I have been given the equation y = axn to investigate the gradient function for varied values of "a" and "n"
Firstly I will choose n = 1 for varied values of a
When n = 1 then the graph will be a straight line
The general equation of a straight line is y = ax
I am going to keep n constant and use varied values for "a".The gradient for a straight line remains the same, therefore I am going to calculate the gradient by taking the difference in the y co-ordinates divided by the difference in x co-ordinates.
In order to calculate the gradient for a straight line we will have to go through the process of calculating the gradient by taking the difference in y co-ordinates divided by the difference in the x co-ordinates.
I would then choose n = 2 for varying values of a where a = 1,2,3,4
The equations would be ; y = x2 , y = 2x2 , y = 3x2 , y = 4x2
These equations would get us curved graphs and the gradient function at each point in a
curve is different therefore I will draw tangents using a capillary tube and calculate the
gradient by taking the change in the value of the y co ordinates divided by the change in t
value of the x co ordinates.
I would then choose n = 3 for varying values of "a".
The equations would be ; y = x3, y = 2x3,y = 3x3.
Again these equations would get me curved graphs and I will establish a relationship
Between the curve and the gradient.
In order to calculate the gradient in a curve , we know we have to draw the tangent for a
given point x and then calculate the gradient.
The gradient can be calculated by the change in the y co-ordinates divided by the change in the x co-ordinates.
I will now draw graphs for the equations I stated above:
Using a capillary tube I am going to draw tangents.
In order to draw the tangents using a capillary tube we have to keep the capillary tube on the curve and keep it at different points to check if it is straight and if it is we draw the normal and perpendicular to the normal we draw tha tangent and calculate the gradient.
Graph 1)y=x
Graph 2)y=2x
Graph3)y=3x
Graph 4)y=4x
Graph 5)y=x2
Graph 6)y=2x2
Graph 7)y=3x2
Graph 8)y=4x2
Graph 9)y=x3
Graph 10)y=2x3
Graph 11)y=3x3
Graph 12)y=4x³
The tables for the gradients of the graphs are as shown below
For the equations when n = 1 the gradients were as follows:
a =
2
3
4
Curve
y=x
y=2x
y=3x
y=4x
Gradient function
2
3
4
I have found that the gradient for the graph ;
y=x is 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.
At (4,20)(5,25)
Gradient function = ?y / ?x
=25-20
5-4
= 5
the table for the equations when n = 2 are as shown below
A
2
3
Curve
y=x2
y=2x2
y=3x2
Tangent at x =
2
3
4
2
3
4
2
3
4
Gradient function
4.2
5.5
8
8.4
1.67
6
1.6
7.69
24.28
I would like to establish a relationship between the curve and the gradient,therefore I will be rounding the answers I have obtained from the graphs to the nearest whole number as it would be a tedious process to do the calculations with decimals.
Curve
y = x2
y = 2x2
y = 3x2
Tangent at x =
2
3
4
2
3
4
2
3
4
Gradient from curve =
4.2
5.5
8
8.4
1.67
6
1.6
7.69
24.28
?
4
6
8
8
2
6
2
8
24
I would like to establish a relationship between the curve and the gradient
When y = x2 and a = 1 ,the gradients are as follows:0
Tangent at x =
2
3
4
Gradient function
4
6
8
From the table I can see that the gradient function is twice the point at which we draw the tangents .
Gradient function = 2x
For the curve y = 2x2,the gradients are as follows,
Tangent at x =
2
3
4
Gradient function
8
2
6
As we can see from the table above the gradient function is 4 multiplied by the point at which we draw the tangent at.Thus I can say that the gradient function for the curve of y=2x² is 4x.
For the curve y=3x²
Tangent at x =
2
3
4
Gradient function
2
8
24
I can see that the gradient function is 6 multiplied by the point at which we draw the tangent.
Thus I can conclude from this table that the gradient function for the curve = 6x
After looking at the tables for the various values of "a" and how they are related to the gradient function I have come to the conclusion that the gradient function for the general equation y = ax² is 2ax
Testing:-
y = 7x2
Gradient function for the tangent x = 2 is 2(7*22) is equal to 56.
Firstly I will choose n = 1 for varied values of a
When n = 1 then the graph will be a straight line
The general equation of a straight line is y = ax
I am going to keep n constant and use varied values for "a".The gradient for a straight line remains the same, therefore I am going to calculate the gradient by taking the difference in the y co-ordinates divided by the difference in x co-ordinates.
In order to calculate the gradient for a straight line we will have to go through the process of calculating the gradient by taking the difference in y co-ordinates divided by the difference in the x co-ordinates.
I would then choose n = 2 for varying values of a where a = 1,2,3,4
The equations would be ; y = x2 , y = 2x2 , y = 3x2 , y = 4x2
These equations would get us curved graphs and the gradient function at each point in a
curve is different therefore I will draw tangents using a capillary tube and calculate the
gradient by taking the change in the value of the y co ordinates divided by the change in t
value of the x co ordinates.
I would then choose n = 3 for varying values of "a".
The equations would be ; y = x3, y = 2x3,y = 3x3.
Again these equations would get me curved graphs and I will establish a relationship
Between the curve and the gradient.
In order to calculate the gradient in a curve , we know we have to draw the tangent for a
given point x and then calculate the gradient.
The gradient can be calculated by the change in the y co-ordinates divided by the change in the x co-ordinates.
I will now draw graphs for the equations I stated above:
Using a capillary tube I am going to draw tangents.
In order to draw the tangents using a capillary tube we have to keep the capillary tube on the curve and keep it at different points to check if it is straight and if it is we draw the normal and perpendicular to the normal we draw tha tangent and calculate the gradient.
Graph 1)y=x
Graph 2)y=2x
Graph3)y=3x
Graph 4)y=4x
Graph 5)y=x2
Graph 6)y=2x2
Graph 7)y=3x2
Graph 8)y=4x2
Graph 9)y=x3
Graph 10)y=2x3
Graph 11)y=3x3
Graph 12)y=4x³
The tables for the gradients of the graphs are as shown below
For the equations when n = 1 the gradients were as follows:
a =
2
3
4
Curve
y=x
y=2x
y=3x
y=4x
Gradient function
2
3
4
I have found that the gradient for the graph ;
y=x is 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.
At (4,20)(5,25)
Gradient function = ?y / ?x
=25-20
5-4
= 5
the table for the equations when n = 2 are as shown below
A
2
3
Curve
y=x2
y=2x2
y=3x2
Tangent at x =
2
3
4
2
3
4
2
3
4
Gradient function
4.2
5.5
8
8.4
1.67
6
1.6
7.69
24.28
I would like to establish a relationship between the curve and the gradient,therefore I will be rounding the answers I have obtained from the graphs to the nearest whole number as it would be a tedious process to do the calculations with decimals.
Curve
y = x2
y = 2x2
y = 3x2
Tangent at x =
2
3
4
2
3
4
2
3
4
Gradient from curve =
4.2
5.5
8
8.4
1.67
6
1.6
7.69
24.28
?
4
6
8
8
2
6
2
8
24
I would like to establish a relationship between the curve and the gradient
When y = x2 and a = 1 ,the gradients are as follows:0
Tangent at x =
2
3
4
Gradient function
4
6
8
From the table I can see that the gradient function is twice the point at which we draw the tangents .
Gradient function = 2x
For the curve y = 2x2,the gradients are as follows,
Tangent at x =
2
3
4
Gradient function
8
2
6
As we can see from the table above the gradient function is 4 multiplied by the point at which we draw the tangent at.Thus I can say that the gradient function for the curve of y=2x² is 4x.
For the curve y=3x²
Tangent at x =
2
3
4
Gradient function
2
8
24
I can see that the gradient function is 6 multiplied by the point at which we draw the tangent.
Thus I can conclude from this table that the gradient function for the curve = 6x
After looking at the tables for the various values of "a" and how they are related to the gradient function I have come to the conclusion that the gradient function for the general equation y = ax² is 2ax
Testing:-
y = 7x2
Gradient function for the tangent x = 2 is 2(7*22) is equal to 56.
