解:(1)拋物線的對(duì)稱軸為直線x=-1,
把C(0,-3)代入y=(x+1)
2+k得-3=1+k,
∴k=-4;
(2)連接AC,交對(duì)稱軸于點(diǎn)P,如圖1,
對(duì)于y=(x+1)
2-4,令y=0,則(x+1)
2-4=0,解得x
1=1,x
2=-3,
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∴A點(diǎn)坐標(biāo)為(-3,0),B點(diǎn)坐標(biāo)為(1,0),
設(shè)直線AC的關(guān)系式為:y=mx+b,
把A(-3,0),C(0,-3)代入y=m x+b得
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,解得
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,
∴直線AC的關(guān)系式為y=-x-3,
當(dāng)x=-1時(shí),y=1-3=-2,
∴P點(diǎn)坐標(biāo)為(-1,-2);
(3)連接OM,如圖1,設(shè)M點(diǎn)坐標(biāo)為(x,(x+1)
2-4)
S
四邊形AMCB=S
△AMO+S
△CMO+S
△CBO=
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×AO×|y
m|+
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×CO×|x
m|+
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×OC×BO
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=
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[4-(x+1)
2]+
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×3×(-x)+
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×3×1
=-
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x
2-
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x+6
=-
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(x+
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)
2+
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,
當(dāng)x=-
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時(shí),S最大,最大值為
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;
此時(shí)M點(diǎn)坐標(biāo)為(-
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,-
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);
(4)存在.點(diǎn)F的坐標(biāo)為(-1,-4)、(3,12)、(-5,12).
當(dāng)以AB為對(duì)角線,如圖2,
∵四邊形AFBE為平行四邊形,
而EA=EB,
∴四邊形AFBE為菱形,
∴點(diǎn)F也在對(duì)稱軸上,即F點(diǎn)為拋物線的頂點(diǎn),
∴F點(diǎn)坐標(biāo)為(-1,-4);
當(dāng)以AB為邊時(shí),如圖3,
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∵四邊形AFBE為平行四邊形,
∴EF=AB=4,即F
2E=4,F(xiàn)
1E=4,
∴F
1的橫坐標(biāo)為3,F(xiàn)
2的橫坐標(biāo)為-5,
對(duì)于y=(x+1)
2-4,
當(dāng)x=3時(shí),y=16-4=12;
當(dāng)x=-5時(shí),y=16-12,
∴F點(diǎn)坐標(biāo)為(3,12)或(-5,12).
分析:(1)根據(jù)拋物線的頂點(diǎn)式即可得到拋物線的對(duì)稱軸為直線x=-1,然后把C點(diǎn)坐標(biāo)代入解析式可求出k=-4;
(2)令y=0得到(x+1)
2-4=0,解得x
1=1,x
2=-3,可確定A點(diǎn)坐標(biāo)為(-3,0),B點(diǎn)坐標(biāo)為(1,0),再利用待定系數(shù)法確定直線AC的關(guān)系式為y=-x-3,由于使得PA+PC的值最小的點(diǎn)P為直線AC與對(duì)稱軸的交點(diǎn),把x=-1代入y=-x-3即可確定P點(diǎn)坐標(biāo);
(3)連接OM,設(shè)M點(diǎn)坐標(biāo)為(x,(x+1)
2-4),利用S
四邊形AMCB=S
△AMO+S
△CMO+S
△CBO可得到S
四邊形AMCB=-
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x
2-
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x+6,配方得到S=-
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(x+
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)
2+
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,然后根據(jù)二次函數(shù)的最值問題得到當(dāng)x=-
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時(shí),S最大,最大值為
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;同時(shí)可得到M點(diǎn)坐標(biāo);
(4)討論:當(dāng)以AB為對(duì)角線,利用EA=EB和四邊形AFBE為平行四邊形得到四邊形AFBE為菱形,則點(diǎn)F也在對(duì)稱軸上,即F點(diǎn)為拋物線的頂點(diǎn),所以F點(diǎn)坐標(biāo)為(-1,-4);當(dāng)以AB為邊時(shí),根據(jù)平行四邊形的性質(zhì)得到EF=AB=4,則可確定F的橫坐標(biāo),然后代入拋物線解析式得到F點(diǎn)的縱坐標(biāo).
點(diǎn)評(píng):本題考查了二次函數(shù)綜合題:二次函數(shù)y=ax
2+bx+c(a、b、c為常數(shù),a≠0)的圖象為拋物線,其頂點(diǎn)式為y=a(x-
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)
2+
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,拋物線的對(duì)稱軸為x=-
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,當(dāng)a>0,y
最小值=
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;當(dāng)a<0,y
最,大值=
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;拋物線上的點(diǎn)的橫縱坐標(biāo)滿足拋物線的解析式;對(duì)于特殊四邊形的判定與性質(zhì)要熟練運(yùn)用.