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• Level: GCSE
• Subject: Maths
• Word count: 1964

# Investigate the gradients of the graphs Y=AXN

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Introduction

GCSE Math’s Coursework

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.

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

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

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