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Introduction

Jatinder Minhas

Maths Coursework

1. Draw a graph of y = x2 for values of x from 0 to 4. Obtain the gradient of the tangent at different points. Record the results.

To conduct this question I first had to obtain the values of y = x2 for values of x from 0 to 4. This is shown below. I decided to use values more accurate and precise values of x to enable me to obtain a more accurate curve and thus enabling me to obtain a more accurate gradient. I took the values of y = x2 and then plotted them on the graph. I rounded the values to two decimal places to enable me to plot the point as accurately as possible on my scale.

 y y = x2 0.0 0.0 0.25 0.0625 0.5 0.25 0.75 0.5625 1.0 1.0 1.25 1.5625 1.5 2.25 1.75 3.0625 2.0 4.0 2.25 5.0625 2.5 6.25 2.75 7.5625 3.0 9.0 3.25 10.5625 3.5 12.25 3.75 14.025 4.0 16.0

I then joined up the points with a flexi-curve.

Middle

4

2

4

6

8

Even though I used an accurate method, which was the tangent method I felt that I could have, improved and found the gradients using a method, which could check and maybe even improve values of the gradients. This method is called the small increment method and is shown below. This method is calculated first by taking a point on the x axis and then square rooting it. You then take another point that is close to the original point and you then square root the x axis point and you follow the method below. Though one rule is that the original point must always remain the same.

Using the small increment method the results for my calculations were;

 Point on line 1 2 3 4 Gradient 2 4 6 8

These results were the same as the results found using the tangent method.

Conclusion

"1">

11.390625

2.5

2.5

2.5

15.625

2.75

2.75

2.75

20.796875

3.0

3.0

3.0

27.0

3.25

3.25

3.25

34.328125

3.5

3.5

3.5

42.875

3.75

3.75

3.75

52.734375

4.0

4.0

4.0

64.0

I then used these values and I plotted a graph. I then joined the points up to create a curve. I then drew the tangents as accurately as possible and worked out the gradients.

y = x1

 Point on line 1 2 3 4 Gradient 1 2 3 4

y = x3

 Point on line 1 2 3 4 Gradient 1 12 27 48

I then used the small increment method for the calculation of the gradient and I acquired the results;

y = x1

 Point on line 1 2 3 4 Gradient 1 2 3 4

y = x3

 Point on line 1 2 3 4 Gradient 3 12 24 42

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

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