8)-x^2+8x-17≡a(x+b)^2 +c
(a)求a,b.+c的值。
由此．寫下-x^2+8x-17的極大值。
(b)求-3/(-x^2 +8x-17)的範圍。
A-maths!
2007-06-10 8:45 am
回答 (1)
2007-06-10 8:56 am
✔ 最佳答案
(a) -x^2 + 8x - 17= -(x^2 - 8x + 16) - 1
= -(x - 4)^2 - 1
Hence a=-1, b=-4, c=-1
It attains its maximum = -1 when x=4.
(b) -x^2 + 8x - 17 <= -1
Thererfore -3 / (-x^2 + 8x - 17) >= 3
2007-06-10 04:28:10 補充：
Sorry, (b) part should be:-x^2 + 8x - 17 <= -11 > = -1 / (-x^2 + 8x - 17)3 >= -3 / (-x^2 + 8x - 17) > 0
收錄日期: 2021-04-23 20:34:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070610000051KK00232