maths~~~~~~~~~~~~~~~~~~~~~~~~
回答 (2)
2009-04-15 6:08 am
✔ 最佳答案
let y=(x+2)^2+c , sub (-1,-1) into equation ,
(-1)=(-1+2)^2+C
=>C=-2//
so the equation is y=(x+2)^2-2=x^2+4x+2//
2009-04-15 8:30 am
設該二次函數為 f(x) = a(x + 2)^2 - k。
f(-1) = a(-1 + 2)^2 - k = -1
a - k = -1 ___ (1)
f(x) = 0
a(x + 2)^2 - k = 0
x = -2 +/- √k/a
在x軸上截得的弦長 = 2√k/a = 2√2
k = 2a ___ (2)
解 (1) 及 (2) 得 a = 1, k = 2
該二次函數為 (x + 2)^2 - 2。
f(-1) = a(-1 + 2)^2 - k = -1
a - k = -1 ___ (1)
f(x) = 0
a(x + 2)^2 - k = 0
x = -2 +/- √k/a
在x軸上截得的弦長 = 2√k/a = 2√2
k = 2a ___ (2)
解 (1) 及 (2) 得 a = 1, k = 2
該二次函數為 (x + 2)^2 - 2。
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