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# Math IA Type 1 In this task I will investigate the patterns in the intersection of parabolas and the lines y = x and y = 2x.

Extracts from this document...

Introduction

Nikhil Lodha

Math Internal Assessment Type 1 – Investigation

Parabola Investigation

Math HL

Mr. Hogatt

17 March, 2009

Introduction

In this task I will investigate the patterns in the intersection of parabolas and the lines y = x and

y = 2x. Based on some of these patterns I will make conjectures and attempt to prove them. Certain conditions need to be held constant while proving the conjecture and each time I will try to broaden the scope of the investigation by attempting to find the effect of varying the conditions that were initially held constant such as, the placement of the vertex or the slope and y-intercept of the intersecting lines. Then after I have proven my conjectures I will look to apply them to cubic polynomial and try to obtain a general term or a modified conjecture which applies to other higher order polynomials. In this process it is very possible that I might need to reference some key mathematical theorems and formulas from a range of resources. One key mathematical theorem for this investigation would be Vieta’s theorem, which analyzes the roots of polynomials, especially 2nd and 3rd degree polynomials. I will also use the program Graphmatica for illustrating the graphs for the various functions. This step will be very important as the position and behavior of the intercepts or roots of the function is very important. Also I will look at the behavior of the functions and intersections in various quadrants, therefore graphing and illustrating would be an integral part of this investigation and I will start off by showing the graph of the intersection followed by the calculation to find D.

1. First I will the graph the functions

Middle

[In red]

Again, the parabola of the function   does not intersect the lines y = x or y = 2x and therefore has no real roots and a value for D cannot be found.

Now I will look analyze the parabola when in intersects the lines y = x and y =2x  with its vertex in quadrant 2.

Example 2

[In blue]

[In black]

[In pink]

Therefore, this results is in accordance with my conjecture  that:

I have found that D = 1 with the help of the graph of the function .  The value for a is -1 and therefore  when a is substituted into the conjecture, then the results is

D=  1

Parabolas with vertices in Quadrant 3.

Example 1

[In red]

[In black]

[In pink]

Example 2

[In red]

[In black]

[In pink]

For examples 1 and 2 here, no solutions or roots can be found because the linear and quadratic function do not intersect and it is therefore impossible to find the roots and therefore D is not defined for these graphs.

Example 3

[In red]

[In black]

[In pink]

When the vertex is in quadrant 3, again the trend continues that

• The placement of the vertex is not important as long as the linear functions intersect the parabola twice each.

Example 1

[In blue]

[In black]

[In pink]

Example 2

[In green]

[In black]

[In pink]

This example shows how there are no solutions x1, x2, x3, x4 if the parabola does not intersect the given lines, y = x and y = 2x.

Therefore, through analysis of the above examples I have found that it does not matter, where the vertex is placed or what the value of a is,

Even this conjecture has some restrictions though, these conditions are  that:

• The lines intersecting the function must be y = x and y = 2x.
• The lines should intersect the parabola in 2 places each.

Proof  showing that the  conjecture[ ] is true.

 The equation of any parabola is . It is given that parabola intersects the lines y=x and y=2x. Therefore in order to find D, their intersections must be found. Their intersection can be found be equation the function of the parabola and the function of the line.The intersection of the functions y = x and  must the roots of Therefore the 2 intersections of the parabola with the first line[y=x] are: X2 = X3 = Now the roots of the intersection of the second line must be found. In order to use the formula,  where x2 and x3 are intersections of the first line with the parabola and x1 and x4 are intersections of the second line with the parabola. Thus I will now find the intersections of the second line with the parabola.Therefore the 2 intersections of the parabola with the first line[y=x] are: X1 = X4  = Now substituting values of  x1, x2, x3, x4 into the formula: →Therefore D =

Even though this conjecture has been proven, still there are some limitations

• The lines intersecting the function must be y = x and y = 2x.
• The lines should intersect the parabola in 2 places each.

Conclusion

Where x2, x3, x5 and x7 are intersections of  one linear function and the quartic function  and x1, x4, x6 and x8 are intersections of another linear function and the quartic function.

D = 0

An easier way to this would be to find the sum of the roots of the first line intersecting the quartic and then subtract that from the sum of the roots of the second line intersecting the quartic.

Where Sn  is the sum of the roots of the first linear function intersecting the quartic function and Sm  is the sum of the roots of the second linear function intersecting the quartic function.

I will look at another higher order polynomial to see if they support the conjecture that D=0.

Example 2

Functions in graph:

y = [x-2][x-1][x+3][x+1][x-4][in red]

y = 0.5x[in black]

y = 2.5x[in blue]

D = 0

Again D = 0 and therefore the conjecture holds even for higher order polynomials. This happens because the sum of the roots is always  where a and b are the coefficients of the xn and xn-1 terms of the polynomial respectively.

D =

Where Sn  is the sum of the roots of the first linear function intersecting the quartic function and Sm  is the sum of the roots of the second linear function intersecting the quartic function.

D =

D =

Therefore D = 0 for all other higher order polynomials!

Bibliography

"ViÃ¨te's formulas -." Wikipedia, the free encyclopedia. 16 Mar. 2009 http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas>.

"Vieta's Formulas -- from Wolfram MathWorld." Wolfram MathWorld: The Web's Most Extensive Mathematics Resource. 16 Mar. 2009 <http://mathworld.wolfram.com/VietasFormulas.html>.

"Math Forum - Ask Dr. Math." The Math Forum @ Drexel University. 16 Mar. 2009 <http://www.mathforum.org/library/drmath/view/61024.html>.

[1] "ViÃ¨te's formulas -." Wikipedia, the free encyclopedia. 16 Mar. 2009 http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas>.

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