# 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.

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

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

This student written piece of work is one of many that can be found in our GCSE Gradient Function section.

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