Determine a mathematical model/equation for the graph below, and describe a method to validate results saying the degree of accuracy of...

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changchengliang | Elementary School Teacher | (Level 2) Adjunct Educator

Posted on

My equation for the best fit curve for H is: H=-5(t-5)^2+125

(Check out the link provided below at, how my equations had evolved from a simple x^2 curve to the orange curve.)


My thinking process:

The graph resembles an inverted quadratic curve of y=x^2.  Hence the maximum to the graph you have given would correspond to the minimum of the standard x^2 curve.


1. A perfect x^2 curve (see blue curve) would have the minimum at x=0.

2. To have the minimum at x=5, substitute "x-5" into "x" of the curve to get y=(x-5)^2.  (see red curve)

3. However, notice that with this, the y-intercept is at 25.  We want it to be 126.1 so that when we "invert" it later, the maximum will take this value. So we multiply the graph with a factor of 126.1/25 which is about 5.044 or just 5, for simplicity to get y=5(x-5)^2.  (see green curve)

4. Finally, we need to invert the curve and add 125 to get y=-5(x-5)^2+125   or   simply  y= 125 - 5(x-5)^2  .  (see orange curve)


Hope this clears the air...