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Quadratic Polynomials. Real and Imaginary components

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Introduction

MATHS IA                                                                                                             Alex Chen

PART A (Quadratic Polynomials)

The investigation is to find out if the zeros and to determine the real and imaginary components of the complex zeros of image00.png

.

From the function given, image01.png

The coordinates of the vertex is image02.png

image01.png

image03.png

by using  the Quadratic equation:image04.png

where image05.png

image06.png

image07.png

image08.png

image09.png

image10.png

image11.png

image12.png

Hence, image00.png

has zeros image13.png

, and image14.png

By subbing in different numbers of image15.png

into the equation: image01.png

For: image16.png

, it is given that image14.png

, which is equal to image17.png

image18.png

:                                                              image19.png

:

Value of a

value of b

value of y1

1

1

image20.png

2

2

image21.png

3

3

image22.png

4

4

image23.png

Value of a

value of b

value of y1

1

1

image24.png

2

2

image25.png

3

3

image26.png

4

4

image27.png

For: image28.png

image29.png

:                                                                image30.png

:

Value of a

value of b

value of y1

1

1

image31.png

2

2

image32.png

3

3

image33.png

4

4

image34.png

Value of a

value of b

value of y1

1

1

image35.png

2

2

2

image36.png

3

3

image37.png

4

4

image38.png

...read more.

Middle

y1 and y2 is shown below when a= 3 and b=5,

image125.png

We know that image00.png

has zeros image13.png

, while image42.png

 has opposite concavity to image00.png

,which is in the form image43.png

.

From the graph, it can be seen that, image42.png

is a reflection of image00.png

,

Therefore, the equation of the quadratic image42.png

 is :

image44.png

.

When

image45.png

image46.png

image47.png

image48.png

image126.png

As the graph shown above,

image00.png

 and image42.png

 have point of intersection, The shadow generating function is image49.png

.

Therefore, the shadow generating function for this quadratic is image50.png

To express image42.png

 in terms of image00.png

,

image51.png

,

image52.png

.

By substitution into this function:

image53.png

image44.png

As image54.png

image55.png

image56.png

image57.png

Coordinates of the vertex,image58.png

Shadow function:image59.png

Shadow root:image60.png

Complex zeros:image61.png

The reflectionimage62.png

As the above results shown, in the complex zeros:image63.png

the imaginary components is image64.png

PART B (Cubic Polynomials)

The investigation

...read more.

Conclusion

image116.png

.

image117.png

image118.png

It can be written in the below format:

image119.png

For the generating function of image49.png

Firstly, it’s slope needs to be found and the intersection point of image41.png

,

By using the equation: image111.png

And the generating formula is calculated as:

image114.png

image129.png

As the above graph shown, the cubic function image00.png

 and image42.png

, image120.png

 cuts through between the intersection points of

image00.png

 and image42.png

,

image100.png

image116.png

image121.png

 fits the result.

In general, the cubic function image42.png

 is a reflection of image00.png

.

Considering the derived function,

image51.png

,

image122.png

 the functionimage123.png

image130.png

Again, considering the derived function,

image51.png

,

As the above diagram shows,

from point E to D and point F to D is the distance of the quadratic function image124.png

,

therefore the derived function image51.png

works in a quadratic function.

image131.png

As the above diagram shows,

from point a to b and point c to d is the distance of the image124.png

,

therefore the derived function image51.png

works in cubic function.

...read more.

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