如图，已知抛物线y=x2＋bx＋c与x轴交于A、B两点（A点在B点左侧）， 如图，已知抛物线y=x2＋bx＋c与x轴交于A、B两点（A点...
\u5982\u56fe\uff0c\u5df2\u77e5\u629b\u7269\u7ebfy=x2\uff0bbx\uff0bc\u4e0ex\u8f74\u4ea4\u4e8eA\u3001B\u4e24\u70b9\uff08A\u70b9\u5728B\u70b9\u5de6\u4fa7\uff09\uff0c\u4e0ey\u8f74\u4ea4\u4e8e\u70b9C\uff080\uff0c\uff0d3\uff09\uff0c\u5bf9(1) (x+b/2)²+c-b²/4 \u4e0ey\u8f74\u4ea4\u4e8e\u70b9C\uff080\uff0c\uff0d3\uff09\u5219:c= -3
\u5bf9\u79f0\u8f74\u662f\u76f4\u7ebfx=1\uff0c\u5219\uff1a1+b/2=0 b= -2
\u629b\u7269\u7ebf\u7684\u51fd\u6570\u8868\u8fbe\u5f0f\uff1ay=x²-2x-3
\uff082\uff090=x²-2x-3 A\uff08-1,0\uff09 B\uff083,0\uff09 AB=4
BC\u7684\u51fd\u6570\u8868\u8fbe\u5f0f:y=x-3 \u6545D\uff081\uff0c-2\uff09
(3)\u2460PQ=0.75AB \u65f6\uff0cPQ=3 3/2+1=2.5
\u6545PQF \u4e09\u70b9\u7eb5\u5750\u6807\uff1ay=2.5²-2*2.5-3= -1.75 E\u70b9\u7eb5\u5750\u6807\uff1a3-2*1.75= -0.5 \u5373\uff1aE\uff080\uff0c-0.5\uff09
tan\u2220CED =1/[-0.5-(-2)]=2/3
\u2461\u5f53\u4ee5\u70b9C\u3001D\u3001E\u4e3a\u9876\u70b9\u7684\u4e09\u89d2\u5f62\u662f\u76f4\u89d2\u4e09\u89d2\u5f62(\u2220CED\u4e3a\u76f4\u89d2\uff09 \u65f6\uff0c\u70b9P\u7684\u5750\u6807(0,-2.5)
(x+b/2)²+c-b²/4
\u4e0ey\u8f74\u4ea4\u4e8e\u70b9C\uff080\uff0c\uff0d3\uff09\u5219:c=
-3
\u5bf9\u79f0\u8f74\u662f\u76f4\u7ebfx=1\uff0c\u5219\uff1a1+b/2=0
b=
-2
\u629b\u7269\u7ebf\u7684\u51fd\u6570\u8868\u8fbe\u5f0f\uff1ay=x²-2x-3
\uff082\uff090=x²-2x-3
A\uff08-1,0\uff09
B\uff083,0\uff09
AB=4
BC\u7684\u51fd\u6570\u8868\u8fbe\u5f0f:y=x-3
\u6545D\uff081\uff0c-2\uff09
(3)\u2460PQ=0.75AB
\u65f6\uff0cPQ=3
3/2+1=2.5
\u6545PQF
\u4e09\u70b9\u7eb5\u5750\u6807\uff1ay=2.5²-2*2.5-3=
-1.75
E\u70b9\u7eb5\u5750\u6807\uff1a3-2*1.75=
-0.5
\u5373\uff1aE\uff080\uff0c-0.5\uff09
tan\u2220CED
=1/[-0.5-(-2)]=2/3
\u2461\u5f53\u4ee5\u70b9C\u3001D\u3001E\u4e3a\u9876\u70b9\u7684\u4e09\u89d2\u5f62\u662f\u76f4\u89d2\u4e09\u89d2\u5f62(\u2220CED\u4e3a\u76f4\u89d2\uff09
\u65f6\uff0c\u70b9P\u7684\u5750\u6807(0,-2.5)
则 c=-3
y=(x+b/2)²-3-b²/4
对称轴 x=1=-b/2 得b=-2
抛物线的函数表达式：y=x²-2x-3
(2) 令y=0，
即 x²-2x-3=(x-3)(x+1)=0
A(-1，0)、B(3，0）
设直线BC的函数表达式：y=kx+b
将B(3，0)、C(0，-3)代入，得
0=3k+b
-3=b
得 k=1、b=-3
直线BC的函数表达式：y=x-3
(3) AB=4
PQ=0.75AB=3
设E(0，y0)
抛物线顶点D(1,4)
PQ直线：y=(4-y0)/2+y0=2+y0/2
首先这是一道2011;沈阳的题
(1)y=x²+bx+c与y轴交于点C(0，-3) 则 c=-3
y=(x+b/2)²-3-b²/4
对称轴 x=1=-b/2 得b=-2
抛物线的函数表达式：y=x²-2x-3
(2) 令y=0，
即 x²-2x-3=(x-3)(x+1)=0
A(-1，0)、B(3，0）
设直线BC的函数表达式：y=kx+b
将B(3，0)、C(0，-3)代入，得
0=3k+b
-3=b
得 k=1、b=-3
直线BC的函数表达式：y=x-3
(3) AB=4
PQ=0.75AB=3
设E(0，y0)
抛物线顶点D(1,4)
PQ直线：y=(4-y0)/2+y0=2+y0/2(1)y=x²+bx+c与y轴交于点C(0，-3)
则 c=-3
y=(x+b/2)²-3-b²/4
对称轴 x=1=-b/2 得b=-2
PQ直线：y=(4-y0)/2+y0=2+y0/2
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