Y=x²
This reprents the graph Y=x². Now I have to use a bit on the curve which is the straightest at the point I want (the circled point). Then I will find the gradient of that straighest bit, but it will not be 100% accurate as that line will have a slight curve.
At (1,1) this will be 1.0201 – 1 = 0.0201 = 2.01
1.01 – 1 0.01
this is correct because to 1sf it is 2
At (2,4) 4 .0401 – 4 = 0.0401 = 4 .01
2.01 – 2 0.01
At (3,9) 9.0601 – 9 = 0.0601 = 6.01
3.01 – 3 0.01
At (4,16) 9.0601 – 16 = 0.0801 =8.01
4.01 – 4 0.01
At (5,25) 25.1001 –25 = 0.1001 =10.01
5.01 – 5 0.01
These results have proved that this method is more accurate than the tangent of the gradient method, but not perfect, as it only seems to be 0.01 out.
I will now use this method to find out the results for y=x³.
Compared to the tangent method these seem more accurate. I now know that the tangent for (1,1) and (2,8) were correct and that the tangent for (3,27) and (4,64) were incorrect, this proves that the small increments method is more accurate.
I will now use the small increment method to find out the results for y=x4.
I can definitely tell that the results of the tangent are incorrect and these are more correct. But they aren’t entirely correct as the numbers start to get bigger they start to get more inaccurate. So to solve this inaccuracy I will go to a thousandth instead of a hundredth.
So I will do the results for y=x4 again but going to a thousandth.
These are more accurate results.
I will now work out the results for y=x5.
These are quite accurate but when it gets to at (4,1024) and (5,3125) it starts to get inaccurate again so I will go to 0.0001 instead of 0.001.
I will do (4,1034) and (5,3125) again, going to 0.0001.
These results are more accurate.
I will also use this method for y=x1, though I know that the answer is one all the time, just to prove that this method is more accurate.
These are very accurate results.
Overall results.
These are the results of the small increments method. The results will be to 1 the nearest whole integer.
I will now compare the results of the tangent of the gradient method and the small increments method. (TG= gradient method, S.I small increments method)
The highlighted ones are the anomalous results. This shows me that the small increments method is more accurate. I did not use the gradient method for x=5 or y=x5.
There must be an easier way to find out what the results should be. I will now try and find a formula for each one using the small increment results.
Y = x1.
1 1 1 1 1 n= x0
1 1 1 1
Y=x2
2 4 6 8 10 n=2x1
2 2 2 2
Y=x3
3 12 27 48 75
9 15 21 27 n=3x2
6 6 6
Y=x4
4 32 108 256 500
32 76 148 244 n=4x3
48 72 96
24 24
Y=x5
5 80 405 1280 3125
75 325 875 1845
250 550 970 n=5x4
-
420
120
This is the table of results for the formulae.
For Y=x5 I did not have enough information to find the formula but as it is in an obvious pattern I have guessed what it may be.
I see a pattern in the formula. The number that x is multiplied by is the same as what the power of x is in the graph, and it is the power to n-1.
I will put all the formulae into a table and work out the overall formula.
Table of results for the formula.
I will now use the overall formula and use it on all the equations to prove that it is correct.
I will predict that the results of y=x6 and then use the small increments method to double-check it.
Formula Method.
Small Increments Method.
These are pretty correct results so this shows that the formula is correct. For x=5 I had to do it to 0.00001 because the number is likely to be incorrect if done to 0.0001.
I think that there may be a formula that I can work for powers that are negative, and I think that the follow the same pattern as the positive powers.
I will try and find out the gradient function of the y=x-1.
I will use the small increments method first.
This seems pretty correct because as the number that x is increases the gradient function decreases and as it is to the power –1, it makes sense, as it will be a smaller number. I know that is a smaller number because y = 1 x
Which is the same as y=x-1, which means the formula will be 1x-2, which is the same as – 1 . This is the formula I will use.
X2
I will now use the formula method to work out the gradient function and compare them to the small increments and they should be more accurate than the small increments results.
These results are more accurate I can tell that the formula is also correct.
I will now compare the formula results and small increment results.
My small increment results are to 2 decimal places whereas the formula results are the calculator display. My results are pretty accurate between both methods, so I can tell that the formula method is correct.
Each of the results are negative, this is because the graph must have a negative correlation. It has a negative correlation because 1 is divided by it and it makes a smaller number than 1.
I think that for negative powers the pattern is the same as the positive numbers.
I will now see what happens when a number is to fractional indices.
I am going to use y = x½. This is the same as y = x
I will use the small increments method first.
I will now use the formula method to find out the results. The formula for this is ½x-½.
X=1
½ x 1-½ = 0.5
X=2
½ x 2-½ = 0.35 to 2d.p
X=3
½ x 3-½ = 0.29 to 2d.p
X=4
½ x 4-½ = 0.25
X=5
½ x 5-½ = 0.22 to 2d.p
These results are nearly the same as the small increments method, so the formula must be correct. The graph of this equation must look like this:
Conclusion.
During this investigation I have used three different methods to find the gradient function:
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The Tangent of the gradient – this method included me drawing a graph and plotting tangents through each point, and finding the gradient by drawing a triangle from two points. This proved to be quite an inaccurate method. So I used the….
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Small Increments Method – this method proves to be more accurate than the tangent method, but it still has a level of inaccuracy when the numbers start to get bigger. So from this method I could find formulas and use the…
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Formula Method – this method is very accurate and once you know the overall formula you can work out any equation of a curved graph. The overall formula is Y=nxn-1. This formula works for positive, negative and fractional indices.
Throughout the investigation I have used two of the three methods for each equation. This is so that I can check the results of both methods and compare them to double-check it and to find a more accurate method. My overall conclusion of the methods is that the tangent method is very inaccurate as there are many factors that can make it inaccurate (the curve, the tangent drawn etc), and the formula method is very accurate as it makes it easier to find the correct answer and does not cause inaccuracies unless the formula is worked out the wrong way.
I have found out from the positive indices the numbers increase, whereas the negative indices decrease, as well as the fractional indices.