• 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
• Level: GCSE
• Subject: Maths
• Word count: 1048

Extracts from this document...

Introduction

Introduction

For this project, I will deduce the gradient for several y=axn graphs and with this find any sort of relationships between the x values and the gradient. A gradient is the steepness of a curve at a point. Gradients can prove to be very useful. It usually means something in most graphs for e.g. in distance-time graphs, the gradient indicates the speed. The gradient formula for a straight line is:

Change in x

However, since the y=axn forms a curve rather than a simple straight line, it is much more apprehensive to calculate its gradient. Therefore, another method has to be applied. The gradient for the non-linear graph is the steepness of a curve at one particular point. In order to find this, I will need to draw specific tangents on different x-axis points. A tangent in a non-linear graph is a straight line that essentially touches the curve at one point with two tiny alike angles either side.

Middle

5.5

9.02

And here are the theoretical results:

 x -4 -3 -2 -1 0 1 2 3 4 y 16 9 4 -1 0 1 4 9 16 Gradient -8 -6 -4 -2 0 2 4 6 8

The graph y=x2 evidently shows that there is a strong positive correlation between the x-axis and gradient. Each time the x value is increased, the gradient tends to double. Therefore f(x) = 2x. This formula is also supported by the calculation I have done above. I also managed to spot a pattern with the gradient values: add 2 to the previous gradient to get the subsequent one. Nonetheless, for the negative slope, you have to add minus 2 to the formal gradient in order to get the consequent one. As with my own results, it was fairly good with several values near the actual value than others. Moreover, the results were quite consistent. The mean of the difference of my result to the real result was 1.2.

Graph: y=x3

Theoretical formula for y=x3

P(x, x3) Q(x+h,(x+h)3)

Gradient function = (x+h)3-x3= x3+3x2h+3h2x+h3-x3 = 3x2+3xh+h2 = 3x2

(x+h)-x                    h

As h is relatively a small value, h2 would be even less significant. Thus h and h2 are neglected as it would not have that much of an effect.

Here are my practical results for the graph y=x3 with the x axis ranging from -4 to +4:

 x -4 -3 -2 -1 0 1 2 3 4 y -64 -27 -8 -1 0 1 8 27 64 Gradient 44.45 23.53 13.19 4.5 0 3.72 12.5 25.77 50

Conclusion

s="c3">y

-64

-27

-8

-1

0

1

8

27

64

48

27

12

3

0

3

12

27

48

My results for the graph above were fairly close to the actual one. The mean of the difference was 1.77. This indicates that my results were better for the graph y=x2 than y=x3. Instead of dividing the total difference by 9, I divided it by 8 because 0 can be ignored as the result will be exact, which will give a far more improved mean.

Graph: y=2x2

Theoretical formula for y=2x2:

P(2x,2x2) Q(2x+h,(2x+h)2)

Gradient function = (2x+h)2-2x2 = (2x+h-2x)(2x+h+2x) = 2x+h+2x = 4x

2x+h-2x                 2x+h-2x

Here are my results for the graph y=2x2 with the x axis ranging from -4 to +4:

 x -4 -3 -2 -1 0 1 2 3 4 y 32 18 8 2 0 2 8 18 32 Gradient -19.5 -11.6 -8.17 4.41 0 4.33 8.28 11.47 15.59

And here are the verified results

 x -4 -3 -2 -1 0 1 2 3 4 y 32 18 8 2 0 2 8 18 32 Gradient -16 -12 -8 -4 0 4 8 12 16

For my final graph, my results were moderately close. 0.75The gradient was 4 times the value of its x-axis.

Roshan The Great

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

## Here's what a star student thought of this essay

4 star(s)

### Response to the question

Using 3 different worked examples and explanations is the best part of this coursework which helps the reader completely understand what is being explained here. The student successfully answers the question and provides useful information to justify his answers however ...

### Response to the question

Using 3 different worked examples and explanations is the best part of this coursework which helps the reader completely understand what is being explained here. The student successfully answers the question and provides useful information to justify his answers however they are quite not enough. Using drawn graphs in the coursework would have helped the reader understand the sequence of logic better.

### Level of analysis

I can strongly say that the student has worked hard on this coursework but his/her method is not completely right. Looking at this work I can suggest that this essay is written slightly on a memorising basis. It is not clear why we can reduce the value of h in all 3 examples to zero and neglect them. Since we are trying to find the gradient at one specific point and not between two points like in a straight line, therefore the value of h MUST be neglected to prevent us from virtually drawing a straight line between two point on a curve and find the gradient of that straight line. Nonetheless the coursework is very useful and deserving of a good grade.

### Quality of writing

There is absolutely no problem with the writing part if we neglect a few typographical errors here and there.

Reviewed by alireza.parpaei 03/09/2012

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)

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.

crosses through x = 1.5 on the curve. I have drawn the triangle as I explained on the previous page in yellow. The change in x is 0.9 units. The change in y is 2.7 units. We now have to substitute these values into the equation.

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

So I predict that the gradient formula for y=x4 is: g=4x3. Let's try it out. The co-ordinates: x 1 2 3 4 Y 1 16 81 256 The gradients in y=x4: x g (The Gradient) Increment Method 1 4.040601 2 32.240801 3 108.541201 4 256.96101 g1=4=4�1=4x13 g2=32=4�2�2�2=4x23 g3=108=4�3�3�3=4x33 g4=256=4�4�4�4=4x43 So my prediction works this time!

This is because it has a horizontal asymptote of 0. 3.6 1-1.55e^ (8-0.31x) The difference between the reciprocal function and the logistic function is that the logistic function fits more accurately. However, it is still hard to predict other values from the equations The table below shows the average number

1. ## The Gradient Function Investigation

Also, the power of X in the gradient function (e.g. 3x�) is one less than the power of X in the graph equation (e.g. y = x�). I can therefore summarise the results from these three examples in the following formula: Gradient Function for graph y = xn is nx(n-1)

-2.25 -0.5 4.5 2.6 6.76 -2.76 -0.6 4.6 2.7 7.29 -3.29 -0.7 4.7 2.8 7.84 -3.84 -0.8 4.8 2.9 8.41 -4.41 -0.9 4.9 3 9 -5 -1 5 Power: 2 Coefficient: 1 Fixed point: 2 My second fixed point is 5, 25 x y change in y change in x

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

functions have only one root after all, but with a multiplicity of three, my hypothesis have also proven this correct.) Functions with Two Roots Example 1- Roots = -2, 2 1. Roots -2 & 2 Average of the two roots = Plug the point, (0, -8)

x 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

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