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Gradient Graphs Investigation

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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


12    =  6


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

Here is a table of all of my results:-image01.png

Diff 1                        2           2               2                  2               2

This means that my formula should be:-

 Gradient = 2x

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.


...read more.


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 function is 2x.

        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:-image03.png

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:-


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    0.63image06.png

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

Below is another copy of the graph y = x³.


...read more.


x is:-

Multiply by the index and reduce the index by 1

This means that:-

Grad x  = 5x

        Grad x  = 6x

So my general formula will be:-

Grad x  = nx

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.

        So Grad x = 1

Again this agrees with my formula as:-

Grad x  = 1 × x° = 1        


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.

...read more.

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