如图,已知抛物线y=ax2+bx+c经过点A（1,1）B（6,1）C（0,-2）,与x轴交于E,F两点,点P（m,n）在

(1)
A:a + b + c = 1
B:36a + 6b + c 1
C:c = -2
a = -1/2,b = 7/2
抛物线的解析式:y = -x²/2 + 7x/2 -2
(2) BC的解析式:(y+2)/(x-0) = (1 + 2)/(6-0)
x - 2y -4 = 0,y = x/2 -2
取 y = 0,x = 4
D(4,0)
P(m,-m²/2 + 7m/2 -2),Q(m,m/2 -2)
PQ = -m²/2 + 7m/2 -2 - (m/2 -2) = -m²/2 + 3m
= -(m - 3)²/2 + 9/2
m = 3时,PQ最长,为9/2
P(3,4)
(3) S = S∆OPC + S∆CDP
S∆OPC = (1/2)OC*P的横坐标 = (1/2)*2*m = m
CD = √[(4 - 0)² + (0 + 2)²] = 2√5
P(m,-m²/2 + 7m/2 -2)与CD (x - 2y -4 = 0)的距离h = |m -2(-m²/2 + 7m/2 -2) -4|/√(1² + 2²)
= |m² - 6m|/√5
= (6m - m²)/√5 (易证,0

楼上的最后一问有漏洞。当0

1年前

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