【答案】
分析:(1)先根據(jù)y=kx沿y軸向下平移3個(gè)單位長(zhǎng)度后經(jīng)過(guò)y軸上的點(diǎn)C求出C點(diǎn)的坐標(biāo),再用待定系數(shù)法求出直線(xiàn)BC的解析式,再根據(jù)拋物線(xiàn)y=-x
2+bx+c過(guò)點(diǎn)B,C,把B、C兩點(diǎn)的坐標(biāo)代入所設(shè)函數(shù)解析式即可求出此解析式;
(2)根據(jù)(1)中二次函數(shù)的解析式可求出A、D兩點(diǎn)的坐標(biāo),判斷出△OBC是等腰直角三角形,利用銳角三角函數(shù)的定義可求出∠OBC的度數(shù),過(guò)點(diǎn)A作AE⊥BC于點(diǎn)E,利用勾股定理可求出BE、AE及CE的長(zhǎng),再根據(jù)相似三角形的判定定理可得出△AEC∽△AFP,根據(jù)相似三角形的對(duì)應(yīng)邊成比例可求出PF的長(zhǎng),再點(diǎn)P在拋物線(xiàn)的對(duì)稱(chēng)軸上即可求出點(diǎn)P的坐標(biāo).
解答:
解:(1)∵y=kx沿y軸向下平移3個(gè)單位長(zhǎng)度后經(jīng)過(guò)y軸上的點(diǎn)C,
∴此時(shí)直線(xiàn)的解析式為y=kx-3,令x=0,則y=-3,
∴C(0,-3)(1分)
設(shè)直線(xiàn)BC的解析式為y=kx-3.(1分)
∵B(-3,0)在直線(xiàn)BC上,
∴-3k-3=0解得k=-1.
∴直線(xiàn)BC的解析式為y=-x-3.(1分)
∵拋物線(xiàn)y=-x
2+bx+c過(guò)點(diǎn)B,C,
∴

(2分)
解得

,
∴拋物線(xiàn)的解析式為y=-x
2-4x-3;(1分)
(2)由y=-x
2-4x-3.可得D(-2,1),A(-1,0).(1分)
∴OB=3,OC=3,OA=1,AB=2,
可得△OBC是等腰直角三角形.
∴∠OBC=45°,CB=3

.(1分)
設(shè)拋物線(xiàn)對(duì)稱(chēng)軸與x軸交于點(diǎn)F,
∴AF=

AB=1.
過(guò)點(diǎn)A作AE⊥BC于點(diǎn)E.
∴∠AEB=90°.
可得BE=AE=

,CE=2

,(1分)
在△AEC與△AFP中,∠AEC=∠AFP=90°,∠ACE=∠APF,
∴△AEC∽△AFP.(1分)
∴

=
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,
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=
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,
解得,PF=2,
∵點(diǎn)P在拋物線(xiàn)的對(duì)稱(chēng)軸上,
∴點(diǎn)P的坐標(biāo)為(-2,-2),(-2,2)(不合題意舍去).(2分)
點(diǎn)評(píng):本題考查的是二次函數(shù)綜合題,涉及到用待定系數(shù)法求一次函數(shù)及二次函數(shù)的解析式、等腰直角三角形的判定與性質(zhì)、特殊角度的三角函數(shù)值及相似三角形的判定與性質(zhì),涉及面較廣,難度較大.