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• Level: GCSE
• Subject: Maths
• Word count: 1555

My aim is to investigate the gradient function for all kinds of curves.

Extracts from this document...

Introduction

Aim: to investigate the gradient function for all kinds of curves.

Research: What does “gradient” mean? Generally, the “steepness” of a curve is measured by its gradient. We can look at the figure below:

This is the curve of y=x2. The point P (3, 9) has been marked and the tangent QPM drawn. The gradient of the tangent is QN/MN.

So we can use the “tangent method” to obtain the gradients of graphs of different functions.

First Step: I am going to investigate the gradient of y=x, y=x2, y=x3 first because they are likely to be the simplest equations to solve, and after getting these results easily, by looking at them, the more complex equations will seem easier to discover.

I am going to look at y=x first because it is the easiest.

 x 1 2 3 4 y 1 2 3 4

Please see graph on separate pieces of paper

As we see, the gradients of y=x is very simple, a=1. We even do not need to draw any tangents to obtain the gradients. So the relationship between a and x can be shown in the table below:

 x a (The Gradient) 1 1 2 1 3 1 4 1

So it is extremely obvious that in the graph of y=x, whatever x is, the gradient a stays 1.

Middle

a1=3=3x12

a2=12=3x22

a3=27=3x32

a4=28=3x42

It means that the gradient function for y=x3 is a=3x2, as you see this formula does work. It proves again that the increment method is much more accurate than tangent method. And also from now on the line in the graph is tend to be much steeper and steeper so that it makes it hard and not accurate to draw the tangent, so I am going to use increment method from now on.

Things are not solved yet; I have to look at the formula for these three times to find any possible relationship. In y=x, a=1=x0; in y=x2, a=2x; in y=x3, a=3x2; as we can see, in the three gradient formulas, the number before x and the indices are both increasing as the indices of the equation get higher, and the difference between the number before x and the index is 1 (the index number is smaller than the number before x). So I predict that the gradient formula for y=x4 is: a=4x3.

Let’s try it out.

The co-ordinates:

 x 1 2 3 4 y 1 16 81 256

 x a (The Gradient)Increment Method 1 4.040601 2 32.240801 3 108.541201 4 256.96101

a1=4=4×1=4x13

a2=32=4×2×2×2=4x23

a3=108=4×3×3×3=4x33

a4=256=4×4×4×4=4x43

Therefore, my prediction works this time!

Conclusion

y=x-1/2

 x a(Increment Method) a(a=nxn-1) 1 -0.499 -0.5×1-1.5=0.5 2 -0.176 -0.5×2-1.5=-0.176 3 -0.096 -0.5×3-1.5=-0.096 4 -0.062 -0.5×4-1.5=-0.062

I can strongly say that the gradient function can also work on fractions and negative numbers.

However, this gradient function is only a basic one; we can develop it to make it suit all kind of cases e.g. y=2x3, y=4x2+3x. How to work them out? I will take y=2x3 as an easy example: first, work out y=x3 by using a=nxn-1; second, multiply all the results by 2. That is, to split up the equations.

If you want to work out a more complex equation such as y=4x2+3x, just divide the equation into two piece, y=4x2 and y=3x, then do it one by one as shown below:

y=4x2:

 x a 1 8 2 16 3 24 4 32

y=3x:

 x a 1 3 2 3 3 3 4 3

After you have done this, simply combine the two together.

 x a 1 8+3=11 2 16+3=19 3 24+3=27 4 32+3=35

Using tangent method in my graph I can check that my results are correct.

 x Gradient (Tangent Method) 1 2 3 4

This method can work on any similar type of equations.

In this piece of coursework I have discovered that we can get the gradient of any line at any point by using the formula a=nxn-1, and it work on positives, negatives, fractions and complex equations

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)

I will draw up a table with the results and try to generalise the rule. Type of Graph Formula for Gradient (m) y=ax 1ax� y=ax� 2ax� y=ax� 3ax� y=ax^n nax^n-1 The term in the box should determine the gradient of any point on any graph.

If you work out the sum of the tangent at x = 0.8 x 6 of x, it will be 0.8 x 6 = 4.8. The gradient is actually 4.8, however as I explained at the beginning of the investigation, this method that I am using, tends to be inaccurate.

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

However, this gradient function is only a basic one; we can develop it to make it suit all kind of cases e.g. y=2x3, y=4x2+3x. How to work them out? I will take y=2x3 as an easy example: first, work out y=x3 by using g=nxn-1; second, multiply all the results by 2.

a� - b� = (a - b)(a + b) a� - b� = (a - b)(a� + ab + b�) a4 - b4 = (a - b)(a� + a�b + ab� + b�)

1. Maths Coursework - The Open Box Problem

the cut out and the size of the square and was able to derive a general expression.

my x numeral is -7 and my gradient is -28, if I divide them I get 4. Another example is when my x value is 5 and my gradient is 20, if I divide them I get 4, which means my slope is definitely m = 4x.