I am going to investigate the gradients of different curves and try to work out a pattern that I could use to find the gradient of any curve.
Maths Coursework
Introduction
I am going to investigate the gradients of different curves and try to work out a pattern that I could use to find the gradient of any curve. I will draw graphs of a selection of curves, some by hand, some using Autograph and some using Excel. I will use three methods to investigate the graphs. Firstly, I will draw tangents to the curves at 4 or 5 points and measure the gradients. Secondly, I will draw chords between x = 1 and 4 or 5 points and measure the gradients. Thirdly, I will use algebra to work out a formula for the gradient and see how this matches the first two methods.
At first I split up the coursework into 3 main families (for each family there are additional equations to investigate):
Part One: Curves involving x2
. y = x2
2. y = 2x2
3. y = 3x2
4. y = 4x2
5. y = x2 + 1
6. y = 7x2 + 6
Part Two: Curves involving x2 + x
. y = x2 + x
2. y = x2 + 2x
3. y = 7x2 + 4x + 5
Part Three: Curves involving x3 + x
. y = x3
2. y = 2x3
3. y = 4x3 + 2x - 5
Finally, I will summarise my results in a series of tables and work out an overall formula that I could use to predict the gradient of any curve.
PART ONE: CURVES CONTAINING X2
(1) y = x2
I am investigating the changes in gradient for the curve y = x2. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
6
y
0
4
9
6
25
36
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 1 to 4. The gradient of the tangent will equal the gradient of the curve at that point.
(1,1) Gradient = difference in y values = 1 = 2
difference in x values 0.5
(2,4) Gradient = difference in y values = 2 = 4
difference in x values 0.5
(3,9) Gradient = difference in y values = 2 = 6
difference in x values 0.5
(4,16) Gradient = difference in y values = 2 = 8
difference in x values 0.5
x-axis value
2
3
4
Gradient
2
4
6
8
I predict that the gradient for x = 5 will be 10.
Test: (5,25) Gradient = 5 = 10 I used the formula 2x.
0.5
My prediction was correct.
Chords for x2
I will investigate a series of chords on the y = x2 graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 25 - 1 = 24 = 6
5 - 1 4
Chord AC
Gradient = 16 - 1 = 15 = 5
4 - 1 3
Chord AD
Gradient = 9 - 1 = 8 = 4
3 - 1 2
Chord AE
Gradient = 4 - 1 = 3 = 3
2 - 1 1
Chord AF
Gradient = 2.25 - 1 = 1.25 = 2.5
1.5 - 1 0.5
Table of results can be seen below.
I predict that for the next value, the answer will be 7.
Test
Chord start point
Chord finish point
5
4
3
2
.5
6
Gradient of chord
6
5
4
3
2.5
7
Test: 36 - 1 = 35 = 7 This proves my prediction.
6 - 1 5
From this result and the table, I have noticed that as the finish point of the chord approaches its limit of x = 1, the gradient of the chords approach 2. This is the gradient of the tangent at (1,1).
Using this result and algebra, I can prove that the value of the gradient will be 2x.
y
P2
P1
x
Coordinates of P1 are (x, x2); coordinates of P2 are (x + d, (x + d)2)
Gradient of chord = difference in y coordinates = (x + d)2 - x2
difference in x coordinates x + d - x
= x2 + d2 + 2xd - x2 = d2 + 2xd = d + 2x
d d
= 2x+d
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = x2 , the gradient of any chord is given by 2x+d. If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 2x.
I conclude that the gradient of the curve y = x2 can be found using the formula:
Gradient = 2x
(2) y = 2x2
I am investigating the changes in gradient for the curve y = 2x2. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
6
y
0
2
8
8
32
50
72
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 1 to 4. The gradient of the tangent will equal the gradient of the curve at that point.
(1,2) Gradient = difference in y values = 4
difference in x values
(2,8) Gradient = difference in y values = 8
difference in x values
(3,18) Gradient = difference in y values = 12
difference in x values
(4,32) Gradient = difference in y values = 16
difference in x values
x-axis value
2
3
4
Gradient
4
8
2
6
I predict that the gradient for x = 5 will be 20.
Test: (5,50) Gradient = 20 I used the formula 4x.
My prediction was correct.
Chords for 2x2
I will investigate a series of chords on the y = 2x2 graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 12
Chord AC
Gradient = 10
Chord AD
Gradient = 8
Chord AE
Gradient = 6
Chord AF
Gradient = 5
Table of results can be seen below.
Chord start point
Chord finish point
5
4
3
2
.5
Gradient of chord
2
0
8
6
5
From this result and the table, I have noticed that as the finish point of the chord approaches its limit of x = 1, the gradient of the chords approach 4. This is the gradient of the tangent at (1,1).
Using this result and algebra, I can prove that the value of the gradient will be 4x.
y
P2
P1
x
Coordinates of P1 are (x, 2x2); coordinates of P2 are (x + d, 2(x + d)2)
Gradient of chord = difference in y coordinates = 2(x + d)2 - 2x2
difference in x coordinates x + d - x
= 2x2 + 2d2 + 4xd - 2x2 = 2d2 + 4xd = d + 4x
d d
= 4x + d
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = 2x2 , the gradient of any chord is given by 4x + d. If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 4x.
I conclude that the gradient of the curve y = 2x2 can be found using the formula:
Gradient = 4x
(3) y = 3x2
I am investigating the changes in gradient for the curve y = 3x2. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
6
y
0
3
2
27
48
75
08
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 1 to 4. The gradient of the tangent will equal the gradient of the curve at that point.
(1,3) Gradient = difference in y values = 4
difference in x values
(2,12) Gradient = difference in y values = 12
difference in x values
(3,27) Gradient = difference in y values = 18
difference in x values
(4,48) Gradient = difference in y values = 24
difference in x values
x-axis value
2
3
4
Gradient
4
2
8
24
I predict that the gradient for x = 5 will be 30.
Test: (5,75) Gradient = 30 I used the formula 6x.
My prediction was correct.
Chords for 3x2
I will investigate a series of chords on the y = 3x2 graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 6
Chord AC
Gradient = 5
Chord AD
Gradient = 4
Chord AE
Gradient = 3
Chord AF
Gradient = 2.5
Table of results can be seen below.
I predict that for x = ...
This is a preview of the whole essay
Test: (5,75) Gradient = 30 I used the formula 6x.
My prediction was correct.
Chords for 3x2
I will investigate a series of chords on the y = 3x2 graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 6
Chord AC
Gradient = 5
Chord AD
Gradient = 4
Chord AE
Gradient = 3
Chord AF
Gradient = 2.5
Table of results can be seen below.
I predict that for x = 6, the answer will be 21.
Test
Chord start point
Chord finish point
5
4
3
2
6
Gradient of chord
8
5
2
9
21
Test: (6,108) Gradient of chord = 21
This proves my prediction.
From this result and the table, I have noticed that as the finish point of the chord approaches its limit of x = 1, the gradient of the chords approach 6. This is the gradient of the tangent at (1,1).
Using this result and algebra, I can prove that the value of the gradient will be 6x.
y
P2
P1
x
Coordinates of P1 are (x, 3x2); coordinates of P2 are (x + d, 3(x + d)2)
Gradient of chord = difference in y coordinates = 3(x + d)2 - 3x2 difference in x coordinates x + d - x
= 3x2 + 3d2 + 6xd - 3x2 = 3d2 + 6xd = d + 6x
d d
= 6x + d
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = 3x2 , the gradient of any chord is given by 6x + d. If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 6x.
I conclude that the gradient of the curve y = 3x2 can be found using the formula:
Gradient = 6x
(4) y = 4x2
I am investigating the changes in gradient for the curve y = 4x2. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
6
y
0
4
6
36
64
00
44
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 1 to 4. The gradient of the tangent will equal the gradient of the curve at that point.
(1,4) Gradient = difference in y values = 4 = 8
difference in x values 0.5
(2,16) Gradient = difference in y values = 8 = 16
difference in x values 0.5
(3,36) Gradient = difference in y values = 12 = 24
difference in x values 0.5
(4,16) Gradient = difference in y values = 16 = 32
difference in x values 0.5
x-axis value
2
3
4
Gradient
8
6
24
6
I predict that the gradient for x = 5 will be 40.
Test: (5,100) Gradient = 20 = 40 I used the formula 2x.
0.5
My prediction was correct.
Chords for 4x2
I will investigate a series of chords on the y = 4x2 graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 100 - 4 = 96 = 24
5 - 1 4
Chord AC
Gradient = 64 - 4 = 60 = 20
4 - 1 3
Chord AD
Gradient = 36 - 4 = 32 = 16
3 - 1 2
Chord AE
Gradient = 16 - 4 = 12 = 12
2 - 1 1
Chord AF
Gradient = 9 - 4 = 5 = 10
1.5 - 1 0.5
Table of results can be seen below.
I predict that for the next value, the answer will be 7.
Test
Chord start point
Chord finish point
5
4
3
2
.5
6
Gradient of chord
24
20
6
2
0
28
Test: 144 - 4 = 140 = 28 This proves my prediction.
6 - 1 5
From this result and the table, I have noticed that as the finish point of the chord approaches its limit of x = 1, the gradient of the chords approach 8. This is the gradient of the tangent at (1,4).
Using this result and algebra, I can prove that the value of the gradient will be 8x.
y
P2
P1
x
Coordinates of P1 are (x, 4x2); coordinates of P2 are (x + d, 4(x + d)2)
Gradient of chord = difference in y coordinates = 4(x + d)2 - 4x2
difference in x coordinates x + d - x
= 4x2 + 4d2 + 8xd - 4x2 = 4d2 + 8xd = d + 8x
d d
= 8x + d
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = 4x2 , the gradient of any chord is given by 8x+d. If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 8x.
I conclude that the gradient of the curve y = 4x2 can be found using the formula:
Gradient = 8x
(5) y = x2 + 1
I am investigating the changes in gradient for the curve y = x2 + 1. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
6
y
0
2
5
0
7
26
37
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 1 to 4. The gradient of the tangent will equal the gradient of the curve at that point.
(1,2) Gradient = difference in y values = 2.25 = 2.25
difference in x values 1
(2,5) Gradient = difference in y values = 4 = 4
difference in x values 1
(3,10) Gradient = difference in y values = 8 = 8
difference in x values 1
(4,17) Gradient = difference in y values = 2 = 9
difference in x values 1
x-axis value
2
3
4
Gradient
2.25
4
8
9
I predict that the gradient for x = 5 will be 10.
Test: (5,26) Gradient = 12 = 12 I used the formula 2x.
1
My prediction was close. I noticed that the +1 made little difference to the gradient; it was the same as the y = x2 curve.
Chords for x2 + 1
I will investigate a series of chords on the y = x2 + 1 graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 26 - 2 = 24 = 6
5 - 1 4
Chord AC
Gradient = 17 - 2 = 15 = 5
4 - 1 3
Chord AD
Gradient = 10 - 2 = 8 = 4
3 - 1 2
Chord AE
Gradient = 5 - 2 = 3 = 3
2 - 1 1
Table of results can be seen below.
I predict that for the next value, the answer will be 7.
Test
Chord start point
Chord finish point
5
4
3
2
6
Gradient of chord
6
5
4
3
7
Test: 37 - 2 = 35 = 7 This proves my prediction.
6 - 1 5
From this result and the table, I have noticed that as the finish point of the chord approaches its limit of x = 1, the gradient of the chords approach 2. This is the gradient of the tangent at (1,2).
Using this result and algebra, I can prove that the value of the gradient will be 2x.
y
P2
P1
x
Coordinates of P1 are (x, x2 + 1); coordinates of P2 are (x + d, (x + d)2 + 1)
Gradient of chord = difference in y coordinates = (x + d)2 + 1 - (x2 + 1)
difference in x coordinates x + d - x
= x2 + d2 + 2xd - x2 = d2 + 2xd = d + 2x
d d
= 2x + d
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = x2 , the gradient of any chord is given by 2x + d. If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 2x.
I conclude that the gradient of the curve y = x2 + 1 can be found using the formula:
Gradient = 2x
This is the same as for the curve y = x2. The + 1 makes no difference to the gradient.
(6) y = 7x2 + 6
I am investigating the changes in gradient for the curve y = 7x2 + 6. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
6
y
6
3
34
69
18
81
258
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 1 to 4. The gradient of the tangent will equal the gradient of the curve at that point.
(1,13) Gradient = difference in y values = 14
difference in x values
(2,34) Gradient = difference in y values = 28
difference in x values
(3,69) Gradient = difference in y values = 42
difference in x values
(4,118) Gradient = difference in y values = 56
difference in x values
x-axis value
2
3
4
Gradient
4
28
42
56
I predict that the gradient for x = 5 will be 70.
Test: (5,181) Gradient = 70 I used the formula 14x.
My prediction was correct.
Chords for 7x2 + 6
I will investigate a series of chords on the y = 7x2 + 6 graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 35
Chord AC
Gradient = 42
Chord AD
Gradient = 28
Chord AE
Gradient = 21
Table of results can be seen below.
Chord start point
Chord finish point
4
3
2
.5
Gradient of chord
42
35
28
21
From this result and the table, I have noticed that as the finish point of the chord approaches its limit of x = 1, the gradient of the chords approach 14. This is the gradient of the tangent at (1,1).
Using this result and algebra, I can prove that the value of the gradient will be 14x.
y
P2
P1
x
Coordinates of P1 are (x, 7x2 + 6); coordinates of P2 are (x + d, 7(x + d)2 + 6)
Gradient of chord = difference in y coordinates = 7(x + d)2 + 6 - (7x2 + 6) difference in x coordinates x + d - x
= 7x2 + 7d2 + 14xd + 6 - 7x2 - 6 = 7d2 + 14xd
d d
= 14x + d
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = 7x2 + 6 , the gradient of any chord is given by 14x + d. If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 14x.
I conclude that the gradient of the curve y = 7x2 + 6 can be found using the formula:
Gradient = 14x
I noticed that this would be the same result as the curve y = 7x2.
PART TWO: CURVES CONTAINING X2 + X
. y = x2 + x
I am investigating the changes in gradient for the curve y = x2 + x. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
6
y
0
2
6
2
20
30
42
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 1 to 4. The gradient of the tangent will equal the gradient of the curve at that point.
(1,2) Gradient = difference in y values = 1.5 = 3
difference in x values 0.5
(2,6) Gradient = difference in y values = 2.5 = 5
difference in x values 0.5
(3,12) Gradient = difference in y values = 3.5 = 7
difference in x values 0.5
(4,20) Gradient = difference in y values = 4.5 = 9
difference in x values 0.5
Test
x-axis value
2
3
4
5
Gradient
3
5
7
9
1
I predict that the gradient for x = 5 will be 11. I used the formula 2x + 1 for the gradient. Gradient = 5 x 2 + 1 = 11.
Test: (5,30) Gradient = 5.5 = 11
0.5
My prediction was correct.
Chords for x2 + x
I will investigate a series of chords on the y = x2 + x graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 42 - 2 = 40 = 8
6 - 1 5
Chord AC
Gradient = 30 - 2 = 28 = 7
5 - 1 4
Chord AD
Gradient = 20 - 2 = 18 = 6
4 - 1 3
Chord AE
Gradient = 12 - 2 = 10 = 5
3 - 1 2
Chord AF
Gradient = 6 - 2 = 4 = 4
2 - 1 1
Table of results can be seen below.
Chord start point
Chord finish point
6
5
4
3
2
Gradient of chord
8
7
6
5
4
From this result and the table, I have noticed that as the finish point of the chord approached its limit of x = 1, the gradients of the chords approach 3. This is the gradient of the tangent at point (1,2).
Using this result and algebra, I can prove that the value of the gradient of the curve will be given by 2x + 1.
y
P2
P1
x
Coordinates of P1 are (x, x2 + x); coordinates of P2 are (x + d, (x + d)2 + (x + d))
Gradient of chord = diff in y coordinates = (x + d)2 + (x + d) - (x2 + x) diff in x coordinates x + d - x
= x2 + 2xd + d2 + x + d - x2 - x
d
= 2xd + d2 +d = 2x + 1 + d
d
When d = 0,
Gradient = 2x + 1
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = x2 + x , the gradient of any chord is given by 2x + 1 + d. If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 2x + 1.
I conclude that the gradient of the curve y = x2 + x can be found using the formula:
Gradient = 2x + 1
2. y = x2 + 2x
I am investigating the changes in gradient for the curve y = x2 + 2x. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
6
y
0
3
8
5
24
35
48
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 1 to 4. The gradient of the tangent will equal the gradient of the curve at that point.
(1,3) Gradient = difference in y values = 4
difference in x values
(2,8) Gradient = difference in y values = 6
difference in x values
(3,15) Gradient = difference in y values = 8
difference in x values
(4,24) Gradient = difference in y values = 10
difference in x values
Test
x-axis value
2
3
4
5
Gradient
4
6
8
0
2
I predict that the gradient for x = 5 will be 12. I used the formula 2x + 2 for the gradient. Gradient = 5 x 2 + 2 = 12.
Test: (5,35) Gradient = 12
My prediction was correct.
Chords for x2 + 2x
I will investigate a series of chords on the y = x2 + 2x graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 8
Chord AC
Gradient = 7
Chord AD
Gradient = 6
Chord AE
Gradient = 5
Table of results can be seen below.
Chord start point
Chord finish point
5
4
3
2
Gradient of chord
8
7
6
5
From this result and the table, I have noticed that as the finish point of the chord approached its limit of x = 1, the gradients of the chords approach 4. This is the gradient of the tangent at point (1,3).
Using this result and algebra, I can prove that the value of the gradient of the curve will be given by 2x + 2.
y
P2
P1
x
Coordinates of P1 are (x, x2 + 2x); coordinates of P2 are (x + d, (x + d)2 + 2(x + d))
Gradient of chord = diff in y coordinates = (x + d)2 + 2(x + d) - (x2 + 2x) diff in x coordinates x + d - x
= x2 + 2xd + d2 + 2x + 2d - x2 - 2x
d
= 2xd + d2 +2d = 2x + 2 + d
d
When d = 0, Gradient = 2x + 2
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = x2 + 2x , the gradient of any chord is given by 2x + 2 + d. If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 2x + 2.
I conclude that the gradient of the curve y = x2 + 2x can be found using the formula:
Gradient = 2x + 2
3. y = 7x2 + 4x + 5
I am investigating the changes in gradient for the curve y = 7x2 + 4x + 5. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
6
y
5
6
41
80
33
200
281
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 1 to 4. The gradient of the tangent will equal the gradient of the curve at that point.
(1,16) Gradient = difference in y values = 9 = 18
difference in x values 0.5
(2,41) Gradient = difference in y values = 16 = 32
difference in x values 0.5
(3,80) Gradient = difference in y values = 23 = 46
difference in x values 0.5
(4,133) Gradient = difference in y values = 30 = 60
difference in x values 0.5
Test
x-axis value
2
3
4
5
Gradient
8
32
46
60
80
I predict that the gradient for x = 5 will be 74. I used the formula 14x + 4 for the gradient. Gradient = 14 x 5 + 4 = 74.
Test: (5,200) Gradient = 80
My prediction was very close.
Chords for 7x2 + 4x + 5
I will investigate a series of chords on the y = 7x2 + 4x + 5 graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 200 - 16 = 184 = 46
5 - 1 4
Chord AC
Gradient = 133 - 16 = 117 = 39
4 - 1 3
Chord AD
Gradient = 80 - 16 = 64 = 32
3 - 1 2
Chord AE
Gradient = 41 - 16 = 25 = 25
2 - 1 1
Table of results can be seen below.
Chord start point
Chord finish point
5
4
3
2
Gradient of chord
46
39
32
25
From this result and the table, I have noticed that as the finish point of the chord approached its limit of x = 1, the gradients of the chords approach 18. This is the gradient of the tangent at point (1,3).
Using this result and algebra, I can prove that the value of the gradient of the curve will be given by 14x + 4.
y
P2
P1
x
Coordinates of P1 are (x, 7x2 + 4x + 5); coordinates of P2 are (x + d, 7(x + d)2 + 4(x + d) + 5)
Gradient of chord = diff in y coordinates = 7(x + d)2 + 4(x + d) + 5 - (7x2 + 4x + 5) diff in x coordinates x + d - x
= 7x2 + 14xd + 7d2 + 4x + 4d + 5 - 7x2 - 4x - 5
d
= 14xd + 7d2 +4d = 14x + 4 + 7d
d
When d = 0, Gradient = 14x + 4
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = 7x2 + 4x + 5 , the gradient of any chord is given by 14x + 4 + 7d. If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 14x + 4.
I conclude that the gradient of the curve y = 7x2 + 4x + 5 can be found using the formula:
Gradient = 14x + 4
PART THREE: CURVES CONTAINING X3
. y = x3
I am investigating the changes in gradient for the curve y = x3. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
y
0
8
27
65
25
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 1 to 4. The gradient of the tangent will equal the gradient of the curve at that point.
(1,1) Gradient = difference in y values = 1.5 = 3
difference in x values 0.5
(2,8) Gradient = difference in y values = 6 = 12
difference in x values 0.5
(3,27) Gradient = difference in y values = 13.5 = 27
difference in x values 0.5
(4,64) Gradient = difference in y values = 24 = 48
difference in x values 0.5
Test
x-axis value
2
3
4
5
Gradient
3
2
27
48
75
I predict that the gradient for x = 5 will be 75. I used the formula 3x2 for the gradient. Gradient = 5 x 2 + 1 = 11.
Test: (5,30) Gradient = 37.5 = 75
0.5
My prediction was correct.
Chords for x3
I will investigate a series of chords on the y = x3 graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 31 = 31
1
Chord AC
Gradient = 64 - 1 = 63 = 21
4 - 1 3
Chord AD
Gradient = 27 - 1 = 26 = 13
3 - 1 2
Chord AE
Gradient = 8 - 1 = 7 = 7
2 - 1 1
Chord AF
Gradient = 3.375 - 1 = 2.375 = 4.75
1.5 - 1 0.5
Table of results can be seen below.
Chord start point
Chord finish point
5
4
3
2
.5
Gradient of chord
31
21
3
7
4.75
From this result and the table, I have noticed that as the finish point of the chord approached its limit of x = 1, the gradients of the chords approach 2. This is the gradient of the tangent at point (1,1).
Using this result and algebra, I can prove that the value of the gradient of the curve will be given by 3x2.
y
P2
P1
x
Coordinates of P1 are (x, x3); coordinates of P2 are (x + d, (x + d)3)
Gradient of chord = difference in y coordinates = (x + d)3 - x3
difference in x coordinates x + d - x
= x3 + 3x2d + 3xd2 + d3 - x3
d
= 3x2 + 3xd + d2
When d = 0, Gradient = 3x2
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = x3 , the gradient of any chord is given by 3x2 + 3xd + d2. If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 3x2.
I conclude that the gradient of the curve y = x3 can be found using the formula:
Gradient = 3x2
2. y = 2x3
I am investigating the changes in gradient for the curve y = 2x3. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
y
0
2
6
54
28
250
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 1 to 4. The gradient of the tangent will equal the gradient of curve at that point.
(1,2) Gradient = difference in y values = 5 - 2 = 3 = 6
difference in x values 0.5 0.5
(2,16) Gradient = difference in y values = 16 - 4 = 12 = 24
difference in x values 0.5 0.5
(3,54) Gradient = difference in y values = 78 - 52 = 26 = 52
difference in x values 0.5 0.5
(4,130) Gradient = difference in y values = 175 - 135 = 40 = 80
difference in x values 0.5 0.5
Test
x-axis value
2
3
4
5
Gradient
6
24
52
80
40
I predict that the gradient for x = 5 will be 150. I used the formula 6x2 for the gradient.
(5,250) Gradient = difference in y values = 320 - 250 = 70 = 140
difference in x values 0.5 0.5
Test: (5,250) Gradient = 140
My prediction was very close.
Chords for 2x3
I will investigate a series of chords on the y = 2x3 graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 250 - 2 = 248 = 62
5 - 1 4
Chord AC
Gradient = 128 - 2 = 126 = 42
4 - 1 3
Chord AD
Gradient = 54 - 2 = 52 = 26
3 - 1 2
Chord AE
Gradient = 16 - 2 = 14 = 14
2 - 1 1
Table of results can be seen below.
Chord start point
Chord finish point
5
4
3
2
Gradient of chord
62
42
26
4
From this result and the table, I have noticed that as the finish point of the chord approached its limit of x = 1, the gradients of the chords approach 6. This is the gradient of the tangent at point (1,2).
Using this result and algebra, I can prove that the value of the gradient of the curve will be given by 6x2.
y
P2
P1
x
Coordinates of P1 are (x, 2x3); coordinates of P2 are (x + d, 2(x + d)3)
Gradient of chord = difference in y coordinates = 2(x + d)3 - 2x3 difference in x coordinates x + d - x
= 2x3 + 6x2d + 6xd2 + 2d3 - 2x3
d
= 6x2 + 6xd + 2d2
When d = 0,
Gradient = 6x2
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = 2x3 , the gradient of any chord is given by 6x2 + 6xd + 2d2. If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 6x2.
I conclude that the gradient of the curve y = 2x3 can be found using the formula:
Gradient = 6x2
3. y = 4x3 + 2x - 5
I am investigating the changes in gradient for the curve y = 4x3 + 2x - 5. To plot the curve, I will use the table of values given below.
x
0
2
3
4
5
y
-5
31
09
259
505
I will be working out the gradients of tangents to the curve by using the equation:
Gradient = difference in y values
difference in x values
To help in this, I will draw the tangents to the curve at x values 2 to 5. The gradient of the tangent will equal gradient of the curve at that point.
(2,31) Gradient = difference in y values = 30 = 60
difference in x values 0.5
(3,109) Gradient = difference in y values = 60 = 120
difference in x values 0.5
(4,259) Gradient = difference in y values = 100 = 200
difference in x values 0.5
(5,505) Gradient = difference in y values = 160 = 320
difference in x values 0.5
Test
x-axis value
2
3
4
5
Gradient
60
20
200
320
4
2x2
60
08
92
300
2
Predictions were difficult but from the graphs y = x3 and y = 2x3, I predict that the formula for the gradient will involve 12x2. So for x = 1, the gradient will be 12. I drew the graph again over a smaller scale so I could measure the gradient at x = 1 accurately.
Test: (1,1) Gradient = 1.4 = 14
0.1
My prediction was very close.
Chords for 4x3 + 2x - 5
I will investigate a series of chords on the y = 4x3 + 2x - 5 graph.
Gradient of chord = difference in y coordinates
difference in x coordinates
Chord AB
Gradient = 505 - 1 = 504 = 126
5 - 1 4
Chord AC
Gradient = 259 - 1 = 258 = 86
4 - 1 3
Chord AD
Gradient = 109 - 1 = 108 = 54
3 - 1 2
Chord AE
Gradient = 31 - 1 = 30 = 30
2 - 1 1
Chord AF
Gradient = 11.5 - 1 = 10.5 = 21
1.5 - 1 0.5
Table of results can be seen below.
Chord start point
Chord finish point
505
259
09
31
1.5
Gradient of chord
26
86
54
30
21
From this result and the table, I have noticed that as the finish point of the chord approached its limit of x = 1, the gradients of the chords approach 14. This is the gradient of the tangent at point (1,1).
Using this result and algebra, I can prove that the value of the gradient of the curve will be given by 12x2 + 2.
y
P2
P1
x
Coordinates of P1 are (x, 4x3 + 2x - 5); coordinates of P2 are (x + d, 4(x + d)3 + 2(x + d) - 5)
Gradient of chord = diff in y coordinates
diff in x coordinates
= [4(x + d)3 + 2(x + d) - 5] - (4x3 + 2x - 5) x + d - x
= 4x3 + 12x2d + 12xd2 + 4d3 + 2x + 2d - 5 - 4x3 - 2x + 5
d
= 12x2 + 12xd + 4d2 + 2
When d = 0, Gradient = 12x2 + 2
When a chord starts and finishes at the same point, its gradient will equal the gradient of a tangent drawn at the same point. From algebra I proved that for the graph y = 4x3 + 2x - 5 , the gradient of any chord is given by 12x2 + 12xd + 4d2 + 2 . If the chord starts and finishes at the same point, then d = 0 and the gradient of the tangent is equal to 12x2 + 2.
I conclude that the gradient of the curve y = 4x3 + 2x - 5 can be found using the formula:
Gradient = 12x2 + 2
CONCLUSION
I have concluded some results for my gradients in the table below.
Part One: x2 graphs
Equation
Gradient
Value of a
2ax
y = x2
2x
2x
y = 2x2
4x
2
4x
y = 3x2
6x
3
6x
y = 4x2
8x
4
8x
y = x2 + 1
2x
2x
Test:
y = 7x2 + 6
4x
7
4x
If the number in front of x is called 'a', then I have found that the gradient of this type of curve is always given by 2ax.
Evidence to support this comes from looking at the result for y = 7x2 +6.
Part Two: x2 + x graphs
Equation
Gradient
Value of a
Value of b
2ax + b
y = x2 + x
2x + 1
2x + 1
y = x2 + 2x
2x + 2
2
2x + 2
Test:
y = 7x2 + 4x + 5
4x + 4
7
4
4x + 4
If the number in front of x2 is called 'a' and the number in front of x is called 'b', then I have found that the gradient of this type of curve is always given by 2ax + b.
Evidence to support this comes from looking at the result for y = 7x2 +4x + 5.
Part Three: x3 and x3 + x graphs
Equation
Gradient
Value of a
Value of b
3ax2 + b
y = x3
3x2
0
3x2
y = 2x3
6x2
2
0
6x2
Test:
y = 4x3 + 2x - 5
2x2 + 2
4
2
2x2 + 2
If the number in front of x3 is called 'a' and the number in front of x is called 'b', then I have found that the gradient of this type of curve is always given by 3ax2 + b.
Evidence to support this comes from looking at the result for y = 4x3 + 2x - 5.
I can summarise these results in a single table.
Part of equation
Part of gradient formula
ax3
3ax2
bx2
2bx
cx
c
d
Not present
Prediction
x4
4x3
This means that I can find the gradient of any curve involving x3, x2, x and a number at any value of x.
Sample equation
Formula for gradient
Gradient at x = 1
y = x3 - 2x2 + 4x - 1
3x2 - 4x + 4
3 - 4 + 4
= 3
y = 2x3 - x2 + 5x -5
6x2 - 2x + 5
6 - 2 + 5
= 9
y = 3x3 + 2x2 - 2x + 3
9x2 + 4x - 2
9 + 4 - 2
= 11
y = 4x3 - 2x2 + 6x - 1
2x2 - 4x + 6
2 - 4 + 6
= 14
y = 5x3 + x2 + 2x + 1
5x2 + 2x + 2
5 + 2 + 2
= 19
SUMMARY
The formula for gradient involves taking the power of x and multiplying it by x to one less power.
y = Dxb gradient = bDxb - 1
Aidan Casey 11CS 1
