Difference between revisions of "Applications-of-GeoGebra/C2/Roots-of-Polynomials/English"

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A '''complex number''' is expressed as '''''z = a + ib''''': where ''''''a'''''' is the real part, ‘'''''bi’''''' is '''imaginary '''part, and '''a''' and '''b''' are constants.
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A '''complex number''' is expressed as '''''z = a + bi''''': where ''''''a'''''' = real part, ‘'''''bi’''''' = '''imaginary '''part, and '''a''' and '''b''' are constants.
  
 
'''Imaginary number, ''i'' '''= sqrt{-1}
 
'''Imaginary number, ''i'' '''= sqrt{-1}
  
  
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In the '''XY plane''', '''''a + ib''''' corresponds to the point ('''a, b''').
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In the '''XY plane''', '''''a + bi''''' is point ('''a, b''').
  
  
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In the '''complex plane''', '''x axis''' is called '''real axis''', '''y axis''' is called '''imaginary axis'''.
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In the '''complex plane''', '''x axis''' = '''real axis''', '''y axis''' = '''imaginary axis'''.
 
|  | '''Complex numbers, XY plane'''
 
|  | '''Complex numbers, XY plane'''
  
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A '''complex number''' is expressed as '''''z''''' '''equals''' '''''a''''' '''plus''' '''''ib'''''.
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A '''complex number''' is expressed as '''''z''''' '''equals''' '''''a''''' '''plus''' '''''bi'''''.
  
  
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In the '''XY plane''', '''''a''''' '''plus''' '''''ib''''' corresponds to the point '''''a''''' '''comma''' '''''b'''''.
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In the '''XY plane''', '''''a''''' '''plus''' '''''bi''''' corresponds to the point '''''a''''' '''comma''' '''''b'''''.
  
  
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'''Complex numbers, complex plane'''
 
'''Complex numbers, complex plane'''
  
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'''Argument '''''ϴ''''' '''= angle between '''real axis''' and '''line segment''' connecting '''''z''''' to O '''(0,0)''' in counter-clockwise direction
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'''Argument '''''ϴ''''' '''= angle between '''real axis''' and '''line segment''' connecting '''''z''''' to O '''(0,0)'''; CCW
  
−
'''Polar form''' of '''''z = a + ib''''' is
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'''Polar form''' of '''''z = a + bi''''' is
  
 
'''''z = r (cosϴ + i sinϴ)'''''
 
'''''z = r (cosϴ + i sinϴ)'''''
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'''Polar form''' of '''''z''''' equals '''''a''''' '''plus''' '''''ib''''' is
+
'''Polar form''' of '''''z''''' equals '''''a''''' '''plus''' '''''bi''''' is
  
  
 
'''''z''''' equals '''''r times cos theta plus i sin theta'''''
 
'''''z''''' equals '''''r times cos theta plus i sin theta'''''
  
−
where '''''a''''' is equal to '''''r cos theta''''' and '''''b is r sin theta'''''
+
Where '''''a''''' is equal to '''''r cos theta''''' and '''''b is r sin theta'''''
 
|-
 
|-
 
|  | Show the '''GeoGebra''' window.
 
|  | Show the '''GeoGebra''' window.
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Point to the '''equation h(x)''' appearing in '''Algebra''' view.
 
Point to the '''equation h(x)''' appearing in '''Algebra''' view.
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|  | Drag boundaries to see '''Algebra''' and '''Graphics''' view properly.  
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|  | Drag boundaries to see '''Algebra''' and '''Graphics''' views properly.  
  
 
Observe equation '''h of x''' in '''Algebra''' view.
 
Observe equation '''h of x''' in '''Algebra''' view.
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Click in '''Graphics''' view.  
 
Click in '''Graphics''' view.  
 
|-
 
|-
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|  | Click on '''Move Graphics View''' and move background to see the graph.  
+
|  | Click on '''Move Graphics View''' and move '''Graphics''' background to see the graph.  
 
|  | Click on '''Move Graphics View''' and move '''Graphics''' background to see the graph.  
 
|  | Click on '''Move Graphics View''' and move '''Graphics''' background to see the graph.  
 
|-
 
|-
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Substitute this x in original function to get y co-ordinate
 
Substitute this x in original function to get y co-ordinate
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|  | A '''point of inflection PoI''' on a curve is the point where the '''curve''' changes its direction.
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|  | '''Point of inflection'''
−
 
+
−
 
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A '''point of inflection PoI''' on a curve is the point where the '''curve''' changes its direction.
−
To find the '''co-ordinates''' of '''PoI''' '''''x''''' comma '''''y''''':
+
  
  
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We equate the second '''derivative''' of the given '''function''' to 0.
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To find the '''co-ordinates''' of '''PoI''' '''''x''''' comma '''''y''''', we equate second '''derivative''' of the given '''function''' to 0.
  
  
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'''Summary'''
 
'''Summary'''
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|  | In this tutorial, we have learnt how to:
+
|  | In this tutorial, we have learnt to:
  
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*Plot graphs of '''polynomial functions''' using '''CAS''' view and '''input bar'''.
+
*Plot graphs of '''polynomial functions''' using '''CAS''' view and '''input bar'''
  
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*Find '''real roots, extrema''' and '''inflection point(s)'''.
+
*Find '''real roots, extrema''' and '''inflection point(s)'''
  
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*'''Complex roots''' will be covered in another tutorial.
+
*'''Complex roots''' will be covered in another tutorial
 
|-
 
|-
 
|  | '''Slide Number 13'''
 
|  | '''Slide Number 13'''

Revision as of 12:20, 17 May 2018

Visual Cue Narration
Slide Number 1

Title Slide

Welcome to this tutorial on Roots of Polynomials.
Slide Number 2

Learning Objectives

In this tutorial, we will learn:
  • To plot graphs of polynomial equations
  • About complex numbers, real and imaginary roots
  • To find extrema and inflection points
Slide Number 3

Pre-requisites

www.spoken-tutorial.org

To follow this tutorial, you should be familiar with *GeoGebra interface
  • Basics of coordinate system
  • Polynomials

If not, for relevant tutorials, please visit our website.

Slide Number 4

System Requirement

Here I am using:
  • Ubuntu Linux OS version 14.04
  • GeoGebra 5.0.388.0-d
Let us begin with the binomial theorem.
Slide Number 5

Binomial Theorem

Binomial theorem states that if a, b Єℝ, index n is a positive integer, 0 ≤ r ≤n, then (a + b)n can be expanded as follows:

(a + b)n = nC0 an + nC1 an-1 b1 + nC2 an-2 b2 + … + nCr an-r br + … + nCn bn

Reminder: nC1 = n!/[1! (n-1)!]

a and b are real numbers. 
index n is a positive integer.

r lies between 0 and n.

Binomial theorem states that a plus b raised to n can be expanded as shown.

Slide Number 6

Quadratic Equations and Roots

A second degree polynomial, y = a x2+ b x+ c has roots

x=- b ± sqrt{( b 2-4 ac)/2a }

where ▲= b2-4ac

When ▲< 0, roots are complex

When ▲=0, roots are real and equal

When ▲>0, roots are real and unequal

Quadratic Equations and Roots

A 2nd degree polynomial, y equals a x squared plus b x plus c has roots given by values of x.

x is equal to ratio of minus b plus or minus squareroot of b squared minus 4 a c to 2 a.

Where determinant Delta is equal to b squared minus 4 a c

When Delta is less than 0, roots are complex

When Delta is equal to 0, roots are real and equal

When Delta is greater than 0, roots are real and unequal

Slide Number 7

Quadratic Equations and Roots

When roots are real, ax2+b x+ c =0 has extremum (xv, yv)

xv = - b/2a and yv= a xv2+ b xv+ c

Quadratic Equations and Roots

When roots are real, ax squared plus b x plus c equals 0 has extremum xv comma yv

xv equals minus b divided by 2 a and yv equals axv squared plus bxv plus c

Show the GeoGebra window. I have already opened the GeoGebra interface.
Click on View >> select CAS. Click on View tool and select CAS to open the CAS view.
In line 1 in CAS view, type f(x):=x^2-2x-3 >> press Enter. In line 1 in CAS view, type the following line.

f x in parentheses colon equals x caret 2 minus 2 space x minus 3.

To type caret symbol, hold Shift key down and press 6.

The space indicates multiplication.

Press Enter.

Drag boundary to see Algebra view properly. Drag boundary to see Algebra view properly.
Point to the equation f(x) appearing in Algebra view.

Point to exponent 2 in f(x).

Observe the equation f of x in Algebra view.

The degree of this quadratic polynomial f of x is 2.

Drag boundary to see Graphics view properly. Drag boundary to see Graphics view properly.
Click in Graphics view to see Graphics View toolbar. Click in Graphics view to see Graphics View toolbar.
Move Graphics View>> click on Zoom Out Under Move Graphics View, click on Zoom Out tool.
Click in Graphics view >> minimum vertex of parabola f. Click in Graphics view to see the minimum vertex of parabola f.
Click on Move Graphics View tool >> click in Graphics background. Click on Move Graphics View tool and click in Graphics background.
Hand symbol appears >> drag Graphics view to see parabola f. When hand symbol appears, drag Graphics view so you can see parabola f.
Drag boundaries Drag boundaries to see CAS view properly.
Type Root(f) in line 2 of CAS view >> press Enter. In line 2 of CAS view, type Root f in parentheses.

Press Enter.

Point to roots in CAS view. The roots appear below, in the same box, in curly brackets.
Point to roots in Graphics view. Note that these are the x-intercepts of parabola f in Graphics view.
Type Extremum(f) in line 3 of CAS view >> press Enter. In line 3 of CAS view, type Extremum f in parentheses.

Press Enter.

Point to extremum in CAS view. The extremum appears below, in the same box, in curly brackets.
Note that this is the minimum vertex of parabola f in Graphics view. Note that this is the minimum vertex of parabola f in Graphics view.
In line 4 in CAS view, type g(x):=x^2+5x+10 >> press Enter. In line 4 in CAS view, type the following line.

g x in parentheses colon equals x caret 2 plus 5 space x plus 10.

Press Enter.

Drag boundary to see Algebra view properly.

Point to the equation g(x) appearing in Algebra view.

Drag boundary to see Algebra view properly.

Observe the equation g of x in Algebra view.

Drag boundary to see Graphics view properly. Drag boundary to see Graphics view properly.
Uncheck f of x in CAS view.

Note that this also unchecks it in Algebra view and hides parabola f in Graphics view.

Uncheck f of x in CAS view.

Note that this also unchecks it in Algebra view and hides parabola f in Graphics view.

Click and drag Graphics view so you can see parabola g. Click in and drag Graphics view so you can see parabola g.
Again, drag boundary to see CAS view properly. Again, drag boundary to see CAS view properly.
Type Root(g) in line 5 of CAS view >> press Enter. In line 5 of CAS view, type Root g in parentheses.

Press Enter.

Point to empty curly brackets for roots in CAS view. Empty curly brackets appear below.

Parabola g does not have any real roots as it does not intersect x axis at all.

Roots are said to be complex.

Type Extremum(g) in line 6 of CAS view >> press Enter. In line 6 of CAS view, type Extremum g in parentheses.

Press Enter.

Point to extremum in CAS view. The extremum appears below, in the same box, in curly brackets.
Note that this is the minimum vertex of parabola g in Graphics view. Note that this is the minimum vertex of parabola g in Graphics view.
Point to Evaluate tool.

Point to extremum in form of fractions.

While Evaluate tool is highlighted in CAS View toolbar, the extremum appears as fractions.

Minus five divided by 2 comma 15 divided by 4.

Click on the extremum in line 6 and click on Numeric tool.

Point to extremum in form of decimals.

In line 6, click on the extremum and click on Numeric tool.

The extremum now appears in decimal form.

Minus 2 point 5 comma 3 point 7 5.

Let us look at complex numbers.
Slide Number 8

Complex numbers, XY plane


A complex number is expressed as z = a + bi: where 'a' = real part, ‘bi’ = imaginary part, and a and b are constants.

Imaginary number, i = sqrt{-1}


In the XY plane, a + bi is point (a, b).


In the complex plane, x axis = real axis, y axis = imaginary axis.

Complex numbers, XY plane

A complex number is expressed as z equals a plus bi.


a is the real part; bi is imaginary part;a and b are constants


i is imaginary number and is equal to square root of minus 1.


In the XY plane, a plus bi corresponds to the point a comma b.


In the complex plane, x axis is called real axis, y axis is called imaginary axis.

Slide Number 9

Complex numbers, complex plane

In complex plane, z is a vector with real axis coordinate a and imaginary axis coordinate b


Length of the vector z = |z| = r


r = sqrt (a2+b2) (Pythagoras’ theorem)

Complex numbers, complex plane

In complex plane, z is a vector.


Its real axis coordinate is a and imaginary axis coordinate is b.


The length of the vector z is equal to the absolute value of z and to r.


According to Pythagoras’ theorem, r is equal to squareroot of a squared plus b squared.

Slide Number 10

Complex numbers, complex plane

Argument ϴ = angle between real axis and line segment connecting z to O (0,0); CCW

Polar form of z = a + bi is

z = r (cosϴ + i sinϴ)

where a= r cosϴ, b=r sinϴ

Argument theta is angle between real axis and line segment connecting z to origin.

It is in counter-clockwise direction.


Polar form of z equals a plus bi is


z equals r times cos theta plus i sin theta

Where a is equal to r cos theta and b is r sin theta

Show the GeoGebra window. Let us go back to the GeoGebra interface we were working on.

We will now use the input bar instead of CAS view.

Click and close CAS view. Click and close CAS view.
In Algebra view, uncheck g of x to hide it. In Algebra view, uncheck g of x to hide it.
In input bar, type the following line.

h(x):=x^3-4x^2+x+6 >> press Enter.

In input bar, type the following line.

h x in parentheses colon equals x caret 3 minus 4 space x caret 2 plus x plus 6.

Press Enter.

Drag boundaries to see Algebra and Graphics view properly.

Point to the equation h(x) appearing in Algebra view.

Drag boundaries to see Algebra and Graphics views properly.

Observe equation h of x in Algebra view.

Function h of x is graphed in Graphics view.

Under Move Graphics View, click on Zoom Out tool.

Click in Graphics view.

Under Move Graphics View, click on Zoom Out tool.

Click in Graphics view.

Click on Move Graphics View and move Graphics background to see the graph. Click on Move Graphics View and move Graphics background to see the graph.
In input bar, type Root(h) and press Enter. In input bar, type Root h in parentheses and press Enter.
Point to co-ordinates of three roots (A, B and C) in Algebra view. The co-ordinates of three roots (A, B and C) appear in Algebra view.
Point to three roots mapped on the curve h of x in Graphics view. The three roots are also mapped as x intercepts of the curve h of x in Graphics view.
In input bar, type Extremum h in parentheses and press Enter. In input bar, type Extremum h in parentheses and press Enter.
Point to co-ordinates of two extrema (Dand E) in Algebra view. Co-ordinates of two extrema (D and E) appear in Algebra view.
Point to two extrema mapped on the curve h of x in Graphics view. The two extrema are also mapped on curve h of x in Graphics view.
Slide Number 11

Point of inflection

Point of inflection (PoI) on a curve is the point where curve changes direction.

To find co-ordinates of PoI (x,y)

Equate 2nd derivative of given function to 0

Solve to get x (x co-ordinate of PoI)

Substitute this x in original function to get y co-ordinate

Point of inflection

A point of inflection PoI on a curve is the point where the curve changes its direction.


To find the co-ordinates of PoI x comma y, we equate second derivative of the given function to 0.


Solution of this equation gives us x (x co-ordinate of PoI).


Substitute this x in original function to get y co-ordinate.

Let us find the point of inflection on h(x). Let us find the point of inflection on h of x.
In input bar, type Inf >> choose InflectionPoint ( <Polynomial> ) option from menu. In input bar, type Inf and scroll down menu to choose InflectionPoint Polynomial option.
Instead of highlighted Polynomial, type h >> Press Enter. Instead of highlighted Polynomial, type h and press Enter.
Point to the point of inflection in Algebra view. In Algebra view, point of inflection appears as point F, below the two extrema.
Point to F on h(x) in Graphics view. F is mapped on h of x in Graphics view.
Let us open a new GeoGebra window to use CAS for a cubic polynomial.
Click on View tool and click on CAS to show it. Click on View tool and click on CAS to show it.
Drag boundary to see CAS view properly. Drag boundary to see CAS view properly.
In line 1 of CAS view, type the following line.

i(x):=x^3-6 x^2+4 x+1 >> press Enter.

In line 1 of 'CAS 'view, type the following line.

i x in parentheses colon equals x caret 3 minus 6 space x caret 2 plus 4 space x plus 1.

Press Enter.

Drag boundary to see CAS view properly. Drag boundary to see CAS view properly.
In line 2 of CAS view, type Root(i) >> press Enter. In line 2 of CAS view, type Root i in parentheses and press Enter.
Point to the three roots in CAS view.

Scroll to see them.

Point to Evaluate tool.

The three roots are shown below with square root notations.

Scroll to see them.

Note that the Evaluate tool is highlighted.

In line 2, click on the roots and click on Numeric tool.

Point to the three roots in decimal form.

In line 2, click on the roots and click on Numeric tool.

The roots are now shown in decimal form in the next line.

In line 4 of CAS view, type Extremum(i) >> press Enter. In line 4 of CAS view, type Extremum i in parentheses and press Enter.
Point to the two extrema in CAS view.

Scroll to see them.

Point to Numeric tool and to extrema in decimal form.

The two extrema points are shown below.

Scroll to see them.

As the Numeric tool was clicked, the points appear in decimal form.

Click in and drag Graphics view so you can see i(x). Click in and drag Graphics view so you can see i of x.
In line 5, type Inf >> choose InflectionPoint ( <Polynomial> ) option from menu. In line 5, type Inf and scroll down menu to choose InflectionPoint Polynomial option.
Instead of highlighted Polynomial, type i >> Press Enter. Instead of highlighted Polynomial, type i and press Enter.
Point to the point of inflection in Algebra view. Co-ordinates of point of inflection appear in curly brackets.
Correlate the degree of the polynomials and the number of roots seen so far.
Point to CAS, then Algebra and Graphics views. Observe that functions entered in CAS appear in Algebra and Graphics views.
Point to Algebra and Graphics views, then CAS view. Functions entered in input bar appear in Algebra and Graphics views but not in CAS view.
Let us summarize.
Slide Number 12

Summary

In this tutorial, we have learnt to:
  • Plot graphs of polynomial functions using CAS view and input bar
  • Find real roots, extrema and inflection point(s)
  • Complex roots will be covered in another tutorial
Slide Number 13

Assignment

Assignment:

Plot graphs and find roots, extrema and inflection points for the following polynomials.

d(x)=x2-6x+5

e(x)=3x3-2x2+0.2x-1

f(x)=-2x4-x3+3x2

g(x)=x5-7x4+9x3+23x2-50x+24

h(x)=(4x+3)/(x-1)

i(x)=(3x2-2x-1)/(2x2+3x-2)

Slide Number 14

About Spoken Tutorial project

The video at the following link summarizes the Spoken Tutorial project.

Please download and watch it.

Slide Number 15

Spoken Tutorial workshops

The Spoken Tutorial Project team conducts workshops and gives certificates.

For more details, please write to us.

Slide Number 16

Forum for specific questions: Do you have questions in THIS Spoken Tutorial? Please visit this site. Choose the minute and second where you have the question. Explain your question briefly. Someone from our team will answer them.

Please post your timed queries on this forum.
Slide Number 17

Acknowledgement

Spoken Tutorial Project is funded by NMEICT, MHRD, Government of India.

More information on this mission is available at this link.

This is Vidhya Iyer from IIT Bombay, signing off.

Thank you for joining.

Contributors and Content Editors

Madhurig, Nancyvarkey, Vidhya

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