請問F(x) = 3x^2 -12x + 5點計啊..
請以a(x-h)^2 + k 作答!!
THX +100000000
F.4 MATHS!
2006-10-18 8:53 am
回答 (2)
2006-10-18 5:06 pm
✔ 最佳答案
f(x) = 3x^2 - 12x + 5Let the eq. 3x^2 - 12x + 5 = 0
3x^2 - 12x + 5 = 0
3(x^2 - 4x + 5/3) = 0
3[x^2 - 4x + (4/2)^2 - (4/2)^2 + 5/3] = 0
3[(x-2)^2 - 4 + 5/3] = 0
3[(x-2)^2 - 12/3 + 5/3] = 0
3[((x-2)^2 - 7/3] = 0
3(x-2)^2 - 7 = 0
.'. a = 3 , h = 2 , k = -7
參考: ME
2006-10-18 8:58 am
Solution:
F(x) = 3x^2 -12x + 5
F(x) = 3 (x^2 – 4x) + 5
F(x) = 3 (x^2 – 4x + 4 – 4) + 5
F(x) = 3 (x^2 – 4x + 4) – 12 + 5
F(x) = 3 (x-2)^2 – 7
∴ a = 3, h = 2 and k = -7
2006-10-18 00:59:48 補充：
This method is called completing square method and always be used to find the extreme values of a quadratic function
F(x) = 3x^2 -12x + 5
F(x) = 3 (x^2 – 4x) + 5
F(x) = 3 (x^2 – 4x + 4 – 4) + 5
F(x) = 3 (x^2 – 4x + 4) – 12 + 5
F(x) = 3 (x-2)^2 – 7
∴ a = 3, h = 2 and k = -7
2006-10-18 00:59:48 補充：
This method is called completing square method and always be used to find the extreme values of a quadratic function
參考: myself
收錄日期: 2021-04-12 20:48:34
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https://hk.answers.yahoo.com/question/index?qid=20061018000051KK00190