【答案】
分析:(1)由直線y=-x+4交x軸的正半軸于點(diǎn)A,交y軸于點(diǎn)B,即可求得A,B的坐標(biāo),又由拋物線上有不同的兩點(diǎn)E(k+3,-k
2+1)和F(-k-1,-k
2+1)的縱坐標(biāo)相等,即可求得此拋物線的對(duì)稱軸,利用待定系數(shù)法即可求得解析式;
(2)分別從當(dāng)點(diǎn)P(x,x)在直線AB上時(shí)與當(dāng)點(diǎn)Q(
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,
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)在直線AB上時(shí)分析,即可求得x的取值范圍;
(3)首先求得當(dāng)點(diǎn)E(x,
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)在直線AB上時(shí)x的值,再分別從當(dāng)2≤x<
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時(shí)與當(dāng)
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≤x≤4時(shí)去分析,注意三角形的面積求解方法與二次函數(shù)最大值的求解方法的應(yīng)用.
解答:解:(1)當(dāng)x=0時(shí),y=4,即B(0,4),
當(dāng)y=0時(shí),x=4,即A(4,0),
∵拋物線上有不同的兩點(diǎn)E(k+3,-k
2+1)和F(-k-1,-k
2+1)的縱坐標(biāo)相等,
∴點(diǎn)E和點(diǎn)F關(guān)于拋物線對(duì)稱軸對(duì)稱,
∴對(duì)稱軸x=-
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=
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=1,
把點(diǎn)A,點(diǎn)B代入拋物線解析式中求得a=
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,b=1,c=4,
∴拋物線解析式為y=-
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x
2+x+4;
(2)當(dāng)點(diǎn)P(x,x)在直線AB上時(shí),x=-x+4,
解得x=2,
當(dāng)點(diǎn)Q(
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,
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)在直線AB上時(shí),
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=-
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+4,
解得x=4.
所以,若正方形PEQF與直線AB有公共點(diǎn),則2≤x≤4.
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(3)當(dāng)點(diǎn)E(x,
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)在直線AB上時(shí),(此時(shí)點(diǎn)F也在直線AB上)
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=-x+4,
解得x=
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.
①當(dāng)2≤x<
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時(shí),直線AB分別與PE、PF有交點(diǎn),設(shè)交點(diǎn)分別為C、D,
此時(shí),PC=x-(-x+4)=2x-4,
又PD=PC,
所以S
△PCD=
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PC
2=2(x-2)
2,
從而:S=
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x
2-2(x-2)
2=-
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x
2+8x-8=-
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(x-
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)
2+
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.
∵2≤
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<
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,
∴當(dāng)x=
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時(shí),S
max=
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.
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②當(dāng)
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≤x≤4時(shí),直線AB分別與QE、QF有交點(diǎn),設(shè)交點(diǎn)分別為M、N,
此時(shí),QN=(-
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+4)-
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=-x+4,
又QM=QN,
∴S
△QMN=
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QN
2=
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(x-4)
2,
即S=
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(x-4)
2.
其中當(dāng)x=
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時(shí),S
max=
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.
綜合①②得,當(dāng)x=
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時(shí),S
max=
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.
點(diǎn)評(píng):此題考查了待定系數(shù)法求二次函數(shù)的解析式,函數(shù)自變量的取值范圍的確定、二次函數(shù)最大值的確定以及三角形面積的求解等知識(shí).此題綜合性很強(qiáng),注意數(shù)形結(jié)合思想與分類討論思想的應(yīng)用.