• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Gradient Function.

Extracts from this document...

Introduction

Gradient Function Coursework

Gradient Function Coursework

My task is to investigate the relationship between the gradients of tangents on the curves of graphs, such as y=x2. To do this, I will first find the gradient of tangents on the graph y=x2 by drawing the graph.

I have labelled the tangents a-g. They go from x=-3 to x=4. Below are the calculations for their gradients. (I am using the formula (y2-y1)/(x2-x1)to calculate the gradient of the line.

a= (12-6)/(-3.5--2.5) = 6/-1 = -6 b= (6-2)/(-2.5--1.5) = 4/-1 = -4

c= (2-0)/(-1.5--0.5) = 2/-1 = -2 d= (2-0)/(1.5-0.5) = 2/1 = 2

e= (6-2)/(2.5-1.5) = 4/-1 = 4 f= (12-6)/(3.5-2.5) = 6/1 = 6

g= (20-12)/(4.5-3.5) = 8/1 = 8

As you can see, the results I have obtained are good round numbers. These results however are not accurate to the tangents I drew on the graph.

...read more.

Middle

12

3

27

3.0000001

27.0000027

27

4

64

4.0000001

64.0000048

48

5

125

5.0000001

125.0000075

75

6

216

6.0000001

216.0000108

108

The first numbers I saw were the first and third numbers. I noticed these because the first number was 3, the power by which I was multiplying the values. The third number was 27, which is 33. This struck me as strange - only this number was cubed, none of the others. To explain this, I tried cross analysing these results with the results I obtained from the y=x2 graph. I saw that the first value was the value of the power by which I multiplied the values. The third value however was not 32, but the second value was 22. Moving back to the results for y=x3, I tried dividing all the results by 3, and was left with the x1 values squared. Therefore, the formula for the gradient of a tangent on the graph of y=x3 was 3x2.

...read more.

Conclusion

n class="c5">2

y2

(y2-y1)/(x2-x1)

4x3

1

1

1.0000001

1.0000004

4.0000006

4

2

16

2.0000001

16.0000032

32.00000238

32

3

81

3.0000001

81.0000108

108.0000054

108

4

256

4.0000001

256.0000256

256.0000097

256

5

625

5.0000001

625.00005

500.0000144

500

6

1296

6.0000001

1296.000086

864.0000213

864

The formula for the gradient of a tangent on the graph of y=xn is nx(n-1)

As I thought, the two sets of results are very close. This shows that the formula is correct. I now need to find a formula which link the different formulae together. As the first number in the formulae is always the power being multiplied by x, and the power in the formulae is always 1 less I can determine the linking formula.

...read more.

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

See related essaysSee related essays

Related GCSE Gradient Function essays

  1. Peer reviewed

    The Gradient Function Coursework

    5 star(s)

    The following graph shows the results: Graph x-coordinate of Point y-coordinate of Point Gradient (m) y=x� 1 1 3 y=x� 2 8 12 y=x� 3 27 27 y=2x� 1 2 6 y=2x� 2 16 24 y=2x� 3 54 54 When I looked at this, it took a while until I

  2. The Gradient Function

    To try and find a more accurate answer to the part of the investigation that I performed above, I will use this method now. I am going to start by investigating the tangent at x=0.8. First of all, you must take the point on the graph of 0.8.

  1. The Gradient Function Investigation

    I can see from these results that in all three cases, the X co-efficient of the gradient function (e.g. 9x�) is the same as the X co-efficient of the graph equation (e.g. 3x�) multiplied by the power of X of the graph equation (e.g.

  2. Maths Coursework - The Open Box Problem

    1999.279 5.12 19.76 19.76 1999.143 5.13 19.74 19.74 1998.995 5.14 19.72 19.72 1998.835 5.15 19.7 19.7 1998.664 5.16 19.68 19.68 1998.48 5.17 19.66 19.66 1998.286 5.18 19.64 19.64 1998.079 5.19 19.62 19.62 1997.861 5.2 19.6 19.6 1997.632 I will zoom in again to increase the degree of accuracy in finding out what length of cut out gives the largest volume.

  1. The Gradient Function

    = 11.33 rounded to 2.d.p -1.5-(-2.25) Gradient at point X = -3 Y2-Y1 X2-X1 = -17-(-29.5) = 25 -2.6-(-3.1) I have now found the gradients of the different graphs, which I had planned to investigate using the tangent method.

  2. The Gradient Function

    If the graph is of form Y= axn where a and n are positive, negative, or fractional, then G = anxn-1. Graphs of form y=axn + bxm I can extend my work further by looking at graphs of the form y=axn + bxm for example: Y = x2 + x3

  1. Investigate the elastic properties of a strip of metal (hacksaw blade) and use the ...

    For the experiment the following apparatus will be required: 1. A work bench 2. 2 Wooden Blocks 3. G-clamp 4. Hacksaw blade 5. Thread 6. Small mass 7. Mass (M) 8. A micrometer 9. A ruler 10. A card with graph paper attached to it 11.

  2. I have been given the equation y = axn to investigate the gradient function ...

    Therefore I will be rounding the gradients to the nearest whole numbers, When y = x3 and a = 1 the tangents and their respective gradients are as follows: Tangent at x = 2 3 4 Gradient function 12 27 48 Tangent at x = 2 Gradient function 12 The

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work