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Introduction

## The Gradient Function

The aim of this investigation is to discover the gradient function for the graphs y = ax where a and n are constants.

I will do this by beginning with the simplest cases, as I believe that these will be the most simple equations to solve.  I am doing this in the hope that discovering the equations for these simple cases will aid me in discovering the more complex formulas.

Firstly I will construct the graphs of: y=x, y=2x, y=3x, y=4x.  And attempt to find a general equation.

Y=x

 X 1 2 3 4 Y 1 2 3 4

Y=2x

 X 1 2 3 4 Y 2 4 6 8

Y=3x

 X 1 2 3 4 Y 3 6 9 12

Y=4x

 X 1 2 3 4 Y 4 8 12 16

(Graph2)

Using the rule stated on the candidate sheet I can calculated the gradient for each of the above straight lines.

 Equation of line Gradient Function Y=x 1 Y=2x 2 Y=3x 3 Y=4x 4

Doing this I have discovered that the co-efficient of x is the gradient of the line.

I will now go on to construct the graph of y=x2.  This is because I know that the graph of x2 will be a curve and it is curves that I am investigating

 X 1 2 3 4 Y 1 4 9 16

To find the gradient of this line, I will use the tangent method.  If I have a point on the curve, I will draw a tangent so only the point on the curve is touching the tangent.  I will then draw a right-angled triangle with the tangent.

Middle

4

3

9

6

4

16

8

Y = 2x2

X=1, Y=2

= 4.2

= 4.02

= 4.002

X=2, Y=8

= 8.2

= 8.02

= 8.002

X=3, Y=18

= 12.2

= 12.02

= 12.002

X=4, Y=32

= 16.2

= 16.02

= 16.002

 x Y Small increment 1 2 4 2 8 8 3 18 12 4 32 16

Y = 3x2

X=1, Y=3

= 6.3

= 6.03

= 6.003

X=2, =12

= 12.3

= 12.03

= 12.003

X=3, Y=27

= 18.3

= 18.03

=18.003

X=4, Y=48

= 24.3

=24.03

=24.003

 X Y Small increment 1 3 6 2 12 12 3 27 18 4 48 24

Y = x3

X = 1, Y=1

= 3.31

= 3.0301

= 3.003001

X = 2, Y=8

= 12.61

=12.0601

= 12.006001

X=3, Y=27

=27.91

= 27.0901

= 27.009001

X=4, Y=64

= 49.21

= 48.1201

=48.012001

 x Y Small increment 1 1 3 2 8 12 3 27 27 4 64 48

Y = 2x3

X =1, Y=2

= 6.62

= 6.0602

= 6.006002

X=2, Y=16

= 25.22

=24.1202

= 24.012002

X=3, Y=54

= 55.82

= 54.1802

=54.018002

 x y Small increment 1 2 6 2 16 24 3 54 54 4 128 96

Y= 3x3

X = 1, Y=3

Conclusion

Small increment method to find the graph of y=x1/2

Where x=4, y=2

=0.248457

=0.2498439

=0.25

Gradient using small increment method of curve y=x1/2 at point x=4, y=2

=0.25

I will then check this using my equation:

½ x (41/2-1)

=0.25

I can now conclude that this formula is correct for graphs where the power of x is a fraction.

Finally I will check if the formula works for a compound graph.

I will take the graph y=x2+4x.  To calculate the gradient of this graph I will use algebra in the hope of

y=x2+4x

Using the formula I have discovered I can prediction that the gradient function of this curve will be= 2x+4

(x+h)2+4(x+h)-(x2+4x)/x+h-x

(x+H)(x+H)+4x+4h-(x2+4x)/h

x2+2hx+h2+4x+4h-x2-4x/h

2hx +h2+4h/h

2x+4+h

Therefore h=0

I can therefore say that my prediction is correct and can conclude that the equation works for compound graphs.

Overall I can say that for all the different types of graphs I have tried, my equation has been correct.  I can put forward the suggestion that the equation naxn-1 gives the gradient of any curve.  However I would need to extend this investigation to insure this is conclusive and try it with many more graphs to check there are no anomalies.

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

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5 star(s)

m=0.5 The function appears to be correct so far, but I will have to use many more examples in order to show that the gradient function I developed is generally correct. I will show working out the gradient for the calculation below by using the formula: m=nax^n-1 => m=0.5*3.31*1^-0.5 =>

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