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Introduction

Part 1

Investigate the gradient function for the set of graphs: -  y=axn where a and n are constant.

For Part 1 of my investigation I will be looking at many different types of y=axn graph    such as:

y = x2, y = 2x2, y = 3x2, y = 4x2,

y = x3, y = 2x3, y = 3x3, y = 4x3

y = x4, y = 2x4, y = 3x4, y = 4x4

y = x2, y = x3, y= x4, y= x5

y = 2x2 y = 2x3 y =2x4 y = 2x5

I will be trying to find the gradients of certain points on the graph which can lead me to the gradient function for that curve.

Gradient: measures the steepness of  a curve. The gradient of any particular point on                      a curve is defined as the gradient of the tangent drawn to the curve at that                         point.

Tangent: is a straight line which touches but does not cut the given curve at a particular                     point.

The gradient of any tangent is vertical

horizontal

Gradient function: is a common function for the gradients of a set of tangents on a particular curve. When this function is obtained, tangents do not need to be drawn to work out the gradient.

Prediction: I predict that there will be a gradient function to every graph that is y=axn                       then from those gradient function a common function can be found.

Middle

1.5

2

3

-2

8

12

16

24

-16

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.

 Graph Gradient Function y=x2 2x y=2x2 4x y=3x2 6x

Using my gradient functions for these curves I can predict the gradient function for y=ax2

 y=ax2 2ax

For the curve y=10x2, using the gradient function 20x I think the gradients will be:-

 Tangent at x 1 1.5 2 3 -2 Gradient 20 30 40 60 -40 Method (2ax) a=10, x=12 x 10 x 1= 20 a=10, x=1.52 x 10 x 1.5= 30 a=10, x=22 x 10 x 2= 40 a=10, x=32 x 10 x 3= 60 a=10, x=-22 x 10 x -2= -40

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:-

 Tangent at x 1 1.5 2 3 -2 Gradient 2 3 4 6 -4

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:-

 Tangent at x 1 1.5 2 3 -2 x2 1 2.25 4 9 4 Gradient 3 6.75 12 27 12

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:-

Conclusion

For the curve y=x5, using the gradient function 5x4 I think the gradients will be:-

 Tangent at x 1 1.5 2 3 -2 Gradient 5 25.3 80 405 80

For the curve y=2x5, using the gradient function 10x4 I think the gradients will be:-

 Tangent at x 1 1.5 2 3 -2 Gradient 10 50.6 160 810 160

For the curve y=3x5, using the gradient function 15x4 I think the gradients will be:-

 Tangent at x 1 1.5 2 3 -2 Gradient 15 76 240 1220 240

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:-

 Tangent at x 1 1.5 2 3 -2 Gradient 20 30 40 60 -40

For the curve y=10x3, using the gradient function 30x2 I think the gradients will be:-

 Tangent at x 1 1.5 2 3 -2 Gradient 30 67.5 120 270 120

For the curve y=10x4, using the gradient function 40x3 I think the gradients will be:-

 Tangent at x 1 1.5 2 3 -2 Gradient 40 135 320 1080 -320

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.

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

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1.  5 star(s)

the function will determine the correct gradient for graphs with fractional powers. To ensure this, I will test the function on the range of graphs with formula y=ax^0.5. For the example below I will show how the gradient is worked out by using the formula: m= nax^n-1 => m=0.5*1*1^0.5 =>

2.  4 star(s)

Gradient function = (x+h)3-x3 = x3+3x2h+3h2x+h3-x3 = 3x2+3xh+h2 = 3x2 (x+h)-x h As h is relatively a small value, h2 would be even less significant. Thus h and h2 are neglected as it would not have that much of an effect.

1. Aim: To find out where the tangent lines at the average of any two ...

cubic function, the second one being a horizontal translation of the first function, and the third being a negative function and translated both horizontally and vertically from the first one, I have found out that their tangent line always goes through the root of the original function, and since the

2. Investigate gradients of functions by considering tangents and also by considering chords of the ...

I am going to look at y=x first because it is the easiest. x 1 2 3 4 y 1 2 3 4 The gradients of y=x is very simple, g=1. We even do not need to draw any tangents to obtain the gradients.

1. I have been given the equation y = axn to investigate the gradient function ...

1, y=2x is 2, y=3x is 3, y=4x is 4, Therefore it is clear here that the gradient of the graph of y=ax1 is a Testing:Graph 12 For the graph of y = 5x the gradient function according to the relation I have established should be 5.Let us try drawing the graph and finding out the gradient.

2. 12=3 3 X 22=12 3 X 32=27 3 X 42=48 As you can see my formula does work, they give almost exactly the same answers for the gradient as my increment results, which means it must be right, also the increment method must be more accurate.

1. and Q(3,3() =3(-2( 3-2 =19 P(2,2() and Q(2.75,2.75() =2.75(-2( 2.75-2 =17.1 P(2,2() and Q(2.5,2.5() =2.5(-2( 2.5-2 =15.25 P(2,2() and Q(2.25,2.25() =2.25(-2( 2.25-2 =13.6 P(2,2() and Q(2.01,2.01() =2.01(-2( 2.01-2 =12.1 P(2,2() and Q(2.001,2.001() =2.001(-2( 2.001-2 =12.01 P(2,2() and Q(2.0001,2.0001() =2.0001(-2( 2.0001-2 =12.001 From these results, I can conclude that on a y=x(, when x=2, the gradient of the curve is 12.

2.   