# A Study On Polynomial Graph Biology Essay

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## 1.0 INTRODUCTION

## 1.1 Research Question

Does the series of curves of fn(x) = nx2+n is has the same properties of midpoint differences between curves for all n=1, 2, 3... , 10 and is it the same as the horizontal lines is increases to infinity?

## 1.2 Background Research

There are many various type of polynomial graph that could be constructed through the Cartesian plane according to the formulae. This many polynomials graphs have certain properties that exhibit constantly with the series of that of polynomials. But the properties of the polynomials only occur to the same type of graphs. It makes me wonder if there is a property that can be hold for all type of polynomials graph that can be made. If there is a property exists whereby all the polynomials can be refer to, attempt for an addition to the knowledge of mathematic will be successful. The property of the series of graph of these polynomials is not the basic property which is, true; exist in all the series of polynomial such as that the polynomial curves is in the range of ]-∞,∞[ .

## 1.3 The Algebra of Polynomials Graphs

Polynomials series is a function built up from powers of x which is in the form of y=axn+bxn-1+cxn-2+...+ex0. The constant is known as the coefficient of x. For this essay, the coefficient of x is n whereas n is a real integer. In this case, the n will be increases by one unit per curves.

## 1.4 Justification of Research

The reason for me to do this research is to find out about the relationship between the series of curves which when it intersect with a given horizontal line, it can be determine by using the midpoint formulae. This is because if this property retains real for this series of quadratic curve, it may too be real for higher order of polynomials such as cubic or quatic. The reason why I used the quadratics curves is that it was the lower positive order of polynomials which means that this quadratics is the fundamental for the polynomials.

## 2.0 METHODOLOGY

## 2.1 The Cartesian Coordinate System (CCS)

CCS is the system used to plot a series of data whether it is continuous or a discrete series of data. Various polynomials can be drawn on this Cartesian plane whether it is in two - dimensional plane or in three - dimensional plane. In this essay, all the polynomials will be referred to the CCS for calculation and estimation of specified relationship. Since the data are represents in an equation, obviously it is continuous data; this CCS can plot it and show the relationship between the polynomials series clearly.

## 2.2 Midpoint

Midpoint calculation is one of the basic calculations that are used in CCS. The basic formula is:

M (x, y) =

In this essay, this basic formula is widely used since the plotted polynomials series shows strong midpoint behaviour. This midpoint can be used to define a point on a curve but not as whole which means that to find a midpoint between two points can be done but to find a midpoint between two curves is, although can be calculated, hard. This is because a curve is made up of many points that connected to each other and most importantly the position of the points is different to each other.

## 2.3 Expected Relationship between the Series of Curves

## 2.3.1 Main Hypothesis

Let's take fx is a curve in the Cartesian plane. fy is another curve that is larger by one unit in its coefficient of fx and fz is another curve that is one unit larger in the coefficient compare to fy. I try to find the relationship between these curves where when there are lines is drawn at any value of y, the intersections point of the line y with the curve fy will be the centre of the point between the intersections of line y and curve fx and fz. Since the equation of curves that I tested is

f(x) = nx2+n, the curve will be in quadratic form, so what I wrote before is also true for the negative side of the curve.

## Figure shows the series of curves with the intersecting line of y =c

My problem statement is, does the series of curves of fn(x) = nx2+n is has the same properties of midpoint differences between curves for all n=1, 2, 3... , 10 and is it the same as the horizontal intersecting lines is increases to infinity?

## 2.3.2 Subsequent Hypothesis

The horizontal intersecting lines are better to be used as intersecting lines rather than slanting intersecting lines.

## 3.0 DATA COLLECTION

## 3.1 The Hypothesis Test

## Figure shows all of the series of curves that will be tested in this essay.

Take for example when I draw a line at y = 10 (because the series of curve has a minimum at y = 10), there are two sets of curves that shows the properties stated earlier.

## Figure shows the series of curves with an intersection line at y = 10

The first set of curve is the curve f1(x), f2(x) and f5(x). These curves have intersection points when the line of y = 10 is drawn. The coordinates of these points are:

A = (3.000, 10.000)

B = (2.000, 10.000)

C = (1.000, 10.000)

By using the basic midpoint calculation to find the midpoint between the intersection point of A and C:

M (x, y) =

So:

M (x, y) =

= (2.000, 10.000)

The coordinate obtained is same as the coordinate of the intersection at B. So I can say that B is the midpoint of the line made by joining point A and point C.

Another series of equation that shows the same properties as the series of curve f1(x), f2(x) and f5(x), are the series of curve f5(x), f8(x) and f10(x). The intersections coordinates are (from the right):

A = (1.000, 10.000)

B = (0.500, 10.000)

C = (0.000, 10.000)

By using the earlier midpoint calculation to find the midpoint between the intersection point of A and C:

M (x, y) =

= (0.500, 10.000)

The coordinates of midpoint calculated is the same as the coordinates of B. So two series of curve in the range of n ≤ 10 that can have the properties of midpoint are curve f1(x): f2(x):f5(x) and f5(x):f8(x):f10(x).

Now let's try to increase the value of y to 17 and due to before calculations that only two series of curves that can have this properties, I only use the two of them to test if it can retain at y = 17.

## Figure shows the series of curves with the intersection line at y = 17

The intersection coordinates of the series of curve f1(x): f2(x):f5(x) and the line y = 17 are (from the right): A = (4.000, 17.000)

B = (2.739, 17.000)

C = (1.549, 17.000)

By using midpoint equation to find the midpoint between the intersection point of A and C:

M (x, y) =

= (2.776, 17.000)

M (2.775, 17.000) ≠ B (2.739, 17.000)

The value obtained from the midpoint calculation is not the same as the coordinates B that are supposed to be the midpoint of A and C. The difference in value is:

## Δ x = |M - B|

= 0.036

This conflict of value also happens with the second series of curve that is f5(x):f8(x):f10(x). The value obtained for midpoint is (1.193, 17.000) while the value of B is (1.061, 17.000). The difference in value is:

## Δ x = |M - B|

= 0.132

The Δ x for the second series of curve is much bigger compare to the series of first curve. This I will explain after this section. So my inference is this property only retain when y = 10. It cannot be hold anymore when y 10. To prove it, let's try if the line of y is at 10.0000001 (only this section that the integer is extended to seven decimal points) and refer to the first series of curve that is f1(x): f2(x):f5(x). The intersection coordinates with the line y=10.0000001 are:

A = (3.00000002, 10.0000001)

B = (2.00000001, 10.0000001)

C = (1.00000001, 10.0000001)

The midpoint that supposes to be obtained is 2.000000015 but the value is not the same as value of B. So even when I change the value of y to the millionth, the value of M still not the same as B even when we round up the value of M. In this case the value of M will be 2.00000002.

After my failure in the value of M if y10, I tried to change the intersection line to adjust the value of y so that it can hold back the properties. The new equation of y will be:

## Y = |m x|+c

So the rough idea should be like this:

## Figure shows the series of curves intersects with the intersecting line of y =|0.1x|+c

When I let the graph to be like above, I found out another series of curves that shows this midpoint property that is f2(x): f3(x):f5(x). I try to increase the M for only a small value that is 0.0000001 and the value of c is 10 because the minimum point where the curve intersects is 10 and f10 have a minimum at 10. Let's try with the first series of curve which is f1(x): f2(x):f5(x). The intersection coordinates of y = |0.0000001x| + 10 are:

A = (3.00000005, 10)

B = (2.00000003, 10)

C = (1.00000001, 10)

From the calculations of midpoint, I obtain that the value of M is equal to B. Let's try if the value of m in the equation of y = |m x| + c to 0.1 and the value of c is still 10. The intersection coordinates of the curve with y = |0.1x| + 10 and the first series of curve are:

A = (3.050, 10.305)

B = (2.025, 10.203)

C = (1.010, 10.101)

The midpoint obtained is (2.025, 10.203). The value of this M is not equivalence to B for Δ x = 0.005. This value is smaller than the other previous values. When I round up the value of M and B (for x to the nearest hundredth) the value obtained is the same which is (2.03, 10.203). So the property of midpoint is still holding up until to the nearest hundredth.

Before releasing any possible equation, let's try if this equation holds up if the c is equal to exact number larger than 10. I will try if c equal to 15 and the m is equal to 0.1 because if I use m = 0.0000001, the value is approximately to 15 and the result can be forecasted. So the coordinates of the intersections point of y = |0.1x|+15 (refer appendix A) are:

A = (3.792, 15.379)

B = (2.575, 15.258)

C = (1.424, 15.142)

The M obtained is (2.608, 17.261) and the value is still more than B with Δ x = 0.041 and Δ y = 0.004. The value of M and B, if rounded up(x to the nearest tenth and y to the nearest hundredth), the value will be the same which is (2.61, 17.28). The value is near but it is actually have a big uncertainty.

When I tried c with a bigger number such as c = 1000 and with the same m, the results are:

Y = -|0.1x|+ 10

f1(x): f2(x):f5(x)

f2(x): f3(x):f5(x)

F3(x):f5(x):f8(x)

f5(x):f8(x):f10(x)

Midpoint, M

(22.889, 1002.289)

Midpoint, M

(18.240, 1001.824)

Midpoint, M

(0.510, 9.951)

Midpoint, M

(0.510, 9.951)

Coordinate of B

(22.363, 1002.236)

Coordinate of B

(18.247, 1001.825)

Coordinate of B

(0.494, 9.951)

Coordinate of B

(0.494, 9.951)

Difference in x, Δx

0.005

Difference in x, Δx

0.008

Difference in x, Δx

0.007

Difference in x, Δx

0.007

Difference in y, Δy

0.000

Difference in y, Δy

0.003

Difference in y, Δy

0.000

Difference in y, Δy

0.000

The results of Δx obtained still exist and although the value is small, it is still significant, So I tried another intersection line that is y = -|0.1x|+c, where the intersection lines should be like below and the c is equal to 10:

## Figure shows the series of curves intersects with the intersecting line of y =|-0.1x|+c

The results are as follow (the difference of x and y is calculated by using the formula of

Δ = |M - B|):

Y = -|0.1x|+ 10

f1(x): f2(x):f5(x)

f2(x): f3(x):f5(x)

f5(x):f8(x):f10(x)

Midpoint, M

(1.970, 9.803)

Midpoint, M

(1.483, 9.852)

Midpoint, M

(0.510, 9.951)

Coordinate of B

(1.975, 9.803)

Coordinate of B

(1.511, 9.849)

Coordinate of B

(0.494, 9.951)

Difference in x, Δx

0.005

Difference in x, Δx

0.008

Difference in x, Δx

0.007

Difference in y, Δy

0.000

Difference in y, Δy

0.003

Difference in y, Δy

0.000

## Table shows the data calculated for the series of curves

And then I try if the c is increase to 15.

Y = -|0.1x|+ 15

f1(x): f2(x):f5(x)

f2(x): f3(x):f5(x)

f5(x):f8(x):f10(x)

Midpoint, M

(2.838, 14.716)

Midpoint, M

(1.964, 14.804)

Midpoint, M

(1.053, 14.895)

Coordinate of B

(2.525, 14.748)

Coordinate of B

(1.983, 14.802)

Coordinate of B

(0.929, 14.907)

Difference in x, Δx

0.313

Difference in x, Δx

0.019

Difference in x, Δx

0.601

Difference in y, Δy

0.032

Difference in y, Δy

0.002

Difference in y, Δy

0.012

## Table shows the data calculated for the series of curves

The result obtain is quite undependable because the difference of x-coordinates and y-coordinates is significantly large. This will be discussed in the following section.

## 3.2 The Efficiency Test

Another thing is I tried to calculate the efficiency of using either the intercept as straight line or slanting line. So I draw all the test line within the same value of c and test it. The curves intercept as the following page. (This is done by assuming that at line y = 100, the differences of midpoint and coordinate of b is already normally distributed at this point)

## A

## B

## C

Fz

Fx

Fy

## Figure shows the series of curves with all the intersecting lines with it.

I tested it with the c of at least five units larger than the maximum vertex and in this test I used c=170 and the series of curves that I tested is f2(x): f3(x):f5(x), f2(x):f4(x):f9(x) and f3(x):f5(x):f9(x). I found out the results is as the other page:

the test line is y = 100

Series of curves

f2(x): f3(x):f5(x)

f2(x):f4(x):f9(x)

f3(x):f5(x):f9(x)

Midpoint, M

(5.680, 100.000)

(5.090, 100.000)

(4.433, 100.000)

Coordinate of B

(5.686, 100.000)

(4.900, 100.000)

(4.359, 100.000)

Difference in x, Δx

0.006

0.190

0.074

Difference in y, Δy

0.000

0.000

0.000

## Table shows the data at the test line of y = 100

the test line is y = |0.1x|+100

Series of curves

f2(x): f3(x):f5(x)

f2(x):f4(x):f9(x)

f3(x):f5(x):f9(x)

Midpoint, M

(5.697, 100.570)

(5.105, 100.511)

(4.444, 100.445)

Coordinate of B

(5.703, 100.570)

(4.912, 100.491)

(4.369, 100.437)

Difference in x, Δx

0.006

0.193

0.075

Difference in y, Δy

0.000

0.020

0.008

## Table shows the data at the test line of y = |0.1x|100

the test line is y = -|0.1x|+100

Series of curves

f2(x):f3(x):f5(x)

f2(x):f4(x):f9(x)

f3(x):f5(x):f9(x)

Midpoint, M

(5.662, 99.433)

(5.075, 99.493)

(4.422, 99.558)

Coordinate of B

(5.667, 99.433)

(4.886, 99.511)

(4.349, 99.565)

Difference in x, Δx

0.005

0.189

0.073

Difference in y, Δy

0.000

0.018

0.007

## Table shows the data at the test line of y = |-0.1x|100

To find the efficiency, which determines which test line is suitable to find the midpoint of a series of curve, I used the formula of mean by which the value nearest to the mean will be considered as efficient. I calculated the mean for all series of curves. In this case, I assume that y-coordinates of B are always the same as the y-coordinates of M. So I considered that the difference in x-coordinates presents their respective test line. The mean formula I used in this calculation is as below:

So the means I obtain are:

Test Line

Mean

f2(x):f3(x):f5(x)

f2(x):f4(x):f9(x)

f3(x):f5(x):f9(x)

y = 100

0.006

0.191

0.074

y = |0.1x|+100

y = -|0.1x|+100

## Table shows the mean value for the midpoint of the differences in Δx

After I collected these data, I tried to find out the degree of efficiency by using the formula below:

The results are as follow:

Test Line

Degree of Efficiency

y = 100

y = |0.1x|+100

y = -|0.1x|+100

## Table shows the result of efficiency test

By observing the value, I found out that the line y = 100 has the degree of efficiency that is near to 100% which only differs for 0.175% compare to other method that differ to 0.800% and 6.355%.

## 4.0 DATA ANALYSIS

Since I chose the intersecting test line where y = c due to the efficiency test done on the section before, I have to ensure that using this equation, y = c, will give the same result of differences between midpoint and b-coordinate even if it is place on the other point on the y-axis.

So I will use Chi - Square test statistic to test to test whether the midpoints of the intersection between the curves have a uniform difference with their value of B. I will test the hypothesis at 5% level of significance.

H0: the data is distributed uniformly.

H1: the data does not follow uniform distribution.

I assume that all the series of curves will have the same differences (due to the same efficiencies). So in this test, I will use the series of curve f2(x):f3(x):f5(x).

I take 30 samples of data (differences between the midpoint obtained and the value of B) by using the software Microsoft Excel 2007. So the degree of freedom is (as I want to test it as it were distributed normally, so the restrictions will be 1):

υ = number of class (n) - number of restrictions (k)

=30 - 1

=29

y

A

B

C

m

Δx

100

4.3589

5.6862

7.0000

5.6794

0.0068

110

4.5826

5.9722

7.3485

5.9655

0.0066

120

4.7958

6.2450

7.6811

6.2385

0.0065

130

5.0000

6.5064

8.0000

6.5000

0.0064

140

5.1962

6.7577

8.3066

6.7514

0.0063

150

5.3852

7.0000

8.6023

6.9937

0.0063

160

5.5678

7.2342

8.8882

7.2280

0.0062

170

5.7446

7.4610

9.1652

7.4549

0.0062

180

5.9161

7.6811

9.4340

7.6750

0.0061

190

6.0828

7.8951

9.6954

7.8891

0.0061

200

6.2450

8.1035

9.9499

8.0974

0.0061

210

6.4031

8.3066

10.1980

8.3006

0.0060

220

6.5574

8.5049

10.4403

8.4989

0.0060

230

6.7082

8.6987

10.6771

8.6926

0.0060

240

6.8557

8.8882

10.9087

8.8822

0.0060

250

7.0000

9.0738

11.1355

9.0678

0.0060

260

7.1414

9.2556

11.3578

9.2496

0.0060

270

7.2801

9.4340

11.5758

9.4280

0.0060

280

7.4162

9.6090

11.7898

9.6030

0.0060

290

7.5498

9.7809

12.0000

9.7749

0.0060

300

7.6811

9.9499

12.2066

9.9439

0.0060

310

7.8102

10.1160

12.4097

10.1100

0.0060

320

7.9373

10.2794

12.6095

10.2734

0.0060

330

8.0623

10.4403

12.8062

9.2496

0.0060

340

8.1854

10.5987

13.0000

9.4280

0.0060

350

8.3066

10.7548

13.1909

9.6030

0.0060

360

8.4261

10.9087

13.3791

9.7749

0.0060

370

8.5440

11.0604

13.5647

9.9439

0.0060

380

8.6603

11.2101

13.7477

10.1100

0.0060

390

8.7750

11.3578

13.9284

10.2734

0.0060

total of Δx

0.1839

## Table shows the data for the test line of y=c at various sample.

So the next step is to find the value. The value of E (expected frequency) is:

E =

=0.00613

O(obtained frequency)

E(expected frequency)

O - E

(O - E)2

0.0068

0.00613

0.00067

4.489E-07

7.323E-05

0.0066

0.00613

0.00047

2.209E-07

3.60359E-05

0.0065

0.00613

0.00037

1.369E-07

2.23328E-05

0.0064

0.00613

0.00027

7.29E-08

1.18923E-05

0.0063

0.00613

0.00017

2.89E-08

4.71452E-06

0.0063

0.00613

0.00017

2.89E-08

4.71452E-06

0.0062

0.00613

7E-05

4.9E-09

7.99347E-07

0.0062

0.00613

7E-05

4.9E-09

7.99347E-07

0.0061

0.00613

-3E-05

9E-10

1.46819E-07

0.0061

0.00613

-3E-05

9E-10

1.46819E-07

0.0061

0.00613

-3E-05

9E-10

1.46819E-07

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

0.0060

0.00613

-0.00013

1.69E-08

2.75693E-06

Total of

0.000207341

## Table shows the calculation to find the sum of

Since we stated that the significance level, α = 0.05, the critical value is χa and when υ = 29, the value of χa is ≈42.56. The graph of chi - square is like below with the rejection and non - rejection region:

Since the value of = 0.000207341 is far too small than 42.56

0.000207341 < 42.56, we accept the H0 and reject the alternative hypothesis. Thus showing that, the differences between the midpoint and the value of B is uniformly distributed.

From the test, I assume that all the series of curves have the same value for differences in midpoint and value of B regardless the position of the intersection lines. However, the sample data is just taken from 90 points away from the maximum vertex which is ten. This will be explained in the next section.

(All of the calculation is done by using Microsoft Excel 2007, refer Appendix B)

## 5.0 RESULTS AND FINDINGS

## 5.1 The Acceptance of a Series

What I discovered was, I cannot test those series of curves as the test line, which is the horizontal line, intersect with the vertices of either one of the curves. The first reason is if I tried to place the test line on the value of less than the maximum c of the series of curve, which is 10 in this case, I cannot test the hypothesis. This is because; the test line will not intersect with the curve that has a larger c than the test line. The second reason is if we plot a tangent along the curves, we can see that as the tangent approaches to the vertex, the tangents' values will decreases and approaches to zero. The problem is like below.

## Figure shows various tangents that can be obtained near the vertex.

I cannot test the hypothesis with the line that is near to the vertices because the gradient of tangents are approaching zero and have significance different between each other. So to test the hypothesis, I need to test it with the horizontal line that is at least 90 units more than the maximum vertex to ensure that the value gradient is close to each tangent. Not just the value is near the tangent, the tangent is also nearly parallel to the curves. The image on the next page will clearly shows the tangent that nearly parallel when it is 90 units away.

Tangent at y ≥ 100

Tangent at y < 100

## Figure shows the difference between tangent under y 100 and after y=100

I take 90 units because by using the first principle of derivative which is to find a derivative (tangent on a curve) we can determine that as the point of a gradient is approaching the other point of the gradient, the gradient is approaching to its tangent. As far as I see, from 90 points of the maximum vertex onwards, the value of the gradient is approaching the tangent for each point. That is my justification of why I used to find the gradient 90 units away the maximum vertex.

We may use the formula below to prove it.

Therefore, all calculation that involving the y-intercept that is within the range of under 90 units away from maximum vertex is not valid. Further observation shows that the series of curves that actually inhibits the midpoint properties are:

f1(x): f2(x):f5(x)

f2(x): f3(x):f5(x)

f2(x): f4(x):f9(x)

f3(x): f5(x):f9(x)

f4(x): f6(x):f9(x)

f5(x): f7(x):f10(x)

f7(x): f8(x):f9(x)

f8(x): f9(x):f10(x)

(Please refer appendix A for the graph)

This series of curves is chosen by roughly estimating it visually. The calculation is made after the observation has been made.

## 5.2 Subsequent Hypothesis

From the efficiency test before, it is proven that the horizontal line is much better to use than the slanting lines. This is because, when using slanting lines as the intersecting lines for this tests, the coordinates of midpoints actually are not easy to be determine because there will be another variable to control which is the gradient. The gradient which is can be any integer will be a problem because it will be much to test and might need another essay to discuss about it. So for these tests in this essay, the horizontal lines are the best intersecting lines to be used. Another important thing is, this series of test is worked best if the n is tested in this equation;

f (x) = nx2+n

The fact is, that equation is used in this test is because his is because it will be easier for the calculation of the midpoints because we only calculate at the positive side and as the negative side have the same value, only it is negative, the calculation of it is only referred to the positive side's calculation.

## 6.0 CONCLUSION

For the conclusion, I have found that my main hypothesis that is, Fx, Fy and Fz shows a relationship where when there are line is drawn at any value of y, the intersections point of the line y with the curve fy will be the centre of the point between the intersections of line y and curve fx and fz, is not accepted. This is because there still at least to a thousandth differences for the coordinate of B to be the midpoint of the point A and point C. To accept the hypothesis, there must not be even slightest difference in the calculations or in other words, the differences between the midpoint calculated and coordinates B is zero.

If we were to accept the hypothesis, the main hypothesis, the best intersecting line is the straight line passing through the y - axis. This is proven by its efficiency. The other two intersecting lines are only worked on certain range of value but for the straight line, it is applicable from the 90 units away from maximum vertex to infinity. This is proven by the Chi - Square test that shows that after the 90 units away, the differences of the midpoint and coordinate of B is uniformly distributed.

Then again, the subsequent hypothesis, the horizontal lines is much suitable to use in proving this relationship rather than using the slanting lines which we needed to consider about its gradient. Furthermore, the equation is used for easy calculation on both sides of the quadratic curves.