using completing square rewrite the equation y=ax^2+bx+c in the form
y=a(x-h)^2+k, where h and k are constants in terms of a,b and c
f.4 maths
2011-01-08 3:50 am
回答 (1)
2011-01-08 4:22 am
✔ 最佳答案
y=
圖片參考：http://www3.wolframalpha.com/Calculate/MSP/MSP139219di81fef08ahee000005465iaia9gdb01b9?MSPStoreType=image/gif&s=13&w=106&h=37
=
圖片參考：http://www3.wolframalpha.com/Calculate/MSP/MSP288719di7d87fah2i9b300000i65988e718ff9ab?MSPStoreType=image/gif&s=10&w=252&h=44
=
圖片參考：http://www3.wolframalpha.com/Calculate/MSP/MSP171819di7ici75dbdchc0000698dfci3fgch4h2b?MSPStoreType=image/gif&s=7&w=174&h=44
=
圖片參考：http://www3.wolframalpha.com/Calculate/MSP/MSP212319di7h077fcai26d00004fdea0g69i9ie3a6?MSPStoreType=image/gif&s=11&w=185&h=44
=
圖片參考：http://www3.wolframalpha.com/Calculate/MSP/MSP155719di7ici75a9c159000059f80hf888a4idbe?MSPStoreType=image/gif&s=7&w=139&h=40
2011-01-07 20:28:22 補充：
= a(x - -b/(2a))² + c - b²/(4a)
So
h = - b/(2a)
k = c - b²/(4a)
2011-01-07 23:37:20 補充：
y
= ax² + bx + c
= a(x² + bx/a + c/a)
= a(x² + 2bx/(2a) + b²/(2a)² + c/a - b²/(2a)²)
= a((x + b/(2a))² + c/a - b²/(2a)²)
= a(x + b/(2a))² + c - b²/(4a)
= a(x - -b/(2a))² + c - b²/(4a)
So
h = - b/(2a)
k = c - b²/(4a)
收錄日期: 2021-04-21 22:19:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110107000051KK00973