The Gradient Function Maths Investigation

increase in x

Firstly I am going to draw out the graph y=x² by hand for the values of 1 to 4 (see graph 1). I am then going to work out the gradient for the values of x=1,2, and 3 using the formula increase in y

increase in x

This method of drawing graphs by hand is very time consuming and is not very accurate. I am now going compare these results with the ones I have found using Omnigraph (see graph 2). These results are a lot more accurate than the ones I found from the hand drawn graph. I can now draw a more conclusive table of results from them:

X

Gradient

2

2

4

3

6

4

8

n

2n

Therefore the formula to find the gradient of the graph y=x² is:

Gradient= 2x

I am now going to find gradients on 2 other graphs to investigate this formula further and to test it.

For the graph y=x³ (see graph 3) I predict that the formula for finding the gradient will be 3x. I will now test this using my results:

X

Gradient

3

2

2

3

27

4

48

n

3n²

My prediction was incorrect. From this table I have worked out that the formula for the graph y=x³ is:

Gradient= 3x²

For the graph y=x4 (see graph 4) I predict that the formula for finding the gradient will be 4x³ because although the formulas for the previous graphs seem different a similar one to 3x² can be used for the graph y=x².

For the graph y=x²:

Gradient= 2x'

I will now test this using my results:

X

Gradient

4

2

32

3

10

4

260

n

4n³

My prediction was correct as the formula for the graph y=x4 is:

Gradient= 4x³

I am now going to put the formulas I have discovered so far into a table to make them easier to interpret:

Equation of graph

Gradient Function

y=x²

2x1

y=x³

3x²

y=x4

4x³

n

nx(n-1)

From the formulas I have found using my graphs I can see the general rule and from them I have worked out this formula:

Gradient= nx(n-1)

This formula is correct for all the graphs I have done so far but I am now going on other graphs. Firstly I am going to see if it works for negative numbers on the graph y=x -1 (see graph 5). These are the results I have found using this graph. I have also tested them to see if the formula Gradient= nx(n-1) works:

x

Gradient=nx(n-1)

Actual Gradient

= -1x1(-1-1)

= -1

-1

2

= -1x2(-1-1)

= -0.25

-0.25

3

= -1x3(-1-1)

= -0.11

-0.11

4

= -1x4(-1-1)

= -0.0625

-0.063

This graph is evidence of my formula for graphs with equations with negative integers but not with all numbers.

I am now going to try it with a graph with a fraction integer, y=x1/2 (see graph 6). These are the results I have found using this graph. I have also tested them to see if the formula Gradient= nx(n-1) works:

x

Gradient=nx(n-1)

Actual Gradient

= 1/2x1(1/2-1)

= 0.5

0.5

2

= 1/2x2(1/2-1)

= 0.35

0.35

3

= 1/2x3(1/2-1)

= 0.29

0.29

4

= 1/2x4(1/2-1)

= 0.25

0.25

I tested these figures using my formula and they seem to be correct when rounded to two significant figures. This is probably because it is very hard to find a graph that is 100% accurate and also because the gradients calculated using the program Omnigraph are only correct to two significant figures.

I am now going to try to algebraically prove my formula because although I have found evidence to support it I have not yet proven it.

To prove the formula for the graph y= x² I am going to use a method called differentiation, which I have found in an A level textbook. I have sketched a tiny part of the graph y= x² (graph 7) and on it I have labelled the two points P and Q. I have zoomed in on a small area that would be used using the small increment method, because of this the line PQ follows almost the same path as the curve and the gradient would equal K.

H

Therefore the coordinates of point P are (x, y) and the coordinates of point Q are (x + H, y + K).

Firstly I must work out the value of K:

Since Q (x + H, y + K) lies on the curve it must satisfy the equation y= x²:

(y + K) = (x + H)²

y + K = x² + 2xH + H²

But as y= x² I can take that out of the equation. Therefore

K = 2xH + H²

I can now try to reach the formula for gradient of the graph y= x² which is 2x.

Gradient of PQ=K

H

=2xH + H

H

=2x + H

I was stuck here for some time because I could not work out where H should go, as I want to get rid of it so that I am left with 2x. After again consulting the A level textbook I have found that as H is worth such a tiny amount that it does not matter. This is because the gradient of Q is on the line of PQ and is therefore equal to the gradient of P. This means H is worth virtually nothing. Therefore:

Gradient of tangent at P =2x

dy =2x

dx

These workings proves that my formula, Gradient= nx (n-1) is correct for the graph y= x². I am now going to try to prove my formula for the graph y=x³ using this method. I have drawn a sketch graph (graph 8) that is very similar to the graph previously used. The symbols all represent the same values as before.

(y + K) = (x + H)³

(y + K) = x³ + 3Hx² + 3H²x + H³

K = 3Hx² + 3H²x + H³

Gradient of PQ = K

H

= 3hx² + 3H²x + H³

H

= 3x² + 3Hx + H²

The Hs can now be ignored leaving:

dy =3x²

dx

As I predicted this method shows that the gradient of the graph y=x³ is 3x².

I have discovered that the gradient of all the graphs I have used can be found using the formula nx (n-1). This formula works for positive, negative and fractional integers. I have also gone some way to proving my formula algebraically by using differentiation.

Sarah Shea 11A

Maths Investigation