【答案】
分析:(1)利用待定系數(shù)法求二次函數(shù)解析式、一次函數(shù)解析式;
(2)根據(jù)兩點(diǎn)之間線段最短作N點(diǎn)關(guān)于直線x=3的對稱點(diǎn)N′,當(dāng)M(3,m)在直線DN′上時,MN+MD的值最��;
(3)需要分類討論:①當(dāng)點(diǎn)E在線段AC上時,點(diǎn)F在點(diǎn)E上方,則F(x,x+3)和②當(dāng)點(diǎn)E在線段AC(或CA)延長線上時,點(diǎn)F在點(diǎn)E下方,則F(x,x-1),然后利用二次函數(shù)圖象上點(diǎn)的坐標(biāo)特征可以求得點(diǎn)E的坐標(biāo);
(4)方法一:過點(diǎn)P作PQ⊥x軸交AC于點(diǎn)Q;過點(diǎn)C作CG⊥x軸于點(diǎn)G,如圖1.設(shè)Q(x,x+1),則P(x,-x
2+2x+3).根據(jù)兩點(diǎn)間的距離公式可以求得線段PQ=-x
2+x+2;最后由圖示以及三角形的面積公式知S
△APC=-
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(x-
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)
2+

,所以由二次函數(shù)的最值的求法可知△APC的面積的最大值;
方法二:過點(diǎn)P作PQ⊥x軸交AC于點(diǎn)Q,交x軸于點(diǎn)H;過點(diǎn)C作CG⊥x軸于點(diǎn)G,如圖2.設(shè)Q(x,x+1),則P(x,-x
2+2x+3).根據(jù)圖示以及三角形的面積公式知S
△APC=S
△APH+S
直角梯形PHGC-S
△AGC=-

(x-

)
2+

,所以由二次函數(shù)的最值的求法可知△APC的面積的最大值;
解答:
解:(1)由拋物線y=-x
2+bx+c過點(diǎn)A(-1,0)及C(2,3)得,

,
解得

,
故拋物線為y=-x
2+2x+3
又設(shè)直線為y=kx+n過點(diǎn)A(-1,0)及C(2,3)得

,
解得
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故直線AC為y=x+1;
(2)如圖1,作N點(diǎn)關(guān)于直線x=3的對稱點(diǎn)N′,則N′(6,3),由(1)得D(1,4),
故直線DN′的函數(shù)關(guān)系式為y=-

x+

,
當(dāng)M(3,m)在直線DN′上時,MN+MD的值最小,
則m=-

×
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=
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;
(3)由(1)、(2)得D(1,4),B(1,2),
∵點(diǎn)E在直線AC上,

設(shè)E(x,x+1),
①如圖2,當(dāng)點(diǎn)E在線段AC上時,點(diǎn)F在點(diǎn)E上方,
則F(x,x+3),
∵F在拋物線上,
∴x+3=-x
2+2x+3,
解得,x=0或x=1(舍去)
∴E(0,1);
②當(dāng)點(diǎn)E在線段AC(或CA)延長線上時,點(diǎn)F在點(diǎn)E下方,
則F(x,x-1)
由F在拋物線上
∴x-1=-x
2+2x+3
解得x=

或x=
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∴E(
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,
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)或(
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,
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)
綜上,滿足條件的點(diǎn)E的坐標(biāo)為(0,1)、(

,

)或(
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,
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);
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(4)方法一:如圖3,過點(diǎn)P作PQ⊥x軸交AC于點(diǎn)Q,交x軸于點(diǎn)H;過點(diǎn)C作CG⊥x軸于點(diǎn)G,設(shè)Q(x,x+1),則P(x,-x
2+2x+3)
∴PQ=(-x
2+2x+3)-(x+1)
=-x
2+x+2
又∵S
△APC=S
△APQ+S
△CPQ=

PQ•AG
=

(-x
2+x+2)×3
=-

(x-

)
2+

∴面積的最大值為

.
方法二:過點(diǎn)P作PQ⊥x軸交AC于點(diǎn)Q,交x軸于點(diǎn)H;過點(diǎn)C作CG⊥x軸于點(diǎn)G,如圖3,
設(shè)Q(x,x+1),則P(x,-x
2+2x+3)
又∵S
△APC=S
△APH+S
直角梯形PHGC-S
△AGC=

(x+1)(-x
2+2x+3)+

(-x
2+2x+3+3)(2-x)-

×3×3
=-

x
2+

x+3
=-

(x-

)
2+

∴△APC的面積的最大值為

.
點(diǎn)評:本題考查了二次函數(shù)綜合題.解答(3)題時,要對點(diǎn)E所在的位置進(jìn)行分類討論,以防漏解.