-1

Power Rule –

formula to find the derivative. After finding the first derivative, set it equal to zero and

solve for x. To determine what type of critical points you are dealing with do the Second

Derivative Test where you substitute the x of the critical point into the second

derivative. If f’’(x) > 0, the point is a minimum.

If f’’(x) > 0, the point is a maximum.

If f’’(x) = 0, it is a point of inflection.

Minimum Maximum Point of Inflection

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Chapter 3

This chapter consists of three examples of full graph analysis with explanations. At the end

practice the three given problems.

Example 1 – Quadratic

f(x)=x2+4x-12

1. General Behavior

End Behavior is in 1st

and 2nd

quadrant because the leading coefficient of the equation is

1 and the equation is in the 2nd

degree.

Y-intercept is f(0)= 02+4*0-12

f(0)=12

The graph touches the Y-axis when y=12.

X-Intercepts or zeroes: set f(0) equal to zero.

x2+4x-12=0

This quadratic equation can be factored:

x2+4x-12

x2+6x-2x-12

x(x+6)-2(x+6)

(x-2)(x+6)

(x-2)(x+6)=0

X-2=0 x+6=0

X1=2 x2=-6

The graph touches the x-axis when x=2 and x=-6

2. Critical Points

Identification of points

Use the 1st

derivative rule to find the critical points.

f’(x)=n*xn-1

Power Rule – formula to find the derivative

f’(x)=2x2-1

+1*4x1-1

+0*120-1

f’(x)= 2x+4

Set the derivative equal to zero. This is because the tangent line in critical points has a

slope of zero since it is a horizontal line. So:

2x+4=0

2x=-4

x=-2

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To find the point substitute -2 for x in the original equation:

f(-2)=(-2)2+4(-2)-12

f(-2)=-16

There is a critical Point at (-2, -16)

Classify the point

Find the second derivative of the equation, or the first derivative of the first

derivative.

f’’(x)=2 Then substitute 0 in the second derivative:

f’’(-2)=2

Compare that to 0:

2 > 0 so the point (-2, -16) is a minimum point

3. Graph.

Sketch the graph of this equation by putting in all the information you found. It should look

like this.

Graph 1 - Quadratic

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Example 2: Cubic that can be factored

f(x)=x3 – 7x + 6

1) General Behavior

End behavior: Since the coefficient of the highest term, x3 is 1, the graph will look like

the parent graph of cubic, ending in 1st

and 3rd

quadrant.

Y intercept: f(0)=(0)3 -7(0)+6

f(0)=6

Y intercept is y=6

X – intercept:

Factoring: x3 – 7x + 6 (6*1=6, factors of 6: ±{1,2,3,6}. (-1)+(-6) =-7)

x3-1x-6x+6

x(x2-1) -6(x+1) (Grouping)

x(x+1)(x-1) -6(x+1) (Factor (x+1))

(x+1)[(x(x-1))-6]

(x+1)(x2-x-6) (Simplify)

To find the x intercept set the whole factored expression equal to zero:

(x+1)(x2-x-6)=0 (At least one of the factors must be 0)

1. x+1=0

x0=-1

2. x2-x-6=0

To solve this you can use synthetic division or the quadratic

formula, but for this demonstration, since our equation is a

quadratic, the example shows how you can use the

quadratic formula.

� ��� � √�� � 4

2

a=1, b=-1, c=-6

� ����1� � ���1�� � 4 � 1 � ��6�

2 � 1

X1=��√����

� x2=

��√�����

X1= ����

x2= ����

X1=3 x2=-2

The graph crosses the x-axis three times: x0=-1, x1=3, x2=-2

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2) Critical Points

Identification of points

Use the 1st

derivative rule to find the critical points.

f(x)=x3 – 7x

1 + 6

0

f’(x)=3*x3-1

–(1*7x1-1

) + (0*60-1

)

f’(x)=3x2-7

Set the derivative equal to zero. This is because the tangent line in critical points has

a slope of zero since it is a horizontal line. So:

3x2-7=0

Notice! To solve this quadratic equation we can use any of the methods explained to

find the zeroes. However, since the x appears only once in the expression of the

derivative, we can use basic algebraic skills to find x

3x2=7

x=±√���� x1=√���� x2=-√���� To find the y of the points: substitute x1=√���� and x2=-√���� into the original equation:

f(x)=x3 – 7x + 6

f1(√����)= ( √���� )3

-7(√����� � 6

f1(√����)=3.58 - 10.69 +6

f1(√����)=-1.11

f2(-√����)= ( -√���� )3

-7(-√����� � 6

f2(-√����)=-3.58 + 10.69 + 6

f2(-√����)=13.11

The two Critical Points are (√����, -1.11) and (-√����, 13.11)

Classification of Critical Points

Our next job is to find out what type of critical points they are. To do that, we need

the 2nd

derivative, which is the derivative of the 1st

derivative found above. So:

f’’(x)=2*3x2-1

f’’(x)=6x

If we substitute 0 for x:

f’’(0)= 0

Now we compare the x values of the critical points found earlier: (√����, -1.11) and (-

√����, 13.11) to the f’’(0)

√���� > 0 so the point (√����, -1.11) is a minimum point

-√���� < 0, so the point (-√����, 13.11) is a maximum point.

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As explained in the previous chapter, the 1st

derivative does not give any points of

inflection. We now that in this graph there must be one because a Point of Inflection

is the only way to between a max and a min. To find the POI: Set the 2nd

derivative

equal to 0:

6x=0

x=0

POI is at the point (0, f(0)) or (0,6)

3. Graph

Now that we now the characteristics of these graph we need to sketch it.

Begin with the end behavior and then plot all the points mentioned. The graph looks like

this:

Graph 2 – Quadratic that can be factored

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Example 3 – Cubic that cannot be factored

f(x) = x3 - 5x

2 +3x +9

1. General behavior

a. End Behavior: The equation is in the 3rd

degree and a=1, so the graph is going to

look like the parent graph of cubic, ending in the 1st

and 3rd

quadrant

b. Intercepts:

Y-Intercept: f(0)=03 – 5*0

2 + 3*0 + 9

f(0)= 9

X-intercepts: x3 - 5x

2 +3x +9=0

To solve this we need the expression to be written and factors. It is not factorable so

we can use the synthetic division.

x3-5x

2+3x+9 has three roots since its highest degree is three

Rational Root Theorem for possible roots:

~your constant is 9 and its factors are ±{1,3,9}

~your leading coefficient (q) is 1, and its factors are ±1

* your possible (not necessarily definite) roots are ±{1/1, 3/1, 9/1}

±1, ±3,and ±9

Use the Descartes Rule of Signs to determine what type of roots you have.

P(x)= x3- 5x

2+3x+9 ~ There are 2 sign changes (positive roots)

P(-x)= -x3-5x

2-3x+9 ~ There is 1 sign change (negative roots)

Positive Negative Imaginary Total

2 1 0 3

0 1 2 3

Construct the Synthetic Division Chart

r 1 -5 3 9

-9 1

-3 1 -8 27 -72

-1 1 -6 9 0

0 1 -5 3 9

1 1 -4 -1 8

3 1 -2 -3 0

9 1 4 39 360

The roots therefore are: x=-1, x=3. The last root must be an imaginary on for a total of 3.

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2. Identifying the Critical Points

Remember that critical points have the first derivative zero.

f’(x)=3x2-10x+3=0

The expression can be factored:

3x2-x-10x+3

(x-3)(3x-1)=0

x-3=0 3x-1=0

x=3 x=1/3

f(3)=27-45+9+9=0 First Critical Point is (3, 0)

f(1/3)=1/27-5/9+1+9 Second Critical Point is(1/3, 256/27)

Classify Points:

Use the second derivative:

f’’(x)=6x-10

Substitute the x’s of the points found above into the 2nd

derivative:

f’’(3)=8 8>0 so point (3,0) is a minimum

f’’(1/3)=-8 -8<0 so point (1/3, 256/27) is a maximum

Between a maximum and a minimum point there is always a point of inflection:

f’’(x)=6x-10=0

x=5/3 (You do not need to find the y of the POIs, as it is half way between the

min and max)

2. Graph

Graph 3 - Cubic that cannot be factored.

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Chapter 4 Practice

Perform a complete graph analysis on the following:

1. f(x)=x3 – 2x

2 – 6x - 3

2. p(x)=2x4 + x

3 – x

2

3. g(x)=x2 + 3x - 6

4. f(x)=x3 – x

2 – 4x +4

5. z(x)=3x3 + 7x - 8

Test Perform a complete graph analysis on the following:

1. f(x)= x3 – 6x

2 +9

2. g(x)=x2 – 3x – 3

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Answers to test

1. f(x)= x3 – 6x

2 +9

2. g(x)=x2 – 3x – 3

About the Authors:

Ana Dede, Rebecca Primak, Grace Raheb, Katherine Tran and Andrew Gallant are all students of Doherty Memorial High School, Worcester, Massachusetts. They love mathematics and made this manual to help kids who are struggling in their Pre-Calculus Class. To learn more about the authors go to the electronic version of this

manual. http://graph-analysis.tk/

To learn more about their school Doherty, visit the website:

http://doherty-high.tk/

To order another copy of this manual email: serendipituous@gmail.com