設二次涵數為 y = -2 ( x - k ) ^2 + [ ( k^2 ) - k + 2 ]，其中 k 為一正常數。
(1) 若 y 的極大值為 4 ，求 k 的值
(2) 求當 x = 0 時 y 的值
(3) 由此，繪畫出 y = -2 ( x - k ) ^2 + [ ( k^2 ) - k + 2 ] 的略圖 <-- 可不答，但答會更好
中四容易數學 - 二次涵數
2007-10-16 6:42 am
回答 (2)
2007-10-16 7:18 am
✔ 最佳答案
(1) 極大值為 4k2 - k + 2 = 4
k2 - k - 2 = 0
(k - 2)(k + 1) = 0
k = 2 或 k = -1 (捨棄)
(2) 將 x = 0 代入函數，
y = -2(-k)2 + (k2 - k + 2)
= -2(2)2 + (22 - 2 + 2)
= -4
(3) 如下:
圖片參考：http://i207.photobucket.com/albums/bb173/kkkpopup/graph1.gif
2007-10-16 7:23 am
1/
y 的極大值 = [ ( k^2 ) - k + 2 ]
so, 若 y 的極大值為 4
[ ( k^2 ) - k + 2 ] = 4
k^2 - k - 2 = 0
(k - 2) (k + 1) = 0
k = 2 or k = -1 (rejected, since k 為一正常數)
2/
當 x = 0 and k = 2 (from (1)),
y = -2 ( 0 - 2 ) ^2 + [ ( 2^2 ) - 2 + 2 ]
= -2 * 4 + 4
= -4
3/
from (1), 二次涵數為 y = -2 ( x - 2 ) ^2 + [ ( 2^2 ) - 2 + 2 ]
= -2 ( x - 2 ) ^2 + 4
條 curve 是倒轉的 u 形咁上下樣, curve 的頂點是 (2,4)
y 的極大值 = [ ( k^2 ) - k + 2 ]
so, 若 y 的極大值為 4
[ ( k^2 ) - k + 2 ] = 4
k^2 - k - 2 = 0
(k - 2) (k + 1) = 0
k = 2 or k = -1 (rejected, since k 為一正常數)
2/
當 x = 0 and k = 2 (from (1)),
y = -2 ( 0 - 2 ) ^2 + [ ( 2^2 ) - 2 + 2 ]
= -2 * 4 + 4
= -4
3/
from (1), 二次涵數為 y = -2 ( x - 2 ) ^2 + [ ( 2^2 ) - 2 + 2 ]
= -2 ( x - 2 ) ^2 + 4
條 curve 是倒轉的 u 形咁上下樣, curve 的頂點是 (2,4)
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