• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15
16. 16
16
17. 17
17
18. 18
18
19. 19
19
20. 20
20
• Level: GCSE
• Subject: Maths
• Word count: 4863

Extracts from this document...

Introduction

Michael Layer

10 Clare / 10A1

Part 1

We have been told to investigate the Gradient Function for the set of graphs:

y=axn

Where a and nare constants.

One way to find out the gradient, is to do what I have done for the curve of y=x2. After drawing the tangent of a specific point of x, you draw a triangle, with the x value in the middle of a line on the tangent. You then find the change in y, and the change in x. You then have to apply them to the following formula:

The Change in Y

The Change in X

You simply have to divide the change in y, by the change in x. On all my graphs, there will be small triangles that have been measured out on the tangent. They will be marked with the letters A, B, and C. The line of the triangle running in parallel with the ‘y axis’ is The Change in Y. The line of the triangle running in parallel with the ‘x axis’ is The Change in X. This will give you the gradient. However, this is quite an inaccurate method of working out the gradient. This is because if you draw the co-ordinates for the triangle incorrectly, then it will alter the answer that you will discover. There is also the small increment method that you can apply to the curve. I will go into this in a later graph, as well as using this method to work out the gradient.

Firstly, I thought that the best thing to do, would be to investigate the graph of y = x2. Below I have drawn a table of values, to inform me on where I should plot the points of this graph.

 x -3 -2 -1 0 1 2 3 x2 9 4 1 0 1 4 9 y = x2 9 4 1 0 1 4 9

Middle

1.9632 – 1.92        =        0.0483        =   4.83

The Change in X        =          0.81 – 0.8                =          0.01

As you can see, this is by far a much more accurate approach to finding out the gradient. The figure 4.83 is much closer to 4.8 instead of 5. I am now going to try and find a more accurate result, when the tangent is at -0.5.

Below is my ‘working out’ using the small increment method of x=-0.5:

-0.5 x 3 = -1.5 (-0.5, -1.5)

-0.51 x 3 = -1.53 (-0.51, -1.53)

The Change in Y        =        -1.53 - -1.5        =        -0.03        =        3

The Change in X        =        -0.51 - -0.5        =        -0.01

As it is a negative tangent, you now just need to add a negative sign to the answer. The gradient of the tangent at x=-0.5 is -3. If we work out the gradient, using the theory that the gradient function for the graph y=3x2, we can see that -0.5 x 6 is -3. This is an extremely accurate answer, as well as when the tangent is at x=0.8. As you can see from my work above, the small increment method is a much more accurate method of working out the gradient. This proves that the gradient function for the curve y=3x2, is 6x.

Now I have investigated three different graphs that have been squared, I am going to display my results of the gradient function of these curves in a table for convenience.

 The Curve The Gradient Function y=x2 2x y=2x2 4x y=3x2 6x

As you can see from my above prediction, it was correct – the pattern did continue. Every time I increased the value of ‘a’ from the equation by 1, the gradient function increased by 2x.

However by looking at my results this way, I have happened to spot another pattern. Every time that you increase the value of ‘a’ by 1, the gradient function will rise by two. For example, when

Conclusion

2, with the graph of y=x2. I draw tangents and find the gradients on y=½x2.

Below is a table of values for the equation y=½x2:

 x -3 -2 -1 0 1 2 3 x2 9 4 1 0 1 4 9 ½x2 4.5 2 0.5 0 0.5 2 4.5 y=½x2 4.5 2 0.5 0 0.5 2 4.5

For the table of values for the curve y=x2, you can refer back to PAGE 1. As you can see if you compare them, the curve will be relatively similar. However (obviously) it will be a little different as the equation has changed.

Below I have used Omnigraph for Windows to illustrate the graph:

I am now going to investigate what the gradient will be, when the tangent is at x=2.5:

(0.5*2.5)1 = 1.25

1.25*2 = 2.5

As you can see, the gradient is exactly the same as the tangent. I am now going to investigate the gradient when the tangent is at x=-2.

(0.5*-2)1 = -1

-1*2 = -2

As you can see, the gradient is once again, exactly the same as the tangent. This means:

GF = G

I am now going to investigate the curve of y=½2x2. Below is a table of values for that graph. I am going to compare it with the graph of y=2x2, and you can find the table of values for that graph, on PAGE 3.

 x -3 -2 -1 0 1 2 3 x2 9 4 1 0 1 4 9 2x2 18 8 2 0 2 8 18 ½2x2 9 4 1 0 1 4 9 y=½2x2 9 4 1 0 1 4 9

Below I have used Omnigraph for Windows to illustrate this curve:

Now I have illustrated this graph, I am going to work out the gradient when the tangent is at x=2.5, using the formula that I calculated as a general rule:

2*((0.5*2)*2.5)1 = 5

As you can see, the gradient is twice that of the tangent. I am now going to investigate when the tangent is at x=-2 to see if this will happen again – using the general rule:

2*((0.5*2)*-2)1 = -4

As you can see this has happened again. This means:

GF=2T

As you can see from all the data in this investigation, the formula and general rule for calculating a gradient function for any curve, is:

nx(n-1)

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

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Gradient Function essays

1. ## The Gradient Function Coursework

5 star(s)

The closer together the points are, the more accurate our result will be, but we will only get 100% accuracy when the distance between the two points is zero. Here by using calculus, you can obtain a simple formula that tells you the gradient of the tangent lines at any point on the curve.

4 star(s)

are the theoretical results x -4 -3 -2 -1 0 1 2 3 4 y -64 -27 -8 -1 0 1 8 27 64 Gradient 48 27 12 3 0 3 12 27 48 My results for the graph above were fairly close to the actual one.

3.71 -0.7 -5.3 -2.2 4.84 4.16 -0.8 -5.2 -2.1 4.41 4.59 -0.9 -5.1 -2 4 5 -1 -5 Power: 2 Coefficient: 1 Fixed point: -3 From looking at my tables I can see that my gradient is two times bigger than my x value.

2. ## Aim: To find out where the tangent lines at the average of any two ...

functions have only one root after all, but with a multiplicity of three, my hypothesis have also proven this correct.) Functions with Two Roots Example 1- Roots = -2, 2 1. Roots -2 & 2 Average of the two roots = Plug the point, (0, -8)

To find the gradient on this curve the equation is m=x 2. The gradient is calculated as change in y change in x This change I can write using differentiation dy dx So, when Y=2X2 dy dx Y=3x2 My first fixed point is 3, 27 x y change in y

the value of x:- Tangent at x 1 1.5 2 3 -2 Gradient 6 9 12 18 -12 Now we can predict the gradient function to be 6x as the gradient is always 6 times 'x' For the curve y=4x�, the gradient of the tangent at x is eight times