解:(1)∵拋物線y=ax
2+bx+3經(jīng)過點B(-1,0)、C(3,0),
∴
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,
解得
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.
∴拋物線的解析式為y=-x
2+2x+3.
(2)當(dāng)直角梯形EFGH運動到E′F′G′H′時,過點F′作F′N⊥x軸于點N,延長E′H’交x軸于點P.
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∵點M的坐標(biāo)為(0,1),點A是拋物線與y軸的交點,
∴點A的坐標(biāo)為(3,0).
∵OA=3,OD=4,
∴AD=5.
∵E′H′∥OM,E′H′=OM=1,
∴四邊形E′H′OM是平行四邊形(當(dāng)E′H′不與y軸重合時).
∵F′N∥y軸,N G′∥x軸,
∴△F′N D∽△AOD.
∴
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.
∵直角梯形E′F′G′H′是直角梯形EFGH沿射線DA方向平移得到的,
∴F′D=t,
∴
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.
∴
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.
∵E′F′=PN=1,
∴OP=OD-PN-ND=4-1-
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=3-
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.
∵E′P=
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,E′H′=1,
∴H′P=
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-1.
若平行四邊形E′H′OM是矩形,則∠MO H′=90°,此時H′G′與x軸重合.
∵F′D=t,
∴
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,即
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.
即當(dāng)
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秒時,平行四邊形EHOM是矩形.
若平行四邊形E′H′OM是菱形,則O H′=1.
在Rt△H′OP中,OP
2+H′P
2=OH′
2,即
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得t
2-6t+9=0,解得t
1=t
2=3.
即當(dāng)t=3秒時,平行四邊形EHOM是菱形.
綜上所述,當(dāng)
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秒時,平行四邊形EHOM是矩形,當(dāng)t=3秒時,平行四邊形EHOM是菱形.
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(3)過A作AR⊥KI于R點,則AR=KR=1.
當(dāng)Q在KI左側(cè)時,△ARP∽△PIQ.
設(shè)PI=n,則RP=3-n,
∴
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,即n
2-3n-m+1=0,
∵關(guān)于n的方程有解,△=(-3)
2-4(-m+1)≥0,
得m≥
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;
當(dāng)Q在KI右側(cè)時,
Rt△APQ中,AR=RK=1,∠AKI=45°可得OQ=5.即P為點K時,
∴m≤5.
綜上所述,m的變化范圍為:
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≤m≤5.
分析:(1)把點B、C兩點的坐標(biāo)分別代入拋物線解析式,列出關(guān)于a、b的方程組
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,通過解該方程組可以求得它們的值,從而得到拋物線的解析式;
(2)當(dāng)直角梯形EFGH運動到E′F′G′H′時,過點F′作F′N⊥x軸于點N,延長E′H’交x軸于點P.根據(jù)平行四邊形的判定可得四邊形E′H′OM是平行四邊形,根據(jù)平行線的性質(zhì)可得△F′N D∽△AOD,根據(jù)相似三角形的性質(zhì)可得
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.再分平行四邊形EHOM是矩形,平行四邊形E′H′OM是菱形,求得
t的值;
(3)過A作AR⊥KI于R點,分當(dāng)Q在KI左側(cè)時,當(dāng)Q在KI右側(cè)時,兩種情況討論可得實數(shù)m的變化范圍.
點評:本題考查了待定系數(shù)法求二次函數(shù)解析式,二次函數(shù)的性質(zhì)以及平行四邊形的判定和性質(zhì),矩形和菱形的判定,平行線的性質(zhì),分類思想的運用,綜合性較強,難度較大.