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# parabola investigation

Extracts from this document...

Introduction

HL MATH PORTFOLIO TYPE 1

PARABOLA INVESTIGATION

School : Ted Ankara College Foundation Private High School

Candidate: Nazlı Deniz BALAMIR 11B 53069

PARABOLA INVESTIGATION

Description

In this task, you will investigate the patterns in the intersections of parabolas and the lines y = x and y = 2x. Then you will be asked to prove your conjectures and to broaden the scope of the investigation to include other lines and other types of polynomials

1) Consider the parabola x2 -6x + 11 and the lines y = x and y = 2x

• .Intersection points of the parabola with the line y=x are x2 and x3

x2: 2.38, x3: 4.62

• Intersection points of the parabola with the line y=2x are x3 and x4

x1: 1.76 x4: 6.24

X2-X1 = SL 2.38-1.76= 0.39

X4-X3 = SR

6.24 – 4.62. = 1.39

SR – SL = D

1.39- 0.39 = 1

2)Find the values of D for other parabolas of the form y= ax2 + bx + c, a>0, with vertices in quadrant 1, intersected by the lines y = x and y = 2x. Consider various values of a, beginning with a =1. Make a conjecture about the value of D for these parabolas.

• a= 2, b= -10, c =13

f(x) =  2x2-10x+13 X1: 1.40

X2: 1.70        X2-X1= 0.30

X3: 3.78

X4: 4.58        X4-X3= 0.80

D= 0.80-0.30 = 0.50

When a = 2, D = 0.5

• a=3, b= -9, c = 7

y =  3x2-9x+7 X1 : 0.82

X2: 1

X3 : 2.33

...read more.

Middle

L  x3 – x4 = 6 - 4.73  =  1.27 = SR

D =1.27-0.27

=1

• The conjecture I made previously was D= 1 /a. However it doesn’t seem to work for a= -1 as 1/a is equal to -1 in this case but  D is equal to 1.  So, I will test this conjecture with another reel value of “a”.
• As I have to  test my conjecture for any placement of vertex,  I will be testing it for a parabola with a vertex in the 2nd quadrant:
• f(x) =  -2x2-10x-9 X1:-4.5

X2: -5.12

X3:-0.88

X4: -1

SL= -5.12-(-4.5) = -0.62

SR= -1-(-0.88) = -0.12

D= -0.12-(-0.62) =0.50

According to my conjecture, D value found is not correct as it must have been 1/a which is   -0.50 in this case. It can be concluded that there is a problem with the sign of a in my conjecture as despite “a”’s having a negative sign D is always positive .  If I modify my conjecture as D = 1/ lal the problem  will be solved.

4)Does your conjecture hold if the intersecting lines are changed? Modify your conjecture, if necessary , and prove it.

• Previously I have made a conjecture about the D values of a parabolas intersection of lines y=x and y=2x. To test my cojecture I have worked with several parabolas and modified my conjecture accordingly as D = 1/ lal .  To generalize this conjecture I must  test it’s validity by changing the intersecting lines.
• Let’s test the conjecture for y= 3x and y=4x

I will continue working on my first parabola which was x2 -6x + 11=y X1 : 1.26

X2: 1.46

1.46-1.26 = 0.2 = SL

X3: 7.54

X4: 8.74

8.74 - 7.54 = 1.2 = SR

D= 1.2- 0.2 = 1 = 1/ lal

• As it can be seen above, even tough I have changed the intersecting lines into y=3x and y= 4x, my conjecture still worked.
• To be sure, I will test my conjecture for the intersecting lines y=5x , y=6x and the parabola 5x2 -5x + 4 X1: 0.46

X2: 0.55

0.55- 0.46 = 0.09 = SL

X3: 1.45

X4 : 1.74

1.74-1.45 = 0.29 = SR

D = 0.29-0.09 = 0.20

As 1/ 5 is  equal to 0.20. I proved my conjecture to be working with different intersecting lines than y=x and y=2x.

• Both two trials were made with intersecting lines that have 1 difference between their slopes. Now I will test my conjecture for the intersecting lines y= 2x and y= 4x and the parabola x2 -6x + 11. • X1 : 1.26

X2: 1.76

1.76- 1.26 =0.5= SL

X3: 6.24

X4: 8.72

8.74 – 6.24 = 2.5 = SR

D = 2.5- 0.5 = 2.0

• According to my conjecture D must have been equal to 1 but it is 2 in this case . When I worked with lines which had one difference in their slopes, my conjecture worked, however it doesn’t seem to be working with the lines y= 2x and y=4x. It is obvious that I must modify my conjecture in order to involve slope differences of the lines.
...read more.

Conclusion

1x2− x2x3+xx1x3+xx2x3−x1x2x3

= a(x3−(x1+x2+x3)x2+(x1x2+x2x3+x1x3)x−(x1x2x3))

= ax3−a(x1+x2+x3)x2+a(x1x2+x2x3+x1x3)x−a(x1x2x3)

From the proof we can see what each of the coefficient equals:

a=a

b= −a(x1+x2+x3)

c= a(x1x2+x2x3+x1x3)

d= a(x1x2x3)

The sum of the roots can be obtained by b:

b= −a(x1+x2+x3)

x1+x2+x3= b/−a = −b/a

According to the formula

D=|(x2+x3)−(x1+x4)|

The sum of the roots for a cubic polynomial is  –b/a so

D= |(−b/a)−( −b/a)|=0

6)The conjecture obtained by the cubic polynomials can be applied to higher order polinomials as  the roots will cancel out so D=0 will always be true.

Conclusion

• By this investigation, I had a chance to apply the fundamental formulae we have learned in lesson and this, I think played a very important role in making my knowledge about parabolas more permenant.  I also learned how to make a conjecture and how to empower it by several different methods. This investigation , I think improved my ability to think critically and independently while providing a practical base for the theoretical knowledge I have learned in the lesson.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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