【答案】
分析:(1)由于拋物線y=ax
2+bx+c過點(diǎn)A(-3,0),B(1,0),

,利用待定系數(shù)法即可確定拋物線的解析式;
(2)①由于△ABC繞AB的中點(diǎn)M旋轉(zhuǎn)180°,可知點(diǎn)E和點(diǎn)C關(guān)于點(diǎn)M對稱,由此利用已知條件即可求出E的坐標(biāo);
②四邊形AEBC是矩形.根據(jù)旋轉(zhuǎn)可以得到△ABC≌△AEB,再根據(jù)全等三角形的性質(zhì)得到AC=EB,AE=BC,接著證明AEBC是平行四邊形,而在Rt△ACO中,OC=

,OA=3,由此得到∠CAB=30°,再利用評(píng)選四邊形的性質(zhì)得到∠ABE=30°,最后在Rt△COB中利用三角函數(shù)求出∠CBO=60°,接著就可以證明∠CBE=90°,這樣就可以證明四邊形ABEC是矩形;(3)首先假設(shè)在直線BC上存在一點(diǎn)P,使△PAD的周長最小.由于AD為定值,所以使△PAD的周長最小,就是PA+PD最小,而根據(jù)四邊形AEBC是矩形可以得到A(-3,0)關(guān)于點(diǎn)C(0,

)的對稱點(diǎn)A
1(3,2

),點(diǎn)A與點(diǎn)A
1也關(guān)于直線BC對稱.連接A
1D,與直線BC相交于點(diǎn)P,連接PA,則△PAD的周長最�。又么ㄏ禂�(shù)法求出BC、A
1D的解析式,接著聯(lián)立解析式解方程組即可P的坐標(biāo).
解答:
解:(1)∵y=ax
2+bx+c過C(0,

),
∴

,
又y=ax
2+bx+c過點(diǎn)A(-3,0)、B(1,0),
∴

,
∴

,
∴此拋物線的解析式為

;
(2)①△ABC繞AB的中點(diǎn)M旋轉(zhuǎn)180°.可知點(diǎn)E和點(diǎn)C關(guān)于點(diǎn)M對稱,
∴M(-2,0),C(0,

),
∴E(-2,-

);
②四邊形AEBC是矩形.
∵△ABC繞AB的中點(diǎn)M旋轉(zhuǎn)180°得到四邊形AEBC,
∴△ABC≌△AEB
∴AC=EB,AE=BC
∴AEBC是平行四邊形
在Rt△ACO中,OC=

,OA=3,
∴∠CAB=30°,
∵AEBC是平行四邊形,
∴AC∥BE,
∴∠ABE=30°,
在Rt△COB中,
∵OC=

,OB=1,
∴∠CBO=60°
∴∠CBE=∠CBO+∠ABE=60°+30°=90°
ABEC是矩形;
(3)假設(shè)在直線BC上存在一點(diǎn)P,使△PAD的周長最�。�
因?yàn)锳D為定值,所以使△PAD的周長最小,就是PA+PD最小;
∵AEBC是矩形,
∴∠ACB=90°.
∴A(-3,0)關(guān)于點(diǎn)C(0,

)的對稱點(diǎn)A
1(3,2

).
點(diǎn)A與點(diǎn)A
1也關(guān)于直線BC對稱.
連接A
1D,與直線BC相交于點(diǎn)P,連接PA,則△PAD的周長最�。�
∵B(1,0)、C(0,

)
∴BC的解析式為
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∵A
1(3,2

)、D(-1,

)
∴A
1D的解析式為

.
∴
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∴
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∴P的坐標(biāo)為(
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).
點(diǎn)評(píng):本題是二次函數(shù)的綜合題型,其中涉及到的知識(shí)點(diǎn)有待定系數(shù)法確定函數(shù)的解析式、軸對稱圖形的性質(zhì)的應(yīng)用、中心對稱圖形的性質(zhì)及直線的交點(diǎn)與它們解析式組成方程組的解的關(guān)系,綜合性很強(qiáng),對于學(xué)生的能力的要求比較高,平時(shí)加強(qiáng)訓(xùn)練.