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# Maths Type I

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Introduction

Introduction

This parabola investigation is to investigate the patterns in the intersections of parabolas the lines y = x and y = 2x. The investigation includes graphs for visual aid and a table showing relationships. Each answer includes examples of how the answer was found and proof.

Method

1. Consider the parabola y = (x-3)2 = x2 – 6x +11 and the lines y = x and y = 2x.
• Using technology find the four intersections illustrated on the right

Using technology the three graphs were plotted on the same graph and the intersections were found. Graph one shows the graph and the points of intersection. The points of intersection are represented by blue dots.         Graph 1: y = x2 – 6x + 11, y = x and y = 2x

• Label the x-values of these intersections as they appear from left to right on the x-axis as x1, x2, x3, and x4.
All four points are labelled on graph one above. The x-values for the above points x1, x2, x3, and x4 are 1.764, 2.382, 4.618 and 6.236 respectively.
• Find the values of x2 – x1 and x4 – x3 and name them SL and SR respectively.
SL = x2 − x1
S
R = x4 –x4
SL = 2.382 – 1.764        = 0.618
S
R = 6.236 – 4.618        = 1.618
• Finally, calculate D = I SLSRI.
D is equal to the absolute value of SL – SR, this means D can never be a negative number.
D =
I SL – SRI
D = I 0.618 – 1.618 I
D = I -1 I
D = 1
Therefore the difference between the S
L and SR is one.

Middle

X1

0.7752

0.2341

0.7929

0.4523

0.6396

X2

1

0.3333

1

0.5683

0.7829

X3

2.5

1

1.75

1.2317

1.3838

X4

3.2247

1.4325

2.207

1.5477

1.6937

SL

0.2248

0.0992

0.2071

0.116

0.1433

SR

0.7247

0.4325

0.457

0.316

0.3099

Difference

0.4999

0.3333

0.2499

0.2

0.1666

D

0.5

0.33

0.25

0.2

0.166

Graph 7

Graph 8

Graph 9

Graph 10

Graph 12

A - Value

7

8

9

10

50

X1

0.3236

0.3964

0.3772

0.4

0.1283

X2

0.4061

0.5

0.4444

0.5

0.146

X3

0.8797

0.875

1

0.8

0.274

X4

1.105

1.1036

1.1783

1

0.3117

SL

0.0825

0.1036

0.0672

0.1

0.0177

SR

0.2253

0.2286

0.1783

0.2

0.0377

Difference

0.1428

0.125

0.1111

0.1

0.02

D

0.143

0.125

0.111

0.1

0.02

Table 1: Data from graphs 2 – 10 including value for D

From table 1 it is apparent there is a direct relationship between the value for ‘a’ and the value for d. It is evident that they are inversely proportional i.e. if ‘a’ goes up d goes down. From the data a conjecture can be made. The conjecture is D = I a-1I. Therefore if ‘a’ is 5 then D is 0.2, this is supported in table 1.

1. 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 in parts 1 and 2 by having the intersections of the first line to be x2 and x3 and the intersections with the second line to be x1 and x4.

Graph 12 below of y = -5x2 - 4x + 1 has a vertex in the 3rd quadrant, according to the conjecture the value for D should be 0.2 and indeed this is true for Graph 12. Graph 12: -5x2 – 4x + 1

Graph 13: -3x2 – 3x + 1

Graph 13 also proves the conjecture for a vertex in quadrant 2.

Conclusion

n-1 i.e.
y
1 = c1xn-1 + d1xn-2 + … + m1x1 + k1 and y2 =  c2xn-1 + d2xn-2 + … + m2x1 + k2 the sum of the roots will always be   – b – y1 and – b – y2
a                      a
The two roots will be the
Therefore D will be equal to:
D =
I(Sum of roots in y1) – (sum of roots in y2) I
D =  I( - b – y1) – (- b – y2) I

a                  a
D =
I- b – y1 + b + y2I
a
D =
y2 – y1
a
That is the conjecture for intersecting lines one degree lower than the polynomial, this is the same conjecture as question two, because question two fit the criteria. If the intersecting lines are more than one order apart a different conjecture must be used which is stated in question five. The conjecture is:
D = 0
Absolute value symbols are not needed as zero is the same whether they are there or not.

Conclusion

There is a definite relationship between the intersecting lines and the parabola. This was found through using quadratic equations and visual aids of graphs to show how a value for D was found throughout the investigation. Through the use of proof to justify each answer it could be used as a way to check each answer and each answer could be used to check the proof, using this method increased the likelihood of the conjectures being correct.

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

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