Consider a quadratic curve C:y=-x+(h-9)x+(10h+10),where h is greater than or equal to 0
(a)Find the vertex of C
(b)If the point lies on C,express p in terms of h
(c)If the curve does not intersect with the horizontal line y=100,find th range of possible values of h
amatha
2008-02-21 6:12 am
回答 (1)
2008-02-21 8:04 am
✔ 最佳答案
(a) By completing the squarey = -x^2 + (h - 9)x + (10h + 10)
= - [ x^2 + (h - 9)x + (h - 9)/2 - (h - 9)/2 ] + (10h + 10)
= - [ x + (h-9)/2 ]^2 + (h - 9)/2 + (10h + 10)
= - [ x+ (h-9)/2 ]^2 + (21h + 11)/2
So, the vertex of C is ( -(h - 9)/2 , (21h + 11)/2 )
(b) unclear definition
(c) put y = 100
100 = -x^2 + (h - 9)x + (10h + 10)
x^2 - (h - 9)x + (90 - 10h) = 0
delta = (h - 9)^2 - 4 (90 - 10h) < 0
h^2 - 18h + 81 - 360 + 40h < 0
h^2 + 22h - 279 < 0
(h - 9)(h + 31) < 0
-31 < h < 9
since h >= 0,
the possible range of h is 0 <= h < 9
收錄日期: 2021-04-16 22:53:54
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