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• Level: GCSE
• Subject: Maths
• Word count: 2700

Extracts from this document...

Introduction

Middle

I will use this data to work out the nth term of the gradient in the graph of y=x4. 4 32 108 256 500 +28 +76 +148 +244 +48 +72 + 96 +24 +24 As the differences are equal at the third level, the pattern for the nth term is an3 + b, if the value of a is the difference divided by 3, then the result for n=1 would be 8 which is twice the real value, so the value must be the difference over 6, this applies for all the values I have. This means that the nth term is 4n3; because it already fits, the value of b must be zero. To test this pattern, I can use it to predict the gradient at x=6. Using the nth term pattern it should be 4 x 63 which is 864. Using the small increment method, the answer is (6.0014 -64) divided by 0.001 which rounds to 864. Therefore the nth term is correct. Patterns for graphs of form y=xn In the graphs of this form I have worked out patterns for, there seems to be a global pattern in the nth term of each sequence: This means that for graphs of form y=xn the gradient at any point could be worked out using G=nxn-1. I will test this on the graph y=x5: PREDICTIONS X=1 5 x 14 = 5 X=2 5 x 24 = 80 X=3 5 x 34 = 405 X=4 5 x 34 = 1280 X=5 5 x 34 = 3125 Graphs of form y=ax2 I will extend my work to see what effect the value of a has on my results so far, initially I kept a =1. ...read more.

Conclusion

Using calculus to calculate gradients If I was to simplify (x+f)n it would be the following: xn + nxn-1f + ?xn-2f2... The value of ? is not relevant, but would in fact be a triangular number, because the line in the calculation that relates to this is as follows: G = (x+f)n - xn f This means that if it is simplified it would be: G = xn + nxn-1f + ?xn-2f2... - xn f But as it is subtracted by xn and divided by f it cancels out to: G = nxn-1 + ?xn-2f ... Therefore as f was zero (which it should be) the value of the ? makes no difference as all that is left is: G = nxn-1 Which is in fact the formulae I originally came up with. I have now demonstrated that calculus can be used to derive the gradient function for all graphs of the type y=axn. Conclusion To conclude, I have worked out that a gradient at any point on a graph of form y=axn can be worked out using the following formulae: G = anxn-1 I have tried four different methods to work this out: * Drawing tangents by hand * Using the small increment method * Using a computer program * Calculus From using these, I know: * That hand drawn tangents are inaccurate and so no use for working out a pattern * That the small increment method is not entirely accurate * That computer programs can generate perfectly accurate gradients * That calculus can work out the formulae in a single equation. I also know that in graphs of form y=axn + bxm can be treated as two gradients from graphs of form y=axn added together. ...read more.

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# Related GCSE Gradient Function essays

1. ## The Gradient Function Coursework

5 star(s)

and that the first point is called P and has the coordinates (x, y). The small y-difference of these two points would thus be ?y and the tiny x-difference would be ?x. ?x or ?y are both a single symbol and are often called the increment in x or y.

y= x4 My first fixed point: 3, 81 x y increase in y increase in x gradient 2 16 -65 -1 65 2.1 19.4481 -61.5519 -0.9 68.391 2.2 23.4256 -57.5744 -0.8 71.968 2.3 27.9841 -53.0159 -0.7 75.737 2.4 33.1776 -47.8224 -0.6 79.704 2.5 39.0625 -41.9375 -0.5 83.875 2.6 45.6976 -35.3024

0.3 70.59 4.8 110.592 14.408 0.2 72.04 4.9 117.649 7.351 0.1 73.51 4.99 124.2515 0.748501 0.01 74.8501 4.999 124.925 0.074985001 0.001 74.985 5 125 5.001 125.075 -0.075015 -0.001 75.015 5.01 125.7515 -0.751501 -0.01 75.1501 5.1 132.651 -7.651 -0.1 76.51 5.2 140.608 -15.608 -0.2 78.04 5.3 148.877 -23.877 -0.3 79.59 5.4

If this gradient function, 3ax2 is correct, then the gradient function of 10x3 will be 30x2. To prove that it is 30x2 I am going to draw up a table of values for certain x values. For the curve y=10x3, using the gradient function 30x2 I think the gradients will

= -0.37 rounded to 2.d.p -3.25-(-2.5) L2: Gradient at X = -2 Y2-Y1 X2-X1 = -1.375-(-1.625) = -0.71 -2.15-(-1.8) L3: Gradient at X =2 Y2-Y1 X2-X1 = 1.675-1.3 =-0.75 1.75-2.25 Graph: -2/x The table below shows the points I have used to plot the graph: X Y -4 0.5 -2

2. ## Investigate the gradients of the graphs Y=AXN

16 0.5 32 3 4*34-1 = 108 3,3.001 81,81.108 0.108/0.001 108 81 0.75 108 4 4*44-1 = 256 4,4.001 256,256.256 0.256/0.001 256 256 1 256 The above table shows that both methods agree for my formula for cases in which A=1 and N= a positive integer, so I shall now