# Parabola investigation. The property that was investigated was the relationship between the parabola and two lines that intersected the parabola.

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Introduction

MATHEMATICS PORTFOLIO

PARABOLA INVESTIGATION

Introduction

This task was carried out to investigate one of the properties of a parabola. The property that was investigated was the relationship between the parabola and two lines that investigated the parabola.

An outline of the task

- First, a particular parabola was selected. This parabola had the equation : y = x2-6x+11
- Then the two intersecting lines were drawn. They had the equation : “y = 2x” and “y = x”
- The relation between these graphs were found out by performing a set of steps that would give a specific value that denoted the relationship between the graphs. This relation was named “D”
- Then the value of “a” in the equation of the parabola was varied so as to find a conjecture in the varying values of “D”.
- Then this conjecture was tested on a parabola with different intersecting lines and the conjecture was modified to make it suitable for the second parabola.

The task in detail

The main aim of the task was to find the variations in the value of “D”. The following steps were used to find the value of D for a particular graph. The below figure is that of the graph that is used as an example in demonstrating how the value of D was found out.

The x values of the intersections were named x1, x2,x3 and x4

Middle

1.12

7.14

1.13

7.08

1.14

7.02

1.15

6.96

1.16

6.9

From this a visible pattern can be observed. That is “ with every increase of 0.01 in the value of a, there is a decrease of 0.06 in the value of D “. So a formula can be made for the value of D. This is shown below. Take the graph 1.17x 2 – 6x + 11 for example.

- First find the difference between the value of a and 1.12. This in this case is 0.05.
- Now find how many 0.01s are there in 0.05. This can be done by dividing it by 0.01.
- The answer is 5. Now multiply 5 and 0.06. The answer is 0.3.
- Now subtract 0.3 from 7.14. The answer is 6.84 which is the value of D.
- From this we can write the formula 7.14 - ( (a-1.12)0.01×0.06)

But as we are not sure whether this conjecture holds for all the values of a till 1.46 where both the lines don’t intersect the graph, the conjecture must be tested. The following shows how the conjecture was tested.

The graph of y = 1.17x 2 – 6x + 11 was drawn. Then the value of D for this graph was found out. The graph is shown below.

The value of D was found in the following way.

X1 = 1.911

X2 = 0 D = 6.84

X3 = 0

X4 = 4.93

The conjecture holds true for this graph, but to make sure it holds for the rest of the values of a , the graph of y = 1.19x 2 – 6x + 11 was drawn. The graph is shown on next page.

The value of D for this graph was 6.72 and the conjecture holds true for this graph too. But the conjecture was further tested with the value of a as 1.20. This is shown in the next page.

Graph : y = 1.2x 2 – 6x + 11

The value of D for this graph was 6.66. This denotes that the conjecture does not hold for this graph and any other graphs from this point onwards, so a new conjecture has to be found out which would prove to be true for this graph too. To find this conjecture, a series of steps need to be taken.

A NEW CONJECTURE

As we have seen, the conjecture holds true for all the graphs till the value of a is 1.2. So a conjecture needs to be found that satisfies all the graphs from then on. So the first two conjectures can be kept the same, only the conjecture needs to be modified after the value of a crosses 1.2.

What can be understood from this ?

We can understand that there needs to be two conjectures to understand the patterns of the changes in the value of D when only one line intersects the graph.

So the general rule for any graph and it’s intersecting lines would be that after a certain point the conjecture changes and in this case that is 1.2.

HOW IS THE CONJECTURE FOUND OUT ?

First the graph of y = 1.21x 2 – 6x + 11 was drawn. After this many more graphs were drawn to find the conjecture. All these processes have been described below.

The graph : y = 1.21x 2 – 6x + 11

The value of D for this graph was 6.61 but according to conjecture it should have been 6.6 so there is a difference of 0.01.

The graph : y = 1.22x 2 – 6x + 11

The value of D for this graph is 6.56

The graph : y = 1.23x 2 – 6x + 11

The value of D for this graph was 6.5

This process was repeated for many other graphs including y = 1.23x 2 – 6x + 11, y = 1.24x 2 – 6x + 11 and y = 1.25x 2 – 6x + 11

The values of D for theses graphs are shown below.

The value of a | The value of D |

1.23 | 6.5 |

1.24 | 6.45 |

1.25 | 6.4 |

If we have to find a pattern, we have to relate it with the values that D would have taken it we used the original conjecture. The value of a and the value that D would have taken if the original conjecture was used is shown below.

The value of a | The value of D if original conjecture is used. |

1.2 | 6.66 |

1.21 | 6.6 |

1.22 | 6.54 |

1.23 | 6.48 |

1.24 | 6.42 |

1.25 | 6.36 |

Conclusion

The value of D in this graph is 1. Now the graph of y = 1.01x 2 – 6x + 11 was drawn.

It is shown below.

The value of D for this graph is 0.98.

The conjecture was that for every increase of 0.01 there would be a decrease of a specific value in the value of D. In the above case, when “a” was increased by 0.01, the value of D reduced by 0.02. Hence the first art of the conjecture holds true.

Now to test the second part of the conjecture, again a graph was drawn. The graph was y = 1.37x 2 – 6x + 1. The graph is shown below.

The value of D for this graph was 5.84. To check whether the conjecture holds true for this graph, it was applied. The following shows how. The conjecture has to be modified as it is a different graph. Instead of 7.14 in the first formula, it is 5.84. Instead of 1.23 in the second formula, it is 1.37. These modifications were done and the conjecture was applied. The result was 5.83. So the conjecture holds true.

CONCLUSION

There are two different conjectures for different parts of the graph. One when both the lines intersect and one when only one line intersects. The conjectures have been explained. But the only problem is that when the intersecting lines change, some values in the equation must be changed.

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

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