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Introduction

My task is to investigate the relationship between the gradients of tangents on the curves of graphs when y=axn

Where a is a constant and is not 0, n is equal to 0, 1, 2, 3………

Definition

Gradient of the curve between x1 and x2 is defined as:

When x2 is getting close to x1 the gradient becomes the gradient of the curve at x1.

The gradient to a curve, at a particular point, is given by the gradient of the tangent to the curve at that point.

Method

1. I will draw the curves, y=2x2 and y=ax2 by hand on 5mm graph papers. Next I will draw the tangents and find the gradient of the tangents on the curve when x= -3, -2, -1, 0, 1, 2, 3.

Middle

Δx

y

y+Δy

Δy/Δx

(x a)

(x a)

(x a)

y=ax

-3

0.001

-3

-2.9990000000

1.0000000000

-2

0.001

-2

-1.9990000000

1.0000000000

-1

0.001

-1

-0.9990000000

1.0000000000

0

0.001

0

0.0010000000

1.0000000000

1

0.001

1

1.0010000000

1.0000000000

2

0.001

2

2.0010000000

1.0000000000

3

0.001

3

3.0010000000

1.0000000000

Observation: When n=1, y=ax, the gradient is equal to a for all values of x

 x Δx y y+Δy Δy/Δx (x a) (x a) (x a) y=ax2 -3 0.001 9 8.9940010000 -5.9990000000 -2 0.001 4 3.9960010000 -3.9990000000 -1 0.001 1 0.9980010000 -1.9990000000 0 0.001 0 0.0000010000 0.0010000000 1 0.001 1 1.0020010000 2.0010000000 2 0.001 4 4.0040010000 4.0010000000 3 0.001 9 9.0060010000 6.0010000000

Observation: When n=2, y=ax2, the gradient is approximately equal to 2ax for all values of x

 x Δx y y+Δy

Conclusion

>

3.0030010000

2

0.001

8

8.0120060010

12.0060010000

3

0.001

27

27.0270090010

27.0090010000

Observation: When n=3, y=ax3, the gradient is approximately equal to 3ax2 for all values of x

 x Δx y y+Δy Δy/Δx (x a) (x a) (x a) y=ax4 -3 0.001 81 80.8920539880 -107.9460119990 -2 0.001 16 15.9680239920 -31.9760079990 -1 0.001 1 0.9960059960 -3.9940039990 0 0.001 0 0.0000000000 0.0000000010 1 0.001 1 1.0040060040 4.0060040010 2 0.001 16 16.0320240080 32.0240080010 3 0.001 81 81.1080540120 108.0540120010

Observation: When n=4, y=ax4, the gradient is approximately equal to 4ax3 for all values of x

Pattern observed: When y=axn the gradient of the curve is naxn-1

 x Δx y y+Δy Δy/Δx gradient (x a) (x a) (x a) 10ax9 y=ax10 -3 0.001 59049.0000000000 58852.4649827130 -196535.0172869510 -196830.0000000000 -2 0.001 1024.0000000000 1018.8915046534 -5108.4953465673 -5120.0000000000 -1 0.001 1.0000000000 0.9900448802 -9.9551197903 -10.0000000000 0 0.001 0.0000000000 0.0000000000 0.0000000000 0.0000000000 1 0.001 1.0000000000 1.0100451202 10.0451202103 10.0000000000 2 0.001 1024.0000000000 1029.1315353735 5131.5353734478 5120.0000000000 3 0.001 59049.0000000000 59246.1255075931 197125.5075931400 196830.0000000000

To prove this I have done a table using the small increment method to calculate the gradients of the curves y=ax10 and also using my formula: naxn-1 the two results show above are very close.

Conclusion: The gradient function for y=axn is naxn-1.

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

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