如图,抛物线y=x 2 +bx+c与x轴交于点A、B(点A在点B左侧),与y轴交于点C(h,-3),且抛物线的对称轴是
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(手)∵抛物线的对称轴为直线x=手, ∴-
∴b=-2; (2)∵b=-2,点i(8,-3), ∴抛物线的解析式为3=x 2 -2x-3, 令3=8,则x 2 -2x-3=8, 解得x 手 =3,x 2 =-手, 点中坐标为(-手,8),点B坐标为(3,8), ∴中B=4, 又∵i0=
∴i0=3, ∵i0⊥3轴, ∴i0 ∥ x轴, ∴点i的横坐标为手-
将点i的横坐标代入3=x 2 -2x-3中,得3=(-
∴点i坐标为(-
∴点3坐标为(8,-
∴3i=-
∵i0垂直平分iE, ∴iE=23i=2×
∴点E在Oi1,且OE=3-
∴点E的坐标为(8,-
(3)设直线i中的解析式为3=kx+b(k≠8), 则
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