• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22

Parabola investigation. In this task, we will investigate the patterns in the intersections of parabola and the lines y =x and y=2x.

Extracts from this document...

Introduction

Parabola Investigation – Portfolio HL TYPE I

PARABOLA INVESTIGATION

Description

In this task, we will investigate the patterns in the intersections of parabola and the lines y =x and y=2x. Then we will prove and find the conjectures and to broaden the scope of the investigation to include other lines and other types of polynomials.

1. Consider the parabola image00.png and the line image01.png andimage03.png.

image96.png

  • Using Graphmatica software, we can find the four intersections. Below these points are illustrated.

We find out 2 intersection points of f(x) and h(x) which are (1.7639; 3.5279)

            and (6.2361; 12.4721).                                                                                                                                                                                                                                                                The other 2 intersection points between f(x) and g(x) are (2.382; 2.382) and    (4.618; 4.618).

image103.png

  • The x-values of these intersections as they appear from the left to right on the x-axis as x1, x2, x3, x4.
  • x1≈ 1.764
  • x22.382
  • x3≈ 4.618
  • x4≈ 6.236
  • Find the values of image83.png and image92.png and name them respectively SL and SR.

 SL = x2−x1≈ 2.382−1.764≈ 0.618

 SR = x4−x3≈ 6.236−4.618≈ 1.618

  • Finally, calculateimage21.png.

image21.png

image129.png

image02.png

image11.png

     2. Find values of D for other parabolas of the formimage15.png,image16.png with vertices in quadrant 1, intersected by the lines image01.png andimage03.png. Consider various values of a, beginning withimage39.png. Make a conjecture about the value of D for these parabolas.

...read more.

Middle

 andimage03.png.

image124.png

The x-values of these intersections from the left to the right on the x-axis:

  • x1≈ 2.417
  • x2≈ 3.172
  • x3≈ 8.828
  • x4≈ 11.583

Calculation of SL and SR:

SL = x2−x1≈ 3.172− 2.417≈ 0.755

SR = x4−x3≈ 11.583−8.828≈ 2.755

Calculateimage21.png.

image21.png

image125.png

image126.png

image127.png

Consider parabolaimage12.png, the linesimage01.png andimage03.png.

image128.png

By using Graphmatica software, we can obtain the four intersections of parabola image12.png , the linesimage01.png andimage03.png.

image130.png

The x-values of these intersections from the left to the right on the x-axis:

  • x1≈ 2.591
  • x2≈ 3.285
  • x3≈ 9.315
  • x4≈ 11.809

Calculation of SL and SR:

SL = x2−x1≈ 3.285− 2.591≈ 0.694

SR = x4−x3≈ 11.809−9.315≈ 2.494

Calculateimage21.png.

image21.png

image131.png

image132.png

image133.png

Table shows the values of D for parabolas of the formimage15.png,image16.png with vertices in quadrant 1, intersected by the lines image01.png andimage03.png.

Parabolas

a

b

c

D

image04.png

1

-6

11

1

image05.png

1

-7

13

1

image06.png

2

-8

9

image07.png

image08.png

6

-12

7

image09.png

image10.png

image07.png

-5

14

2

image12.png

image13.png

-6

17

image14.png

As can be seen from the table D is inversely proportional to the value of a.

The values of D for parabolas of the formimage15.png,image16.png with vertices in quadrant 1, intersected by the lines image01.png andimage03.png, are inversely proportional to the values of a.

image17.png

3. Investigate your conjecture for any real value of a and any placement of the vertex.

...read more.

Conclusion

image61.png andimage62.png.

image63.png

The intersection points then can be found by using Graphmatica software.

image64.png

  • x1≈ 1.258
  • x2≈ 1.459
  • x3≈ 7.541
  • x4≈ 8.742

Calculation of SL and SR:

SL = x2−x1≈ 1.459− 1.258≈ 0.201

SR = x4−x3≈ 8.742−7.541≈ 1.201

Calculateimage21.png.

image21.png

image65.png

image02.png

image11.png

  In this case, the conjecture holds true. image26.png

  Consider parabola image66.png and the lines image67.png andimage69.png

.

image70.png

By using Graphmatica software, we can obtain the four intersections of parabola image66.png , the lines image67.png andimage69.png.

image71.png

  • x1≈ 0
  • x2≈ 0.392
  • x3≈ 4.000
  • x4≈ 5.109

Calculation of SL and SR:

SL = x2−x1≈ 0.392− 0≈ 0.392

SR = x4−x3≈ 5.109−4.000≈ 1.109

Calculateimage21.png.

image21.png

image72.png

image73.png

image75.png

Consider parabola image76.pngand the lines image77.pngand image78.png

image79.png

Again, using Graphmatica software, we can obtain the four intersection points.

image80.png

  • x1≈ -2.000
  • x2≈ 0.209
  • x3≈ 1.500
  • x4≈ 4.791

Calculation of SL and SR:

SL = x2−x1≈ 0.209+2.000≈ 2.209

SR = x4−x3≈ 4.791−1.500≈ 3.291

Calculateimage21.png.

image21.png

image81.png

image82.png

image84.png

Consider parabola image85.pngand the line y = 5 and y =-3.

image86.png

By using Graphmatica software, we can obtain the four intersection points.

image87.png

  • x1≈ -6.123
  • x2≈ -5.000
  • x3≈ 1.000
  • x4≈ 2.123

Calculation of SL and SR:

SL = x2−x1≈ (-5.000) − (-6.123) ≈ 1.123

SR = x4−x3≈ 2.123−1.000≈ 1.123

Calculateimage21.png.

image21.png

image88.png

image89.png

Proof:

ax2 + (b− m1) x + c= 0

ax2 + (b− m2) x + c= 0

image90.png

image91.png

D=|SL−SR| = |(x2−x1) – (x4−x3)|= |x2−x1− x4+x3|

= |x2+x3−x1− x4|= |(x2+x3) −(x1+x4)|

x2 +x3 = 2[−(b− m1)/2a]= −(b− m1)/a

x1 +x4 = 2[−(b− m2)/2a]= −(b− m2)/a

image93.png, with a image94.pngR.

Cao Huu Anh Khoa – 5X

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Math IA type 2. In this task I will be investigating Probabilities and investigating ...

    Note: the histogram suggests that values for probability for 0 and 1 point won is 0 but this is not the case, it's just that the values are so small that they cannot be seen in the graph. These values can be observed in Data Table 2.

  2. Ib math SL portfolio parabola investigation

    x 3x 0.068 0.697 0.63 Example 3 = Cubics with negative roots work with the conjecture.

  1. Investigating Parabolas

    To find the conjectures ( a = 4) i. let a = 4, b = -9 and c = 6 ii. y = 4x2-9x+6 Process - x2 - x1 = 1 - 0.75 = 0.25 = SL - x4 - x3 = 2 - 1.5 = 0.5 = SR - So, the ?SL - SR ?

  2. Math IA Type 1 In this task I will investigate the patterns in the ...

    I will try to look for a pattern in these parabolas which are intersected by the lines y = x and y = 2x and have vertices in the first quadrant. I will find different values of D for the parabolas with values of a as 2, 3, 5 and 8.

  1. Mathematics SL Parellels and Parallelograms. This task will consider the number of parallelograms formed ...

    60 100 150 210 6 15 45 90 150 225 315 7 21 63 126 210 315 441 m Table 1.2 See graph 1 to get a more graphical view of how the number of parallelograms formed by m horizontal parallel lines intersected by n parallel transversals are very sequential,

  2. Ib math HL portfolio parabola investigation

    Now let us consider the parabola y= (x-3)2+2 = x2-6x+11 and the lines y=x and y=2x. The vertex of this equation is (3, 2) and the value, a = 1. Thus, this parabola has its turning point in the first quadrant.

  1. Math SL Circle Portfolio. The aim of this task is to investigate positions ...

    0 The 0 is rejected, because according to the graph shown before, the point P' does not lie on the origin. The coordinates of P' is ( , 0). Again, by using the distance formula, the length of can be calculated.

  2. Stellar Numbers. In this task geometric shapes which lead to special numbers ...

    = 71 + 35 =106 For Stage 6: 7Sn = 7Sn-1 + 7n 7S4 = 7S5 + (7x6) = 106 + 42 =148 Finding a general statement: pSn 7S0 7S1 7S2 7S3 7S4 7S5 7S6 Sequence 1 8 22 43 71 106 148 First Difference 7 14 21 28 35

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work