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Extracts from this document...

Introduction

Derek Pollard  5M                        Page

I have been set a task to find the gradient of different graphs ie. The gradient of y = x² or y= x³.

I will start with the graph of y = x². (This is shown below)

I have found the gradients for all of the numbers shown, I did this by drawing a triangle as close to the graph line as I could. I would then find the height and the width of the triangle. Then I would divide the height by the width.

Here is an example:-

Height of triangle 3 = 12

Width of triangle 3  =  2

So.

12    =  6

2

The gradient of point (3,9) is therefore 6.

Here is a table of all of my results:-

Diff 1                        2           2               2                  2               2

This means that my formula should be:-

I will now show the gradients of chords starting at point (2,4) and finishing at various other points along the plotted line ie. 2 to 5 and 2 to 4 etc.

Middle

2x is the gradient function of the curve y = x².

This can also be written as grad x² = 2x.

Here are some examples for when y = x²:-

When x = 5                y = 5² = 25

The gradient = 10        when x = 5

The point is (5,25) and the gradient is 10.

I have checked this result and it is correct.

will now investigate the gradient of y = x³. A graph to show this is below:-

I have found out the gradient for each point on the graph. I did this in the way that I explained on page1. A table shows this below:-

Diff 1                        3           9               15          21               27

Diff 2                              6                  6              6                 6

This means that my formula for the gradient is:-

3x²

Here is a table for the chords:-

Diff 1               5.75         5.25          4.75            4.25     3.75

Diff 1               0.69         0.67           0.65    0.63

As you can see there is a pattern emerging from the chords of these numbers.

Below is another copy of the graph y = x³.

Conclusion

x is:-

Multiply by the index and reduce the index by 1

This means that:-

## So my general formula will be:-

I know from previous class investigations into straight line graphs ie. y = 6 and y = -3 that these graphs have a gradient of zero. So Grad 4 = 0  and  Grad –3 = 0.

So if we differentiate a constant we get 0.

So far this agrees with my general formula as x° =1. We can write this as  Grad 4 = Grad 4x ° = 0 × 4x    = 0.

Again I know that graph y = mx + c has a gradient of m and graph  y = x has a gradient of 1 and graph  y = 6x  has a gradient of 6.

Again this agrees with my formula as:-

Grad x  = 1 × x° = 1

and,

Grad 6x = 6 × grad x = 6 × 1 = 6.

This shows the general fact that if a function has a factor which is constant, that constant remains the same as a factor of the gradient function.

This concludes my coursework.

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

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