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Introduction

In this piece of coursework I am going to do research on the gradient of various graphs at various points, in order to find a function, which will determine the gradient of these points without drawing or using approximations. I will only need to know the coordinates of the point as well as the type of graph I am considering, to submit them into the gradient function and determine the gradient at this point. The formulae I will use and produce will have particular parameters. Now I am going to explain them.

a: this letter will stand for the coefficient of x in the function y=ax^n and

determines how steep the graph will be.

n: this letter will be the power to which x is raised in the function y=ax^n and

determines the shape of the graph.

m: this letter will stand for the gradient at any point of any graph. I can say for

example the gradient at the point P(1;1) of the graph y=x is 1. Therefore here m=1.

The first range of graphs I am going to investigate will have the function y=ax. I will draw three graphs on the next pages and hope to see a pattern between the gradient and the function of the graphs. I do not need to consider the coordinates of the points at which I will determine the gradient, as the gradient is the same at any point on the graph y=ax.

From these three graphs I clearly recognise a pattern. I will show how I noticed it, by presenting the graphs and their gradients:

y=x        m=1

y=2x      m=2

y=3x      m=3

Middle

1

2

6

y=2x³

2

16

24

y=2x³

3

54

54

When I looked at this, it took a while until I found the relationship, but in the end I noticed that for this range of graphs y=ax³ the gradient function will be m=3ax². I think that this is enough research to find a relationship between these three ranges of graphs and their gradients. I will draw up a table with the results and try to generalise the rule.

 Type of Graph Formula for Gradient (m) y=ax 1axº y=ax² 2ax¹ y=ax³ 3ax² y=ax^n nax^n-1

The term in the box should determine the gradient of any point on any graph. I suggest that the gradient function for any graph would be m= nax^n-1. But this may not be true for negative or fractional powers, so I will have to test this function on these ranges of graphs. I will start with y=ax^-1. For this purpose I have to calculate some gradients of random points on two graphs with -1 as their power. Near the calculation I will present the result of my formula. If it equals approximately the calculated gradient, my formula is correct. If it is not, then I will have to change my formula. The result of my formula for the first example below was calculated like this: m=nax^n-1 => m=-1*1*1^-2 => m= -1 I typed it in as a formula, so that Excel will automatically calculate the result of the formula for me. Here the gradient according to my formula is correct, but I will have to test it more than one time to be sure. The gradient calculated by my formula should be the exact number, and the formula calculated by using the chord method is only an approximation of the gradient at this point. So both numbers must be very close in order for the formula to be correct. But the number calculated by using the chord method in Excel is only an approximation itself, so it could be that both numbers are equal, which also proves my formula to be correct. I will also work out the gradient for the calculation on the next page: m= nax^n-1 => m=-1*1*2.5^-2 => m= -0.16.    The function proves to be correct for negative powers. This function may be the gradient function for any graph y=ax^n at any point, but I do not know if the function will determine the correct gradient for graphs with fractional powers. To ensure this, I will test the function on the range of graphs with formula y=ax^0.5. For the example below I will show how the gradient is worked out by using the formula: m= nax^n-1 => m=0.5*1*1^0.5 => m=0.5 The function appears to be correct so far, but I will have to use many more examples in order to show that the gradient function I developed is generally correct.  I will show working out the gradient for the calculation below by using the formula: m=nax^n-1 => m=0.5*3.31*1^-0.5 => m=1.655   I think that this is enough evidence to show that the gradient function is m= nax^n-1.

Now I will use this formula to predict the gradient of two complex functions by combining simpler ones. My first graph to consider will be y=x²+3x+5. To work out the gradient function for this graph I will need to look at each summand separately and use the rule for it. Then I will need to add the gradients together:

Graph:      y=    x²    +         3x     +      5

Gradient:          2x    +         3       +     0

So the gradient for the graph y=x²+3x+5 will be 2x+3. It is as simple as that. The following two calculations will prove that this is correct. Both calculations are based on the graph y=x²+3x+5:

 x1 value y1 value x2 value y2 value Gradient 1 9 1.1 9.51 5.1 1 9 1.01 9.0501 5.01 1 9 1.001 9.005001 5.001 1 9 1.0001 9.00050001 5.0001 1 9 1.00001 9.00005 5.00001 1 9 1.000001 9.000005 5.000001

Conclusion

Conclusion

In this assignment I wrote about a function that I wanted to develop by using the chord and tangent method. I finally found this function to be m=nax^n-1, which proved to be correct for both fractional and negative powers and also of course positive powers. I also showed how you can work out the gradient of complex functions by combining the results of my formula on simpler parts of the function. In the last part I researched into calculus and in particular differentiation, as this relates to my assignment. It showed that the gradient function I developed was actually used in calculus, which proves my findings to be correct. Overall I created a significant piece of coursework, which gave me an initial view of calculus and differentiation, and what I can expect when I attend an A-level Mathematics course in 6th form.

By Kassra Rashidian, 11u

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

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5 star(s)

Response to the question

This piece of coursework superbly explores the gradient of basic polynomial curves. The argument posed is logical, and diagrams and screenshots of excel are used to support this argument.

Response to the question

This piece of coursework superbly explores the gradient of basic polynomial curves. The argument posed is logical, and diagrams and screenshots of excel are used to support this argument.

Level of analysis

The analysis in this piece is strong. I liked how they begin by trying to work out the gradient of a function at a point, and then showing the understanding that this isn't accurate enough. They use plenty of excel screenshots to give evidence for their hypothesis, but what I really liked was the proof by first principles (shown under the more about calculus section). Using and understanding limits at GCSE level shows high level analysis and I imagine showing this technique will help secure top marks in any piece of coursework. This piece of coursework even goes further to understanding how differential equations and integration may be useful in a real-life application. If I were to make one suggestion, it would be to explore the notation a bit further.

Quality of writing

This piece of coursework is structured well, and spelling and grammar are strongly utilised. Although pieces of formal coursework are becoming less common in mathematics at GCSE, I would note that this piece could be made more sophisticated by removing the frequent use of the first person. Other than that, it should be admired!

Reviewed by groat 11/02/2012

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1.  4 star(s)

I will also try to spot any patterns that may occur. Theoretical Method Small increments Method y=x2 y Q P x x+h P(x, x2), Q(x+h,(x+h)2) Gradient = (x+h)2-x2 = (x+h+x)(x+h-x) = x+h+x = 2x+h = 2x (x+h-x) (x+h-x) From the calculation above, it is showing that the gradient function for y=x2 graph is 2x+h.

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

is 0 then after that as been multiplied by the gradient, a2 will also be 0. After these calculations, a small preliminary experiment was carried out to obtain a few pieces of data that would give a rough idea of the Young's Modulus that is to be obtained.

1. approaches zero from above. Reciprocal functions have the equation: 1 . (ax+b) By use of GDC, geogebra i found the equation by moving the line of the curve to fit accurately. 1 � (0.06x+0.012)-0.08) +1.48 Another function that models the data is the logistic function.

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

Therefore, I am going to test it out with a few more cubic functions to see if my hypothesis is true.)

1. Investigate gradients of functions by considering tangents and also by considering chords of the ...

in the graph y= x2, then look for a very close number as x2 to x1, usually 1.01, then use the equation of the line to work out y2, which can be got by square x2 and gives you the value 1.0201.

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

factors for 12 are : 12 X 1 6 X 2 3 X 4 Therefore 4 can be written as 2� and 4 X 3 gives 12,thus I can conclude that the equation is 3x� Tangent at x = 3 Gradient function= 27 The equation I concluded in the earlier

1. On a computer the more you zoom in the more accurate the gradient will be. You then use the same formula as before to work out the gradient. dy/dx This should then give you an accurate gradient; it tends to be more accurate than the other method, although if you

2.   