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Investigate the gradients of the graphs Y=AXN

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Introduction

GCSE Math’s Coursework

Gradient Function

        In this investigation I am going to investigate the gradients of the graphs Y=AXN Where A and N are constants.  I shall then use the information to find a formula for all curved graphs.

        To start the investigation I will draw the graphs where A=1 and N= a positive integer.

Y=X2

X

Height

Width

Gradient

1

1

0.5

2

2

4

1

4

3

9

1.5

6

4

16

2

8

        Looking at the results above I can see that the gradient is twice the X value, the height is X2 and the width is 1/2 the X value.  This shows me that there are several patterns in the graph but there is not enough to make a formula on so I am going to do another graph

Y=X3

X

Height

Width

Gradient

1

1

0.33

3

2

8

0.66

12

3

27

1

27

4

64

1.33

48

        There are some more patterns in this table, the height is now X3 and the width is 1/3 of the X value. I can see no pattern between the Gradient and the X value in this table.

...read more.

Middle

0.015625,0.015613287

0.000011713/0.001

0.011713

Y=X-4

X

Predicted gradient with formula (3dp)

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

-4*1(-4-1) = -4

1,1.001

1,0.99600998

-0.00399002/0.001

-3.99002

2

-4*2(-4-1)= -0.125

2,2.001

0.0625,0.062375156

-0.000124844/0.001

-0.124844

3

-4*3(-4-1) = -0.016460905

3,3.001

0.012345679,0.012329231

-0.000016448/0.001

-0.016448

4

-4*4(-4-1) = -0.00390625

4,4.001

0.00390625,0.003902346

-0.000003904/0.001

-0.003904

In both of the above cases the formula is almost identical to that of the increment method, this shows that my formula works of all cases for when A=1 and N= an integer. I am now going to experiment for when N= a faction.

Y=X1/2

X

Predicted gradient with formula (3dp)

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

1/2*1-1/2 =0.5

1,1.001

1,1.000499875

0.000499875/0.001

0.49987

2

1/2*2-1/2 =0.35355339

2,2.001

1.414213562,1.414567072

0.00035351/0.001

0.35351

3

1/2*3-1/2 =0.288675134

3,3.001

1.732050808,1.732339459

0.000288651/0.001

0.288651

4

1/2*4-1/2 =0.25

4,4.001

2,2.000249984

0.000249984/0.001

0.249984

This table shows that my formula work when X is a fraction and to prove this I will now look at the graph Y=X1/3 and see if the formula still works.

Y=X1/3

X

Predicted gradient with formula (3dp)

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

1/3*1-2/3 =0.3333333

1,1.001

1,1.000333222

0.000333222/0.001

0.333222

2

1/3*2-2/3 =0.209986841

2,2.001

1.25992105,1.260131002

0.000209952/0.001

0.209952

3

1/3*3-2/3 =0.160249952

3,3.001

1.44224957,1.442409802

0.000160232/0.001

0.160232

4

1/3*4-2/3 =0.132283421

4,4.001

1.587401052,1.587533324

0.000132272/0.001

0.132272

The formula works for when n= a fraction as well.  This is good for I now know that my formula works for any cases of N.  I shall now see if the A in the formula ANXN-1 is correct.  I shall do this by keeping N as 2 as it is the lowest number and make A 2 then 3.

 Y=2X2

X

Predicted gradient with formula

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

2*2*11 =4

1,1.001

2,2.004002

0.004002/0.001

4.002

2

2*2*21  =8

2,2.001

8,8.008002

0.008002/0.001

8.002

3

2*2*31 =12

3,3.001

18,18.012002

0.012002/0.001

12.002

4

2*2*41 =16

4,4.001

32,32.016002

0.016002/0.001

16.002

Y=3X2

X

Predicted gradient with formula

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

3*2*11 =6

1,1.001

3,3.006003

0.006003/0.001

6.003

2

3*2*21  =12

2,2.001

12,12.012003

0.012003/0.001

12.003

3

3*2*31 =18

3,3.001

27,27.018003

0.018003/0.001

18.003

4

3*2*41 =24

4,4.001

48,48.24003

0.024003/0.001

24.003

...read more.

Conclusion

N Where a and n are constants.  Now that I have proven this I shall try it out in other sets of graphs and see if it still works.  I am now going to work out the gradient of the graph Y=2X3+4X+9.  I am going to spilt the formula up into smaller fragments to make it easier to apply the formula so Y=(2X3)+(4X)+9.  The numbers before the X (Y=(2X3)+(4X)+9) are the A’s and the number after the X (Y=(2X3)+(4X)+9) is N, this means that the formula should be Gradient= (ANXN-1)+(AX).  The 9 at the end of the equation will make no difference to the gradient at all, it just tells us were the line cuts the Y axis.

Y=2X3+4X+9

X

Predicted gradient with formula From set

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

(2*3*12)+(4)=10

1,1.001

15,15.010006

0.010006/0.001

10.006

2

(2*3*22)+(4)=28

2,2.001

33,33.028012

0.028012/0.001

28.012

3

(2*3*32)+(4)=58

3,3.001

75,75.058018

0.58018/0.001

58.018

4

(2*3*42)+(4)=100

4,4.001

153,153.100024

0.100024/0.001

100.024

The formula still seems to be working for different curves and I feel that it always work for all graph that includeY=AXN

...read more.

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