• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
• Level: GCSE
• Subject: Maths
• Word count: 2064

Extracts from this document...

Introduction

By Laurence Gagen

I am trying to find a formula that will work out the gradient of any line (the gradient function)

I am going to start with the simplest cases, e.g. g=c², g=c, g=c3 etc. as they are probably going to be the easiest equations to solve as they are likely to be less complex and hopefully the formulas to the more complex equations will be easier to discover by looking at the previous formulas.

I am going to look at the line g=c² first.

g=c²

 c 1 2 3 4 g 1 4 9 16

Please see graph on separate piece of paper

One of the most obvious things I notice is that as the co-ordinates increase so does the gradient. Not only can you see that from the results below, but also on the graph you can that the line gets steeper and steeper. This makes sense as the higher the number c is the larger the difference between c² and c.

Another thing that I have noticed is that the larger the co-ordinates the smaller the increase in gradient.

 Point Gradient (tangent) Gradient (Small Increment Method) (1,1) 2 2.01 (2,4) 3.3 3.01 (3,9) 6.36 (2dp) 6.01 (3.5,12.3) 6.4 7.001

As the table above shows there are two methods that I am using for calculating the gradient of line.

Middle

As you can see my formula does work, they give almost exactly the same answers for the gradient as my small increment result which means it must be right and also the small increment method must be more accurate, so from now on I am going to just use that method instead of using tangents as well. This will mean I don't have to draw out the graphs any more.

I am now going to look at the gradient of the line g=c , I think this could have a very similar formula to the previous ones as looking back on the other equations the rule that I have used for g=c³ works for the two previous ones assuming I adapt it, for example in the equation g=c³ the formula was 2c, but I can also use my new formula to get the same results except of course instead of multiplying by three you would times by 2 as the equation is to the power of 2. So I would expect 4*c3 will be the formula that works for the equation g=c4.

g=c
4

 c g Gradient 1 1 4.060401 2 16 32.240801 3 81 108.541201 4 256 256.96101

My prediction was right the formula for this line is 4*c3

I am going to put the formulas that I have discovered so far into a table so that they are hopefully easier to interpret.

 Equation g=c g=c2 g=c3 g=c4 N Gradient Function 1 2c 3c2 4c 3 nc (n-1)

Looking at my previous formulas I have come up with a formula that will work for any equation, although I need to check that it does work for negative, fractions etc. The formula is as follows:

nc (n-1)

This formula has been developed from the other formulas that I have discovered for the equations, as you can see from the table above it is fairly obvious what the formula is going to have to be. The n stands for the power of e.g. in the equation g=c-1, n would equal -1.

I am going to just check that my formula does work with the equation g=c5although I am already convinced it will work as it has worked for all the previous equations.

g=c5

 c g Gradient 1 1 5.10100501 2 32 80.80401001 3 243 407.70902 4 1024 1286.41602

5*1
4=3
5*2
4=80
5*3
4=405
5*4
4=1280

Indeed my rule does work, however now that the gradients are such high numbers it is more noticeable that the small increment method isn't perfect. I am now going to look at the line g=c-1and see if my rule does or doesn't work of that and if not what does, the reason it might not work is because all the other equation I have used have been positive numbers.

g=c-1

 c g Gradient 1 1 -0.99 2 0.25 -0.25 3 0.333 -0.111 4 0.0625 -0.062

-1*1-2=-1
-1*2
-2=-0.25
-1*3
-2=-0.1111
-1*4
-2=-0.0625

As you can see the formula does work however to make sure I am going to look at the line g=c-2.

g=c-2

 c g Gradient 1 1 -2 2 0.25 -0.25 3 0.1111 -0.074 4 0.0625 -0.031

Conclusion

3+5c2, all you do is use the equation nx(n-1) but break up the equation into two so you have two equations, take for example g=2c3+5c2 you would break it up to g=2c3 and g=5c2 . Then you work out the equations as normal as shown below.

g=2c
3
 c Gradient 1 6 2 24 3 54 4 96

g=5c
2

 c Gradient 1 10 2 20 3 30 4 40

What you then do is combine the two together like this:

g=2c3+5c2

 c Gradient 1 6+10=16 2 24+20=44 3 54+30=84 4 96+40=136

Using my graphical calculator I can check that my method is correct and indeed it is. Of course this method will work for any other similar types of equations.

I think that it is extremely likely that the equations will also work for fractions e.g. g=c0.2, g=c-0.5 etc. To check I am going to test both of the examples, notice that the second one is a fraction and a negative so might is less likely to work using the formula.

g=c
0.2

 c Gradient nc (n-1) 1 0.2 0.2*1-0.8=0.2 2 0.1148 0.2*2-0.8=0.1148 3 0.083 0.2*3-0.8=0.083 4 0.0659 0.2*4-0.8=0.0659 5 0.0551 0.2*5-0.8=0.0551

g=c
-0.5

 c Gradient nc (n-1) 1 -0.499 -0.5*1-1.5=-0.5 2 -0.176 -0.5*2-1.5=-1.176 3 -0.096 -0.5*3-1.5=-0.096 4 -0.062 -0.5*4-1.5=-0.062 5 -0.044 -0.5*5-1.5=-0.044

As you can see the formula does work for both fractions and negative fractions.

What I have discovered is, that it is possible to find the gradient of a line at any point by using the equation and the formula nc (n-1), and that this formula works with positives, negatives and fractions.

### By Laurence Gagen

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

# Related GCSE Gradient Function essays

1. ## The Gradient Function Coursework

5 star(s)

This will make the calculated gradient more and more accurate. I will do calculations that deal with differentiation and explain some expressions. Suggest that the second point is called Q and has the coordinates (x+?x; y+?y)

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

In the curve of y=2x� , the tangent draw at 2 gave us the gradient function of 8. Therefore in the equation y=2x�+x the answer for the gradient function we arrive at is 9 thus I can see that as a = 1 , if we add this 1 to

1. ## I am going to investigate the gradients of different curves and try to work ...

Gradient = 64 - 4 = 60 = 20 4 - 1 3 Chord AD Gradient = 36 - 4 = 32 = 16 3 - 1 2 Chord AE Gradient = 16 - 4 = 12 = 12 2 - 1 1 Chord AF Gradient = 9 - 4

Hence, to calculate the gradient normally it is simply Gradient = 2x. * * * I was not able to find a formula for the graph x3, so I decided to use the above method to find out what it was.

1. ## The Gradient Function Investigation

- x6 (x + h) - x = (x + h) (x� + 2xh + h�) - x6 (expand brackets) h = (x + h)�(x� + 3x�h + 3xh� + h�) - x6 (expand and h simplify) = (x + h)�(x + 4x�h + 6x�h� + 4xh� + h )

25.23 1.77 0.1 17.7 2.99 26.8203 0.1797 0.01 17.97 2.999 26.982 0.017997 0.001 17.997 3 27 3.001 27.018 -0.018003 -0.001 18.003 3.01 27.1803 -0.1803 -0.01 18.03 3.1 28.83 -1.83 -0.1 18.3 3.2 30.72 -3.72 -0.2 18.6 3.3 32.67 -5.67 -0.3 18.9 3.4 34.68 -7.68 -0.4 19.2 3.5 36.75 -9.75 -0.5