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Math HL portfolio

Extracts from this document...

Introduction

Math HL

Portfolio

Patrick Vollmer

Description:

In this task you will investigate the patterns in the intersection 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.

The main aim of my investigation is to conclude an answer for the relations between the graph intersections of the graphs y=x and y=2x

Method

1. Consider the parabola y=(x-3)²+2=x²-6x+11 and the lines y=x and y=2x.image00.png

image01.png

Using Technology find the four intersections illustrated on the right.

Using the Microsoft based program autograph® the four intersections between the three given graphs were found.

image23.png

Label all the x-values of these intersections as they appear from left to right on the x-axis as image06.png, image05.png, image07.pngandimage02.png. ( Label on the actual graph)

image23.png

Find the values of image05.png - image06.png and image02.png- image07.pngand name them respectievlyimage09.pngand image03.png.

image06.png= {1,764}                      image05.png - image06.png= 2,382 - 1,764 = 0,618 = image09.png

image05.png= {2,382}                      image02.png- image07.png= 6,236 – 4,618= 1,618 = image03.png

image07.png= {4,618}

image02.png= {6,236}

Finally, calculate D = image04.pngimage09.png- image03.pngimage04.png

D= image04.pngimage09.png- image03.pngimage04.png = image04.png0,618 – 1,618 image04.png= image04.png– 1image04.png= 1                                  D = 1

2. Find Values for 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.

I used several parabolas to derive to a relation or formula.

The first parabola I used in the form of y=ax²+bx+c is

y=2x²-5x+4

image49.png

image06.png= {0,719}                      image05.png - image06.png= 1 - 0.719 =  0.281 = image09.png

image05.png= {1}                              image02.png- image07.png= 2,781 – 2 = 0.781 = image03.png

image07.png= {2}

image02.png= {2,781}                    D= image04.pngimage09.png- image03.pngimage04.png = image04.png(image05.png - image06.png) -  (image02.png- image07.png)image04.png = image04.png0.2808 – 0.781image04.png= 0.5

The third parabola I used in the form of y=ax²+bx+c is

Y= 0.5x²-3x+5

image50.png

image06.png= {1,127}

image05.png= {1,5505}    

image07.png= {6,4495}

image02.png= {8,873}  

D= image04.pngimage09.png- image03.pngimage04.png =image04.png(image05.png - image06.png) -  (image02.png- image07.png)image04.png = image04.png0.4235 – 2.4235image04.png= 2

The fourth parabola I used in the form of y=ax²+bx+c is

                        Y=2x²-3x+1.2

image51.pngimage06.png= {0,26893}

image05.png= {0,36754}    

image07.png= {1,6325}

image02.png= {2,2311}  

D= image04.pngimage09.png- image03.pngimage04.png =image04.png(image05.png - image06.png) -  (image02.png- image07.png)image04.png = image04.png0.09861 – 0.59861image04.png= 0.5

...read more.

Middle

1

1

2

0.5

2.5

0.4

Π

0.31833  

This conjecture is only possible if the vertex is in the 1st quadrant and if the values of b and c are bounded to limitations.

B has to be smaller than zero in mathematical terms b<0

For example see the same quadratic equation just the variable b is changed

Y= x²-4x+5    (a=1, b= -4, c=5)

And Y= x²+4x+5 (a=1, b= 4, c=5)

The Vertex of the first quadratic lies in the First Quadrant

You can complete the square to convert ax2 + bx + c to vertex form, but it's simpler to just use a formula (derived from the completing-the-square process) to find the vertex.

For a given quadratic y = ax2 + bx + c, the vertex (h, k) is found by computing h = b/2a, and then evaluating y at h to find k where  k = (4ac – b2) / 4a.

So the vertex for the first one is is h = image15.png = image16.png

 K = image17.png=image18.png

So the Vertex is (2,1)

The Vertex of the second quadratic does not lie in the first quadrant

Y= x²+4x+5 (a=1, b= 4, c=5)

h = image15.png = image19.png

K = image17.png=image20.png

That the vertex is not in the first quadrant can also be proved by using technology (autograph) which is shown in the graph below:

image21.png

3. Investigate your conjecture for any real value of a and any placement of the vertex. Refine your conjecture as necessary and prove it. Maintain the labelling convention used in parts 1 and 2 by having the intersections of the first line to be image05.pngand image07.png , and the intersections with the second line to be image06.png and image02.png.

This Question implies that I have to investigate and modify my conjecture for any real value for “a” and any placement of the vertex. This means that I am going to change not only “a”, but also the variables “b” and “c” in the parabola. The lines of y = x and y = 2x are still going to be the same. Similarly, I will note the intersections of those lines with the parabola and also calculate D again.

I used several parabolas to derive to a modified conjecture to the case of a vertex in different quadrants.

Y=−x²+4x+5

image22.png

image06.png= {-1,449}

image05.png= {-1.193}    

image07.png= {4,193}

image02.png= {3.449}  

D= image04.pngimage09.png- image03.pngimage04.png =image04.png(image05.png - image06.png) -  (image02.png- image07.png)image04.png = image04.png0.256 – (-0.744)image04.png= 1

image10.png= -1   Dimage24.png-1

My previously conjecture has to be modified from image10.png = D

To image04.pngimage10.pngimage04.png = D

In this case it would be image04.pngimage10.pngimage04.png = image04.png -1image04.png = 1 = D

The second parabola I used in the form of y=ax²+bx+c is

y=-13x²-4x+7

image25.png

image06.png= {-1}

image05.png= {-0,9509}    

image07.png= {0,5663}

image02.png= {0,5385}  

D= image04.pngimage09.png- image03.pngimage04.png =image04.png(image05.png - image06.png) -  (image02.png- image07.png)image04.png = image04.png0.0491 – (-0.0278)image04.png= 0.0769

image04.pngimage10.pngimage04.png =  image04.pngimage26.pngimage04.png = 0.0769 = D

Y = log(6)x² + 3x – 2

image27.png

image06.png = -2.3697

image05.png = -3.3398

image07.png= 0.7696

image02.png= 1.0846

D =image04.pngimage09.png- image03.pngimage04.png =image04.png(image05.png - image06.png) -  (image02.png- image07.png)image04.png = | (-0.9701) – (0.3150) | = | (-1.2851) | = 1.2851

D=image04.pngimage10.pngimage04.png =  image04.pngimage28.pngimage04.png = 1.2851

Y = 5x² - 3x – 2

image29.png

image06.png= -0.3062

image05.png= -0.3483

image07.png= 1.1483

image02.png= 1.3062

D =image04.pngimage09.png- image03.pngimage04.png =image04.png(image05.png - image06.png) -  (image02.png- image07.png)image04.png = | (-0.0421) – (0.1579) | = | -0.2000 | = 0.2

D=image04.pngimage10.pngimage04.png =  image04.pngimage30.pngimage04.png = 0.2

Y = -18x² + 5x +6

image31.png

image06.png= -0.5

image05.png= -0.4768

image07.png= 0.6991

image02.png= 0.6667

D =image04.pngimage09.png- image03.pngimage04.png =image04.png(image05.png - image06.png) -  (image02.png- image07.png)image04.png = | 0.0232 – (-0.0324) | = | -0.0556 | = 0.0556

D=image04.pngimage10.pngimage04.png =  image04.pngimage32.pngimage04.png = 0.0556 regarding only 4 significant figures.

My modified conjecture works so far which is shown in table of results below:

Value for “a”

-18

Value for “D”

0.0556

-13

0.0769

-1

1

Log(6)

1.2851

                                    5

0.2

...read more.

Conclusion

Therefore my new general conjecture for any polynomial is: D = |d2-d1| / |(power-1)*a|,

Where d1 and d2 are the gradients of the lines intersecting the polynomial and a is the gradient of the polynomial and the power is the order of the polynomial.

Now I want to summarize my findings form question 6 and check whether my conjecture is true.

Polynomial

Lines

D

Calculated value for D

Y=(x-1)(x-3)(x-4)(x-5)

Y=x-6 and y=0.5x-6

0.1367

0.166

Y=2(x-1)(x-3)(x-4)(x-5)

Y=4x-19 and y=3x-12

0.2047

0.166

Y=0.7(x-1)(x-3)(x-4)(x-5)

Y=2x-7 and y=x-3

0.439

0.952

Y=(x-1)(x-3)(x-4)(x-5)(x-6)

Y=3x and y=5x

0.1449

0.5

Again the calculated value for D did not come out to be exactly the same as the value for D that I got from drawing the polynomials and intersecting it with the lines. This is due to the fact that the gradient of the curve does not stay constant and so is only approximated. Therefore, my conjecture can only come out to be close, but not exactly the same.

Conclusion

After now having investigated in the patterns of the intersections of parabolas with two lines, I want to conclude that the general conjecture for any polynomial is:

D = (d2-d1) / ((power-1)*a), where d1 and d2 are the gradients of the lines intersecting the polynomial and a is the gradient of the polynomial and the power is the order of the polynomial.

The difficulties in this Project were that it was impossible to find an exact conjecture for D. It was always only an approximate. This is due to the fact that the gradient of the line, a, is approximated and changed throughout the whole curve. Therefore the calculated values for D will always vary slightly from the actual value for D.

However, the conjecture that I found comes out to be very close.

Parabola Investigation

Patrick Vollmer

Math Higher Level

...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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