From this we can predict the gradient to be 2 because the numbers behind the decimal point become insignificant.
This shows us the gradient of only one point but to find the gradient function of that curve you need to find a set of gradient from different points on the graphs.
I found the gradients of more points on the curve y=x2 using the same method.
P=(2,4)
A1 A2 A3 A4 A5 A6
From this we can predict the gradient to be 4 because the numbers behind the decimal point are insignificant.
P=(3,9)
A1 A2 A3 A4 A5 A6
From this we can predict the gradient to be 4 because the numbers behind the decimal point are insignificant.
P=(-2,4)
A1 A2 A3 A4 A5 A6
From this we can predict the gradient to be 4 because the numbers behind the decimal point are insignificant.
We have found the gradient of particular points on the curve y=x2
For the curve y=x² the gradient of the tangent at x is two times the x value:-
Now we can predict the gradient function to be 2x as the gradient is always 2 times of x.
Another way to find the formula for the gradient function is by :
The point B is the point with co-ordinates (x,x2) and the point C is the point near to B with co-ordinates (x+h,(x+h)2).
For the curve y=2x², the gradient of the tangent at x is four times the value of x:-
Now we can predict the gradient function to be 4x as the gradient is always 4 times of x.
For the curve y=3x², the gradient of the tangent at x is six times the value of x:-
Now we can predict the gradient function to be 6x as the gradient is always 6 times ‘x’
For the curve y=4x², the gradient of the tangent at x is eight times the value of x:-
Now we can predict the gradient function to be 8x as the gradient is always 8 times ‘x’
I have found that there is a link between the gradient function for these graphs.
Using my gradient functions for these curves I can predict the gradient function for y=ax2
For the curve y=10x2, using the gradient function 20x I think the gradients will be:-
I checked this on Omnigraph and found that the gradients were the same as I thought they would be so this gradient function 20x clearly works showing that gradient function for ax2 is right.
For the curve y=x3, the gradient function would be 2x so the gradients of certain points are:-
I checked this on Omnigraph but these gradients weren’t right. These were the correct ones below.
For the curve y=x³, the gradient of the tangent at x:-
Therefore I think the gradient function for this curve is 3x2 because the gradient is 3 times x2
For the curve y=2x³, the gradient of the tangent at x:-
Therefore I think the gradient function for this curve is 6x2 because the gradient is 6 times x2
For the curve y=3x³, the gradient of the tangent at x:-
Therefore I think the gradient function for this curve is 9x2 because the gradient is 9 times x2
For the curve y=4x³, the gradient of the tangent at x:-
Therefore I think the gradient function for this curve is 12x2 because the gradient is 12 times x2
I have found that there is a link between the gradient function for these graphs.
Using my gradient functions for these curves I can predict the gradient function for y=ax3
To find the gradient function for these graphs you have to square ‘x’ then multiply the answer by ‘a‘ then ‘3‘.
To check if this function is correct first, without using Omnigraph I will find the gradient of certain points on the graph 10x3 and then I will check to see if these are right using Omnigraph.
If this gradient function, 3ax2 is correct, then the gradient function of 10x3 will be 30x2. To prove that it is 30x2 I am going to draw up a table of values for certain x values.
For the curve y=10x3, using the gradient function 30x2 I think the gradients will be:-
I checked this with Omnigraph and found that the gradients were the same as I thought they would be so this gradient function 30x2 clearly works for this graph showing that the gradient function for y=ax3 is right.
I know that this gradient function 3ax2 only works for y=ax3 graphs as the gradient function for y=ax2 that we found before doesn’t work for any other y=axn graphs. This means I don’t need to test the gradient function 3ax2 on any other graphs like y=ax4 graphs
For the curve y=x4, the gradient of the tangent at x:-
Therefore the gradient function of the curve is 4x3
For the curve y=2x4, the gradient of the tangent at x:-
Therefore the gradient function of the curve is 8x3
For the curve y=3x4, the gradient of the tangent at x:-
Therefore the gradient function of the curve is 12x3
I have found that there is a link between the gradient function for these graphs.
Using my gradient functions for these curves I can predict the gradient function for y=ax3
To find the gradient function for these graphs you have to cube ‘x’ then multiply the answer by ‘a‘ then ‘4‘.
To check if this function is correct first, without using Omnigraph I will find the gradient of certain points on the graph 10x4 and then I will check to see if these are right using Omnigraph.
If this gradient function, 4ax3 is correct, then the gradient function of 10x4 will be 40x3. To prove that it is 40x3 I am going to draw up a table of values for certain x values.
For the curve y=10x4, using the gradient function 40x3 I think the gradients will be:-
I checked this with Omnigraph and found that the gradients were the same as I thought they would be so this gradient function 40x3 clearly works for this graph showing that the gradient function for y=ax4 is right.
I know that this gradient function 4ax3 only works for y=ax4 graphs as the gradient function for y=ax2 and y=ax3 that we found before doesn’t work for any other y=axn graphs. This means I don’t need to test the gradient function 3ax2 on any other graphs like y=ax4 graphs
For the graphs that I have just done I have only been varying ‘a’ and keeping ‘n’ constant so now I will be keeping ’a’ constant and will be varying ‘n’. I have found that there is a link between the gradient functions for these graphs.
To find the gradient function for these graphs, you have find ‘n-1‘, then do ‘x’ to the power of the answer, then multiply that by ‘n‘.
To check if this function is correct first, without using Omnigraph I will find the gradient of certain points on the graph x10 and then I will check to see if these are right using Omnigraph.
If this gradient function, nx(n-1) is correct, then the gradient function of x10 will be 10x9. To prove that it is 10x9 I am going to draw up a table of values for certain x values.
For the curve y=x10, using the gradient function 10x9 I think the gradients will be:-
I checked this with Omnigraph and found that the gradients were the same as I thought they would be so this gradient function 10x9 clearly works for this graph showing that the gradient function for y=nx(n-1) is right.
I know that this gradient function nx(n-1) only works for y=xn graphs as the gradient function doesn’t include ‘a’ like other y=axn graphs.
I have put all the gradient functions that I have found through my investigation together to try and find the gradient function which applies to all y=anx(n-1).
I think anx(n-1) is the gradient function for all y=axn graphs. To test if this is right I will first use this function to work out the gradient of certain points on the graphs. I am going to use the graphs - x5, 2x5, 3x5, 4x5. First I am going to do this without using Omnigraph but with the gradient function that I think applies to all y=axn graphs then I will use Omnigraph to check if the gradients are right.
For the curve y=x5, using the gradient function 5x4 I think the gradients will be:-
For the curve y=2x5, using the gradient function 10x4 I think the gradients will be:-
For the curve y=3x5, using the gradient function 15x4 I think the gradients will be:-
I checked the gradients of these graphs on Omnigraph and found that they were the same as I thought that they would be which shows that I the gradient function for all y=anx(n-1) clearly works. I have only tested this on graphs where ‘n’ is constant and ‘a’ varies so now I will test it on graphs of - 10x2, 10x3, 10x4, 10x5 using the same method as before.
For the curve y=10x2, using the gradient function 20x I think the gradients will be:-
For the curve y=10x3, using the gradient function 30x2 I think the gradients will be:-
For the curve y=10x4, using the gradient function 40x3 I think the gradients will be:-
Again I checked this on Omnigraph and found that the gradients I thought it would be was correct. This means the gradient function anx(n-1) really does work for all y=axn graphs. It worked on graphs with ‘n’ constant and graphs with ‘a’ constant.