【答案】
分析:(1)根據(jù)第三個頂點C在x軸的正半軸上,利用勾股定理求出OC的長,進而求出C點坐標(biāo),應(yīng)用待定系數(shù)法即可求出直線BC的解析式;
(2)由于拋物線解析式關(guān)于y軸對稱,可知一次項系數(shù)為0,利用待定系數(shù)法,設(shè)出一般式,將A(0,1),D(3,-2)代入解析式即可求出二次函數(shù)解析式;根據(jù)軸對稱定義和角平分線的定義,利用特殊角判斷出則符合條件的點P就是直線BC與拋物線y=-

x
2+1的交點.
(3)根據(jù)軸對稱定義和性質(zhì),作出C關(guān)于y軸的對稱點C′,將求PM+CM的取值范圍轉(zhuǎn)化為求PM+C′M的取值范圍.
解答:
解:(1)∵A(0,1),B(0,3),
∴AB=2,
∵△ABC是等腰三角形,且點C在x軸的正半軸上,
∴AC=AB=2,
∴OC=

=

.
∴C(

,0).(2分)
設(shè)直線BC的解析式為y=kx+3,
∴

k+3=0,
∴k=-

.
∴直線BC的解析式為y=-

x+3.(4分)
(2)∵拋物線y=ax
2+bx+c關(guān)于y軸對稱,
∴b=0.(5分)
又拋物線y=ax
2+bx+c經(jīng)過A(0,1),D(3,-2)兩點.
∴

解得
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∴拋物線的解析式是y=-

x
2+1.(7分)
在Rt△AOC中,OA=1,AC=2,易得∠ACO=30°.
在Rt△BOC中,OB=3,OC=

,易得∠BCO=60°.
∴CA是∠BCO的角平分線.
∴直線BC與x軸關(guān)于直線AC對稱.
點P關(guān)于直線AC的對稱點在x軸上,則符合條件的點P就是直線BC與拋物線y=-

x
2+1的交點.(8分)
∵點P在直線BC:y=-

x+3上,故設(shè)點P的坐標(biāo)是(x,-

x+3).
又∵點P(x,-

x+3)在拋物線y=-

x
2+1上,
∴-

x+3=-

x
2+1.
解得x
1=

,x
2=2

.
故所求的點P的坐標(biāo)是P
1(

,0),P
2(2

,-3).(10分)
(3)要求PM+CM的取值范圍,可先求PM+C′M的最小值.
(I)當(dāng)點P的坐標(biāo)是OC=

時,點P與點C重合,
故PM+CM=2CM.
顯然CM的最小值就是點C到y(tǒng)軸的距離為

,
∵點M是y軸上的動點,
∴PM+CM無最大值,
∴PM+CM≥2

.(13分)
(II)當(dāng)點P的坐標(biāo)是(2

,-3)時,由點C關(guān)于y軸的對稱點C′(-

,0),
故只要求PM+MC'的最小值,顯然線段PC'最短.易求得PC'=6.
∴PM+CM的最小值是6.
同理PM+CM沒有最大值,
∴PM+CM的取值范圍是PM+CM≥6.
綜上所述,當(dāng)點P的坐標(biāo)是(

,0)時,PM+CM≥2

,
當(dāng)點P的坐標(biāo)是(2

,-3)時,PM+CM≥6.(15分)
點評:此題考查了對函數(shù)題綜合應(yīng)用和分析解答的能力.(1)(2)小題難度不大,主要應(yīng)用待定系數(shù)法即可解答,(3)要根據(jù)軸對稱的性質(zhì),將折線轉(zhuǎn)化為兩點之間線段最短的問題來解答.