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

Investigating the Gradients of the Graph of the Form ‘y=x2’

Extracts from this document...

Introduction

Investigating the Gradients of the Graph of the Form 'y=x2' Aim To find a formula for the gradient of the graph 'y=x2'. To do this I will draw the graph and measure the gradient at certain whole numbers of 'x'. ...read more.

Middle

Then you form a right-angled triangle on the tangent and find the gradient as for the 'y=ax' graph. (As shown on graph) e.g. 9 9-4= 5 4 2 3-2= 1 3 gradient=y/x =5/1 GRADIENT=5 Results x y Gradient -3 -1 1 2 5 9 1 1 4 25 From these results I believe it is possible to deduce that the gradient of the curve at any point is equal to '2x'. ...read more.

Conclusion

Conclusions I belief that that the gradient function for a 'y=x2' graph is 2x. To test this theory I will try it with 1.5 as the value of 'x'. Results II x y Gradient 1.5 2.25 Conclusions II I can now conclude that the gradient function for a graph of the form 'y=x2' is 2x. G.F. = 2x ...read more.

The above preview is unformatted text

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. Curves and Gradients Investigation

    is applied to the X value: x + h. Once this gradient has been calculated, it will give the hypothetical gradient at the point (x + h)². This gradient (i.e. 7 + h) can be converted to the gradient at the actual point in question by reducing the 'h' value to 0 so the remaining equation is the gradient at this point.

  2. Analysing Triangle Vertices and Bisectors

    Substituting x = 0 into x� + y� - 14x - 2 = 0 gives y� - 2 = 0 which, when factorised, gives y(y - 2) = 0 resulting in y = 0 or y = 2 Since point O has the y co-ordinate of 0, y = 2

  1. The Gradient Function

    Just to check that this is correct, I am now going to work out the gradient, when the tangent is at x=1. 12 = 1 1 x 3 = 3 3 x 2 = 6 If you compare the two graphs of y=x3 and y=2x3, then you can see a pattern with the curve.

  2. The Gradient Function Investigation

    + (3x� + 9x�h + 9xh� + 3h�)] - [2x� + 3x�] h = 4xh + 2h� + 9x�h + 9xh� + 3h� h = 4x + 2h + 9x� + 9xh + 3h� as h tends to 0 GF tends to 4x + 9x� Results Summary From calculating the

  1. The Gradient Function

    -3 -0.33 -0.33 From the table above we can see that the increment method is much more accurate than the tangent method. Thus I will use the results from the increment method to the generalization of the gradient. Generalization: Gradient Function For Graphs Of Y = a/x To make the

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

    18.605 18.1202 18.012 18.0012 18.00012 18.00001 change in y 11.22 10.605 10.1202 10.012 10.0012 10.00012 10.00001 change in x 1.1 1.05 1.01 1.001 1.0001 1.00001 1.000001 Gradient function 10.2 10.1 10.02 10.002 10.0002 10.00002 10 Gradient function at x = 4 for the curve of y = 2x� X 4.1

  1. The Gradient Function

    General Formula For The Gradient of Positive Graphs I used the above method to expand different brackets from (x+d)2 to (x+d)7. In the table below, I have only included the relevant term y = x2 2x y = x3 3x2 y = x4 4x3 y = x5 5x4 y =

  2. Gradient function

    gradient 4 16 9 1 9 4.1 16.81 8.19 0.9 9.1 4.2 17.64 7.36 0.8 9.2 4.3 18.49 6.51 0.7 9.3 4.4 19.36 5.64 0.6 9.4 4.5 20.25 4.75 0.5 9.5 4.6 21.16 3.84 0.4 9.6 4.7 22.09 2.91 0.3 9.7 4.8 23.04 1.96 0.2 9.8 4.9 24.01 0.99 0.1

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