Part of The Student Room Group

  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Investigate the gradients of the graphs Y=AXN

Extracts from this document...

Introduction

GCSE Math’s Coursework

Gradient Function

        In this investigation I am going to investigate the gradients of the graphs Y=AXN Where A and N are constants.  I shall then use the information to find a formula for all curved graphs.

        To start the investigation I will draw the graphs where A=1 and N= a positive integer.

Y=X2

X

Height

Width

Gradient

1

1

0.5

2

2

4

1

4

3

9

1.5

6

4

16

2

8

        Looking at the results above I can see that the gradient is twice the X value, the height is X2 and the width is 1/2 the X value.  This shows me that there are several patterns in the graph but there is not enough to make a formula on so I am going to do another graph

Y=X3

X

Height

Width

Gradient

1

1

0.33

3

2

8

0.66

12

3

27

1

27

4

64

1.33

48

        There are some more patterns in this table, the height is now X3 and the width is 1/3 of the X value. I can see no pattern between the Gradient and the X value in this table.

...read more.

Middle

0.015625,0.015613287

0.000011713/0.001

0.011713

Y=X-4

X

Predicted gradient with formula (3dp)

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

-4*1(-4-1) = -4

1,1.001

1,0.99600998

-0.00399002/0.001

-3.99002

2

-4*2(-4-1)= -0.125

2,2.001

0.0625,0.062375156

-0.000124844/0.001

-0.124844

3

-4*3(-4-1) = -0.016460905

3,3.001

0.012345679,0.012329231

-0.000016448/0.001

-0.016448

4

-4*4(-4-1) = -0.00390625

4,4.001

0.00390625,0.003902346

-0.000003904/0.001

-0.003904

In both of the above cases the formula is almost identical to that of the increment method, this shows that my formula works of all cases for when A=1 and N= an integer. I am now going to experiment for when N= a faction.

Y=X1/2

X

Predicted gradient with formula (3dp)

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

1/2*1-1/2 =0.5

1,1.001

1,1.000499875

0.000499875/0.001

0.49987

2

1/2*2-1/2 =0.35355339

2,2.001

1.414213562,1.414567072

0.00035351/0.001

0.35351

3

1/2*3-1/2 =0.288675134

3,3.001

1.732050808,1.732339459

0.000288651/0.001

0.288651

4

1/2*4-1/2 =0.25

4,4.001

2,2.000249984

0.000249984/0.001

0.249984

This table shows that my formula work when X is a fraction and to prove this I will now look at the graph Y=X1/3 and see if the formula still works.

Y=X1/3

X

Predicted gradient with formula (3dp)

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

1/3*1-2/3 =0.3333333

1,1.001

1,1.000333222

0.000333222/0.001

0.333222

2

1/3*2-2/3 =0.209986841

2,2.001

1.25992105,1.260131002

0.000209952/0.001

0.209952

3

1/3*3-2/3 =0.160249952

3,3.001

1.44224957,1.442409802

0.000160232/0.001

0.160232

4

1/3*4-2/3 =0.132283421

4,4.001

1.587401052,1.587533324

0.000132272/0.001

0.132272

The formula works for when n= a fraction as well.  This is good for I now know that my formula works for any cases of N.  I shall now see if the A in the formula ANXN-1 is correct.  I shall do this by keeping N as 2 as it is the lowest number and make A 2 then 3.

 Y=2X2

X

Predicted gradient with formula

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

2*2*11 =4

1,1.001

2,2.004002

0.004002/0.001

4.002

2

2*2*21  =8

2,2.001

8,8.008002

0.008002/0.001

8.002

3

2*2*31 =12

3,3.001

18,18.012002

0.012002/0.001

12.002

4

2*2*41 =16

4,4.001

32,32.016002

0.016002/0.001

16.002

Y=3X2

X

Predicted gradient with formula

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

3*2*11 =6

1,1.001

3,3.006003

0.006003/0.001

6.003

2

3*2*21  =12

2,2.001

12,12.012003

0.012003/0.001

12.003

3

3*2*31 =18

3,3.001

27,27.018003

0.018003/0.001

18.003

4

3*2*41 =24

4,4.001

48,48.24003

0.024003/0.001

24.003

...read more.

Conclusion

N Where a and n are constants.  Now that I have proven this I shall try it out in other sets of graphs and see if it still works.  I am now going to work out the gradient of the graph Y=2X3+4X+9.  I am going to spilt the formula up into smaller fragments to make it easier to apply the formula so Y=(2X3)+(4X)+9.  The numbers before the X (Y=(2X3)+(4X)+9) are the A’s and the number after the X (Y=(2X3)+(4X)+9) is N, this means that the formula should be Gradient= (ANXN-1)+(AX).  The 9 at the end of the equation will make no difference to the gradient at all, it just tells us were the line cuts the Y axis.

Y=2X3+4X+9

X

Predicted gradient with formula From set

X Co-ordinates

Y Co-ordinates

Increment formula

Increment gradient

1

(2*3*12)+(4)=10

1,1.001

15,15.010006

0.010006/0.001

10.006

2

(2*3*22)+(4)=28

2,2.001

33,33.028012

0.028012/0.001

28.012

3

(2*3*32)+(4)=58

3,3.001

75,75.058018

0.58018/0.001

58.018

4

(2*3*42)+(4)=100

4,4.001

153,153.100024

0.100024/0.001

100.024

The formula still seems to be working for different curves and I feel that it always work for all graph that includeY=AXN

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

or


Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Gradient Function essays

  1. Peer reviewed

    The Gradient Function Coursework

    5 star(s)

    Integral calculus is concerned with area and volume. The basic idea of integral calculus is this: the simplest shape whose area we can compute is the rectangle. The area is the length of the rectangle multiplied by its width. For instance, a "square mile" is a piece of land with

  2. Curves and Gradients Investigation

    It should become obvious that they are getting nearer and nearer to a certain number and this number will be a reasonably accurate estimate of the gradient at the point in question. This method is illustrated in Graphs B2, C2 and D2.

  1. The Gradient Function

    Firstly, as I am trying to find out the gradient, when the tangent is at x=-1, I am going to 'square' 1. (-1)2 = 1 As I am also trying to work in the graph of y=x3, I am now going to multiply my answer, by 3.

  2. Maths Coursework - The Open Box Problem

    I constructed 3 graphs for each ratio. The ratios are 2: 1, 3: 1, 4: 1, and 5: 1. 2:1 These are the 3 cut out sizes I used for the 2:1 ratio rectangle. * 40:20 * 60:30 * 80:40 After deciding on the sizes I put them into an equation.

  1. The Gradient Function

    = -1, -1 (X2, Y2) = -0.999, (1/X2) (X2, Y2) = (-0.999, -1.001) Gradient = Y2-Y1/X2-X1 = -1.001-(-1)/-0.999-(-1) = -1.001 Graph: Y = 3/x The table below shows the increment method performed at the point (2, 1.5) Y =3/x X1 Y1 X2 Y2 Gradient 2 1.5 2.005 1.496259 -0.74812968 2

  2. The Gradient Function

    will test this for test this by predicting results using it and then checking them using a computer program. PREDICTIONS for y=x-2 X=1 2 x 1 + 3 x 12 = 5 X=2 2 x 2 + 3 x 22 = 16 X=3 2 x 3 + 3 x 32

  1. The Gradient Function

    81 81 4 192 144 I can now calculate the gradient function of the curves to see if a pattern occurs. Equation of line y= Gradient Function X2 2x 2x2 4x 3x2 6x X3 3x2 2x3 6x2 3x3 9x2 There are a few patterns in the above table: 1)

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

    In order to avoid this, I am going to use another method to calculate the gradients: increment method. Increment method works like this: you need 4 numbers: x1, x2, y1, y2. If you want to get the gradient of the co-ordination (1, 1), namely (x1, y1)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work