✔ 最佳答案
Your provided data do not exactly match quadratic equations. A relative easy way is to approximate the quadratic equations by plotting the data on a graph.
(1)
http://img215.imageshack.us/img215/2840/graph1g.png
圖片參考:
http://img215.imageshack.us/img215/2840/graph1g.png
The graph indicate a maximum of about 250.5 at x = 45. At x = 80, y is about 87
My estimation is y = a(x – 45)^2 + 250.5
Sub (80,87) into equation yields a = -0.133
Therefore y = -0.133(x – 45)^2 + 250.5
The result obtained by Excel quadratic curve fitting is
y = 0.1352(x – 44.96)^2 + 250.6
(2)
http://img245.imageshack.us/img245/7848/graph2c.png
圖片參考:
http://img245.imageshack.us/img245/7848/graph2c.png
The graph indicate a maximum of about 311 at x = 44.5. At x = 75, y is about 154
My estimation is y = a(x – 44.5)^2 + 311
Sub (75,154) into equation yields a = -0.169
Therefore y = -0.169(x – 44.5)^2 + 311
The result obtained by Excel quadratic curve fitting is
y = 0.1699(x – 44.67)^2 + 310.7
(3)
http://img187.imageshack.us/img187/3365/graph3.png
圖片參考:
http://img187.imageshack.us/img187/3365/graph3.png
The graph indicate a maximum of about 374 at x = 45. At x = 80, y is about 126
My estimation is y = a(x – 45)^2 + 374
Sub (80,126) into equation yields a = -0.202
Therefore y = -0.202(x – 45)^2 + 374
The result obtained by Excel quadratic curve fitting is
y = 0.2015(x – 44.94)^2 + 373.4