- #1
Ocata
- 198
- 5
Suppose you have a given area under a curve, say 250, and want to come up with a function that produces this value. How would you do this?
Although I came up with two basic functions as follows:First: Let y (x) = 5 from 0<x<50 , thus length*width = yx = 5*50 = 250.
Second:
Area of a triangle: 1/2 (base)*(height) = Area
[1/2 (50) (height) = 250] => [25y = 250] => [y = 10]
[y = mx +b] = y(x) = (1/5)x from 0<x<50
I have not been able to figure out how to come up with a "curvy" function like an upside down parabola with roots 0 and 50 on the x-axis with area under the curve being a given value like 250. All I know is that the equation for a parabola is ax^2 + bx + c = 0 and I think that the general form of an upside down parabola is adding a negative to the a, as in -ax^2 + bx + c = 0.
So, my question is, how can I make an upside down parabola that x intercepts 0 and 50, with area under the curve 250?
Thank you
Although I came up with two basic functions as follows:First: Let y (x) = 5 from 0<x<50 , thus length*width = yx = 5*50 = 250.
Area of a triangle: 1/2 (base)*(height) = Area
[1/2 (50) (height) = 250] => [25y = 250] => [y = 10]
[y = mx +b] = y(x) = (1/5)x from 0<x<50
So, my question is, how can I make an upside down parabola that x intercepts 0 and 50, with area under the curve 250?
Thank you