推 liang6159 : 謝謝大大分享 12/12 20:08
※ 引述《liang6159 (liang6159)》之銘言：
: 三次函數，對稱中心(2,5)，f(1)=-1,f(3)=11，試求f(x)=?
y - 5 = a(x - 2)^3 + b(x - 2)^2 + c(x - 2)，a =/= 0
-6 = -a + b - c
6 = a + b + c
=> b = 0， c = 6 - a
y - 5 = a(x - 2)^3 + c(x - 2) = a(x - 2)[(x - 2)^2 + c/a]
= a(x - 2)[(x - 2)^2 + (6/a) - 1]
= a(x - 2)^3 + 6(x - 2) - a(x - 2)
= a(x - 2)(x - 1)(x - 3) + 6x - 12
=> y = a(x - 1)(x - 2)(x - 3) + (6x - 7)，a =/= 0
另解：
y - 5 = a(x - 2)[(x - 2)^2 - k^2]
=> -6 = -a[1 - k^2] => k^2 = 1 - (6/a)
=> y - 5 = a(x - 2)[(x - 2)^2 + (6/a) - 1]
= a(x - 2)^3 + 6(x - 2) - a(x - 2)
=> y = a(x - 1)(x - 2)(x - 3) + (6x - 7)，a =/= 0
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