= 2x+h
As before, as Q gets closer to P, the value of h gets closer to 0 the gradient of the chord PQ also becomes closer to the actual gradient of PQ. Once h=0, you will be left with 2x, which will be the gradient of the curve.
y=x³
This is the graph of y=x³. Once again I will use the three methods to work out the gradient of this curve.
Tangent Method: x=2
Change in y-axis=12
Change in x-axis=1
Gradient=12
Increment Method: x=2
P(2,2³) and Q(3,3³)
=3³-2³
3-2
=19
P(2,2³) and Q(2.75,2.75³)
=2.75³-2³
2.75-2
=17.1
P(2,2³) and Q(2.5,2.5³)
=2.5³-2³
2.5-2
=15.25
P(2,2³) and Q(2.25,2.25³)
=2.25³-2³
2.25-2
=13.6
P(2,2³) and Q(2.01,2.01³)
=2.01³-2³
2.01-2
=12.1
P(2,2³) and Q(2.001,2.001³)
=2.001³-2³
2.001-2
=12.01
P(2,2³) and Q(2.0001,2.0001³)
=2.0001³-2³
2.0001-2
=12.001
From these results, I can conclude that on a y=x³, when x=2, the gradient of the curve is 12.
General Proof:
(x+h)³-x³
x+h-x
=x³+3x²h+3h²x+h³-x³
h
=h(3x²+3hx+h²)
h
=3x²+3hx+h²
=3x²
Therefore we can conclude that the formula for y=x³ is 3x². Therefore when:
x=1, the gradient=3
x=2, the gradient=12
x=3, the gradient=27
x=4, the gradient=48
This is the graph of y=x4 . I will again, use the three methods to try and find a formula to work out the gradient of this curve.
Tangent Method: x=3
Change in y-axis=144
Change in x-axis=1
=144
1
Gradient=144
Increment Method: x=3
P(3,34) and Q(4,44)
=44-34
4-3
=175
P(3,34) and Q(3.75,3.754)
=3.754-34
3.75-3
=155.7
P(3,34) and Q(3.5,3.54)
=3.54-34
3.5-3
=138.1
P(3,34) and Q(3.25,3.254)
=3.254-34
3.25-3
=122.3
P(3,34) and Q(3.01,3.014)
=3.014-34
3.01-3
=108.5
P(3,34) and Q(3.001,3.0014)
=3.0014-34
3.001-3
=108.05
P(3,34) and Q(3.0001,3.00014)
=3.00014-34
3.0001-3
=108.005
From these results, I can conclude that on the graph y=x, at the point x=3, the gradient is equal to 108.
General Proof: (x+h)4-x4
x+h-x
x4+h4+4hx³+6x²h+4xh³-x4
h
=h(4x³+4xh²+6x²h+h³)
h
=4x³+4xh²+6x²h+h³
=4x³
Therefore from the general proof, we can conclude that the formula for the graph y=x4 is 4x³.
Therefore when:
x=1, the gradient=4
x=2, the gradient=32
x=3, the gradient=108
x=4, the gradient=256
y=x5
This is the graph of y=x5. I will continue to investigate the gradients of the graphs y=xn.
Increment Method: x=3
P(3,35) and Q(4,45)
=45-35
4-3
=781
P(3,35) and Q(3.75,3.755)
=3.755-35
3.75-3
=664.8
P(3,35) and Q(3.5,3.55)
=3.55-35
3.5-3
=564.4
P(3,35) and Q(3.25,3.255)
=3.255-35
3.25-3
=478.4
P(3,35) and Q(3.01,3.015)
=3.015-35
3.01-3
=407.7
P(3,35) and Q(3.001,3.0015)
=3.0015-35
3.001-3
=405.3
P(3,35) and Q(3.0001,3.00015)
=3.00015-35
3.0001-3
=405.03
From these results we can conclude that the gradient on an y=x, when x=3 is 405.
Instead of using general proof to work out the equation for this graph, I will use another method, that I have taught myself. It is called Binomial Expansion.
A binomial expansion is the result of a binomial expression, like (x+h) being raised to a power.
The simplest binomial expansion is (x+1), but I will show you the expansion of (x+h)5. So the expansion is:
1 x5 + 5 x h + 10 x h + 10 x h + 5 x h + 1 h
x5 + 5x4h + 10x3h2 + 10x2h2 + 5xh4 + h5
We can now, add the remainder of the formula and simplify this down, like in general proof:
x5+5x4h+10x3h2+10x2h2+5xh4+h5-x5
h
=h(5x4+10x3h+10x2h+5xh3+h4)
h
=(5x4+10x3h+10x2h+5xh3+h4)
=5x4
Therefore when:
x=1, the gradient=5
x=2, the gradient=80
x=3, the gradient=405
x=4, the gradient=1280
From the graphs that I investigated I produced the following table:
From these results I could see a pattern emerging, therefore I created this formula, which applied to all y=xn graphs:
nxn-1
In the next few graphs I will no prove that my formula is correct.
y=x½
This is the graph of y=x½.
Increment Method: x=1
P(1,1½) and Q(2,2½)
=2½-1½
2-1
=0.41
P(1,1½) and Q(1.75,1.75½)
=1.75½-1½
1.75-1
=0.43
P(1,1½) and Q(1.5,1.5½)
=1.5½-1½
1.5-1
=0.45
P(1,1½) and Q(1.25,1.25½)
=1.25½-1½
1.25-1
=0.47
P(1,1½) and Q(1.01,1.01½)
=1.01½-1½
1.01-1
=0.49
P(1,1½) and Q(1.001,1.001½)
=1.001½-1½
1.001-1
=0.49
P(1,1½) and Q(1.0001,1.0001½)
=1.0001½-1½
1.0001-1
=0.49
From these results we can conclude that when x=1 on a y=x½ graph, the gradient is 0.5
Let us now put this graph into the formula nxn-1. This then gives us 0.5x-0.5. With this formula we can work out whether or not when x=1, the gradient is 0.5.
1-0.5
=1
1x0.5=0.5. This answer matches that of the one we got with the increment method.
y=x¼
Increment Method: x=1
P(1,1¼ ) and Q(2,2¼ )
=2¼ -1¼
2-1
=0.19
P(1,1¼ ) and Q(1.75,1.75¼ )
=1.75¼ -1¼
1.75-1
=0.2
P(1,1¼ ) and Q(1.5,1.5¼ )
=1.5¼ -1¼
1.5-1
=0.21
P(1,1¼ ) and Q(1.25,1.25¼ )
=1.25¼ -1¼
1.25-1
=0.23
P(1,1¼ ) and Q(1.01,1.01¼ )
=1.01¼ -1¼
1.01-1
=0.249068
P(1,1¼ ) and Q(1.001,1.001¼ )
=1.001¼ -1¼
1.001-1
=0.2499063
P(1,1¼ ) and Q(1.0001,1.0001¼ )
=1.0001¼ -1¼
1.0001-1
=0.2499906
From these results we can conclude that the gradient, when x=1, on a y=x¼ is 0.25
Let us now put this graph into the formula nxn-1. This then gives us 0.25x-0.25. With this formula we can work out whether or not when x=1, the gradient is 0.5.
1-0.25
=1
1x0.25=0.25. This answer matches that of the one we got with the increment method.
y=x1.5
This is the graph y=x1.5 .
Increment Method:
P(2,21.5 ) and Q(3,31.5 )
=31.5 -21.5
3-2
=2.37
P(2,21.5 ) and Q(2.75,2.751.5 )
=2.751.5 -21.5
2.75-2
=2.31
P(2,21.5 ) and Q(2.5,2.51.5 )
=2.51.5 -21.5
2.5-2
=2.25
P(2,21.5 ) and Q(2.25,2.251.5 )
=2.251.5 -21.5
2.25-2
=2.19
P(2,21.5 ) and Q(2.01,2.011.5 )
=2.011.5 -21.5
2.01-2
=2.1239
P(2,21.5 ) and Q(2.001,2.0011.5 )
=2.0011.5 -21.5
2.001-2
=2.1216
P(2,21.5 ) and Q(2.0001,2.00011.5 )
=2.00011.5 -21.5
2.0001-2
=2.1213
From these results we can conclude that when x=2 on a y=x1.5 the gradient is 2.12.
Let us now put this graph into the formula nxn-1. This then gives us 1.5x0.5. With this formula we can work out whether or not when x=1, the gradient is 0.5.
20.5
=2
1.41x1.5=2.12. This answer matches that of the one we got with the
As we have now found a formula for all graphs of y=xn, I will now go on and investigate with other graphs, a general rule for the gradient of any type of graph.
y=2x²
This is the graph y=2x². In this graph I will use a different method to general proof, which I have taught myself. It is called Differentiation, and is very similar to general proof.
Differentiation is based upon the gradient function, .dy Being the difference in the y-axis, and dx being the difference in the x-axis. This is similar to that in the tangent method.
can also be written as:
Lim is read as ‘the limit as δx tends towards zero’.
dx has the co-ordinates (x,x²) and dy has the co-ordinates (x+h(x+h)²). Similarly δx has the co-ordinates (x,f(x)). dy has the co-ordinates (x+h(x+h)²). Similarly δy has the co-ordinates (x+δx,f(x+δx)).
f(x) is used here to mean the value of y corresponding to x.
With this information we can now go no to find the formula of the graph y=2x².
y=f(x)=2x²
δy=f(x+δx)-f(x)
=(2x+δx)²-2x²
And so
=lim
=lim (2x+δx)²-2x²
δx
=lim 2x²+4xδx+(δx)²-2x²
δx
=lim 4xδx+(δx)²
δx
=lim (4x+δx)
=4x
This then has given us 4x, which is the formula for the graph y=2x².
Therefore when:
x=1, the gradient=4
x=2, the gradient=8
x=3, the gradient=12
x=4, the gradient=16
y=2x-3
y=f(x)=2x-3
δy=f(x+δx)-f(x)
=(3x+δx)-3-2x-3
And so
=lim
=lim (3x+δx)-3-2x-3
δx
=lim 2x-3+-6x-4δx+6x(δx)-4+(δx)-3-2x-3
δx
=lim δx(-6x-4+6x(δx)-3+(δx)-2)
δx
=lim (-6x-4+-6x(δx)-3+(δx)-2)
=-6x-4
This then has given us -6x-4, which is the formula for the graph y=2x-3.
Therefore when:
x=1, the gradient=0.00077
x=2, the gradient=0.000048
x=3, the gradient=0.0000095
x=4, the gradient=0.00000301
y=3x4
y=f(x)=3x4
δy=f(x+δx)-f(x)
=(3x+δx)4-3x4
And so
=lim
=lim (3x+δx)4-3x4
δx
=lim 3x4+12x³δx+18x²h²+12xh3(δx)4-3x4
δx
=lim h(12x³+12xh²+18x²h+h³)
h
=lim 12x³+12xh²+18x²h+h³
=12x³
This then has given us 12x³, which is the formula for the graph y=3x4.
Therefore when:
x=1, the gradient=3
x=2, the gradient=48
x=3, the gradient=243
x=4, the gradient=768.
From the graphs we have looked at from this part of the investigation we can once again form a results table:
From these results I could see a pattern emerging, therefore I created this formula, which is a general rule for all graphs:
naxn-1
In this formula there is one change, this is the inclusion of ‘a’. Here the term ‘a’ is representing the constant (if there is one) in the graph. For example in the graph 3x4, the constant is the 3.
I am now going to further my investigation, and look at graphs which include terms that are added together. In this the formula is slightly different. The difference is that we split the original formula, naxn-1, into two.
For example, we shall take the graph, 2x²+3x³:
As we can see, two terms involved, therefore if we were to use our formula naxn-1, it would not work, as there is more than one term. Therefore we have to add something to the formula, to include the second term. So when there are formulas that have more than one term involved, the formula is:
naxn-1+laxn-1
Here the change is, the inclusion of laxn-1. This of course is the second term. The formula would, change depending on the amount of terms involved.
So therefore form this formula I can work out the gradient of the graph y=2x²+3x³.
Let’s first take the graph 2x². I, previously got the formula 4x. The second graph 3x³, we haven’t looked at before, but with our formula, we can see that it is equal 9x².
This means that to find the gradient of the curve y=2x²+3x³, we use the formula 4x+9x².
Therefore the gradient of the curve when:
x=1 is 4+9=13
x=2 is 8+36=44
x=3 is 12+81=93
x=4 is 16+144=160
If I were to work out a similar graph to, y=2x²+3x³, using differentiation, I would use exactly the same method. I would take each term in turn, and work out their respective formula, thus enabling me to put this into the general formula.
