Power: 2
Coefficient: 1
Fixed point: -3
Observation
For the equation Y=X2, the coefficient is 1 the power is 2.
I have used 3 different fixed points which are
(2, 4), (5, 25) and (-3, 9). From my results I can see that the closer I get to the fixed points, the gradient gets closer to double the value of X co-ordinate. Using the information I can conclude that the gradient on any point on that curve will be two times the X co-ordinate. To find the gradient on this curve the equation is m=x 2.
The gradient is calculated as
change in y
change in x
This change I can write using differentiation
dy
dx
So, when Y=2X2
dy
dx
Y=3x2
My first fixed point is 3, 27
Power: 2
Coefficient: 3
Fixed point: 3
My Second fixed point 6,108
Power: 2
Coefficient: 3
Fixed point: 6
My third fixed point is -5, 75
Power: 2
Coefficient: 3
Fixed point: -5
Observation
For the equation Y=3X2, the coefficient is 3 the power is 2.
I have used 3 different fixed points which are
(3, 27), (6, 108) and (-5, 75). From my results I can see that the closer I get to the fixed points, the gradient gets closer to 3 times the squared value of X. Using the information I can conclude that the gradient on any point on that curve will be three times the squared value of X co-ordinate. Therefore, the gradient on this curve is
m=3x2X
Using differentiation for Y=3X2
dy
dx
=3x2X
The graph for this equation and the graph for Y=X2 looks similar, because I have only changed the coefficient.
Y=X3
My first fixed point is 3, 27
Power: 3
Coefficient: 1*3
Fixed point: 3
My Second fixed point is 5, 125
Power: 3
Coefficient: 1*5
Fixed point: 5
My third fixed point -3, -27
Power: 3
Coefficient: 1*-3
Fixed point: -3
Observation
For the equation Y=X3, the coefficient is 1 the power is 3.
I have used 3 different fixed points which are
(3, 27), (5, 125) and (-3, -27). From my results I can see that the closer I get to the fixed points, the gradient gets closer to 3 times the squared value of X co-ordinate. Using this information I can conclude that the gradient on any point on that curve will be the cubed value of X. Therefore the gradient on this is
m=3X 2
Using differentiation for Y=X3, I can write this gradient as
dy
dx
Y=3X2 +5x
My first fixed point is 3, 32
My second fixed point 4, 53
My third fixed point 5, 80
Observation
I have used 3 different fixed points which is (3, 32), (4, 58) and (5, 80) from my results I can see that the closer I get to the fixed points, the gradient gets closer to
Conclusion
For the equation y=x2, the gradient is m=2x.
For the equation y=x3, the gradient is m=3x.
Therefore, for a general equation of type y=xn the gradient will be m=xn
In the above equation the coefficient is 1. Changing the coefficient to ‘a’ the general equation for gradient will be m=axn
