Re: [代數] 二次函數
※ 引述《jundarn (小眼)》之銘言：
: 設f(x)=ax^2+bx+c (a不為0)
: 若g(x)=-f(100-x) 且g(x)之圖形包含f(x)之頂點
: 其中f(x)及g(x)與x軸交點 x1<x2<x3<x4 且x3-x2=150
: 求x4-x1=?
: 同事問的~聽說是競賽試題 嘗試從根與係數下手但很卡 求解~~謝謝
x_1, x_3 一組
x_2, x_4 一組
f(x) = a(x-x_1)(x-x_3)
頂點x_0 = (x_1 + x_3)/2
g(x) = -a(100-x-x_1)(100-x-x_3)
= -a[x - (100 - x_1)][x - (100 - x_3)]
2頂點x_0' = 100 - (x_1 + x_2)/2
二拋物線對 (50,0) 點對稱
因為x_3 - x_2 = 150
x_3 + x_2 = 100
=> x_3 = 125
x_2 = -25
g(x) = -a[x - (100 - x_1)][x + 25]
= -a[x - x_4][x + 25]
f(x) = a[x - x_1][x - 125]
x_4 + x_1 = 100
x_0 = (x_1 + 125)/2
f(x_0) = -g(100 - x_0)
a[125 - x_1]/2 * [x_1 - 125]/2 = a[75 - 3x_1]/2 * [175 + x_1]/2
=> [x_1 - 125]^2 = 3[x_1 - 25][x_1 + 175]
= 3[(x_1 - 125) + 100][(x_1 - 125) + 300]
= 3[x_1 - 125]^2 + 1200[x_1 - 125] + 90000
=> [x_1 - 125] = -300 +- 150√2
=> x_1 = -175 - 150√2 取負因為要< x_2 = -25
因為x_4 + x_1 = 100
=> x_4 = 275 + 150√2
x_4 - x_1 = 450 + 300√2
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