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For this investigation, I have to find the relationship between a point of any non-linear graph and the gradient of the tangent, which is the gradient function.

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Introduction

For this investigation, I have to find the relationship between a point of any non-linear graph and the gradient of the tangent, which is the gradient function. First of all, I have to define the word, ‘Gradient’. Gradient means the slope of a line or a tangent at any point on a curve. A tangent is basically a line, curve, or surface that touches another curve but does not cross or intersect it. To find a gradient, observe the graph below:

image00.png

All you have to do to find the gradient is to divide the change in X with the change in Y. In this case, on the graph above, AB and you would have gotten the

                                          BC                                        

gradient for that particular point of the graph.                          

 I am going start by finding the gradient function of y=x², y=2x², and then y=ax². I will move on finding the gradient function of y=x³, y=2x³, and finally y=ax³. I will then find the similarities and generalise y=axⁿ where ‘a’ and ‘n’ are constants, and then investigate the Gradient function for any curves of my choice.

...read more.

Middle

1

8

f

3

12

1

12

From the table, you can see that the gradient of the point is always 4 times that of the amount of X. Which means that the gradient function for y=2x² is f ‘(x)=4x.

Now that I have found both the gradient functions for y=x² and y=2x², I shall draw the y=ax² graph (page 5) to show the equation used to find the gradient function for y=x² graphs. I will use a table to present my findings:

Point

X

Change in Y

Change in X

Gradient

a

-3

6a

-1

-6a

b

-2

4a

-1

-4a

c

-1

2a

-1

-2a

d

1

2a

1

2a

e

2

4a

1

4a

f

3

6a

1

6a

As you can see, this information above is very similar to y=x² ‘s information. Its is simply adding an ‘a’ to the Y-axis. But nevertheless, you can see that the gradient function for y=ax² is f ‘(x)=2ax, where ‘a’ is a constant. This gradient function can be used for any graph as long as it is a similar graph to y=x². i.e. y=4x², y=10x² or even y=250x².


        The graph y=x² is a rather simple one, when trying to find the gradient function. That is why I will use the graph y=x³ and y=2x³ to find the gradient function of y=ax³, and hopefully be able to find the gradient function for all graphs, namely y=axⁿ. I will, obviously start with y=x³ and shall use a program called, “autograph”, to draw the graphs for me from now on as the graphs are noticeably more accurate than drawing by hand.

image01.png

x

-3

-2

-1

1

2

3

gradient

27

12

3

3

12

27

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Conclusion

Observe the picture below:image04.png

You should be able to see that the gradient of the chord (in red) will be the gradient of the function, as h tends to zero.

To find the gradient function of any curve, the equation is:

f(x+h) - f(x)

(x+h) - x

= ( f(x+h) - f(x) )

h

Which means that the gradient of the function at x should be the limit, as h tends to zero.

When we substitute x² into the equation, the equation would be:

                                           f ‘(x)=  (x+h)²-x²

       h

= x²+2xh+h²-x²

h

= 2xh+h²    

h

f ‘(x)= 2x+h

        But if h tends to 0, h=0, then it would be

f ‘(x)= 2x +0 = 2X, where y=x²

        If I substituted y=x³ into the equation, I would get f ‘(x)=3x². If I did the same for y=x4, it would be f ‘(x)=4x³. This now shows a very obvious link which would help me generalise y=axn. In the equation, the square would move down in front of ‘x’ and the square of ‘x’ would be a digit lower than the original square. This means that the gradient function of y=axn is f ‘(x)=anx(n-1) .

...read more.

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