【答案】
分析:(1)把y=x
2-5x+4化成頂點式,求出頂點C的坐標(biāo),y=x
2-5x+4化成(x-1)(x-4),求出A、B的坐標(biāo),設(shè)AC直線為y=kx+b,把A、C的坐標(biāo)代入就能求出直線AC的解析式;
(2)設(shè)直線BC的解析式是y=ax+c,把B、C的坐標(biāo)代入就能求出直線BC,點E坐標(biāo)為(4-t,0),點F坐標(biāo)為(
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),求出EF=
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,F(xiàn)G=2t-3,根據(jù)EF=FG,即可求出t的值;
(3)可分以下幾種情況:①點F在BC上時,如圖1重疊部分是△BEF
2,此時
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時,點F坐標(biāo)為(
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),根據(jù)三角形的面積公式即可求出;②I如圖2,EB≤EH時重疊部分是直角梯形EFKB,此時
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<t≤
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,根據(jù)三角形的面積公式即可求出;II如圖3,EB>EH,點G在BC下方時,重疊部分是五邊形EFKMH,此時
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,
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,因為S=S
正方形EFGH-S
△KMG,根據(jù)三角形的面積公式即可求出;Ⅲ.如圖4,點G在BC上或BC上方時,重疊部分是正方形EFGH,此時

≤t<3,
根據(jù)正方形的面積公式求出即可.
解答:(1)解:
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∵y=x
2-5x+4=
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,
頂點C的坐標(biāo)為(
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),
∵y=x
2-5x+4=(x-1)(x-4),
∴點A(1,0),B(4,0),
設(shè)AC直線為y=kx+b,得
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,
解得:k=-
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,b=
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,
∴
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,
答:頂點C的坐標(biāo)為(
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),直線AC的解析式是
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.
(2)解:設(shè)直線BC的解析式是y=ax+c,
把B(4,0),C(
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,-
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)代入得:0=4a+c且-
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=
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a+c,
解得:a=
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,c=-6,
直線BC的解析式為
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,
當(dāng)F在AC邊上,G在BC邊上時,
點E坐標(biāo)為(4-t,0),點F坐標(biāo)為(
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),
得EF=
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,
而EF=FG,
∵拋物線的對稱軸和等腰△ABC的對稱軸重合,
∴FG=
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,
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=2t-3,
∴
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=2t-3,
解得
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,
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答:當(dāng)點F在AC邊上,G在BC邊上時t的值是
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.
(3)解:點E坐標(biāo)為(4-t,0)隨著正方形的移動,重疊部分的形狀不同,可分以下幾種情況:
①點F在BC上時,如圖1重疊部分是△BEF,
此時

時,點F坐標(biāo)為(
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),
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=
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,
②點F在AC上時,點F坐標(biāo)為(
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)又可分三種情況:
Ⅰ.如圖2,EB≤EH時重疊部分是直角梯形EFKB(設(shè)FG與直線BC交于點K),
此時
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<t≤
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,
∴
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,
Ⅱ.如圖3,EB>EH,點G在BC下方時,重疊部分是五邊形EFKMH(設(shè)FG與直線BC交于點K,GH與直線BC交于點M),
此時
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,
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,
點H坐標(biāo)為(
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),點M坐標(biāo)為(
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),
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,
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,
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,
∴S=S
EFGH-S
△KMG=(
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)
2
,
=
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,
Ⅲ.如圖4,點G在BC上或BC上方時,重疊部分是正方形EFGH,此時

≤t<3,
∴
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=
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t
2-
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t+
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,
答:動點E從點B向點A運動過程中,S關(guān)于t的函數(shù)關(guān)系S=
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t
2(0<t≤
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)或S=-
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t
2+9t-
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(
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<t≤
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)或S=-
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t
2+
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t-
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(
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<t<
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)或S=
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t
2-
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t+
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(
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≤t<3).
點評:本題主要考查對二次函數(shù)與X軸的交點,用待定系數(shù)法求一次函數(shù)的解析式,解二元一次方程組,三角形的面積,用十字相乘法分解因式,二次函數(shù)圖象上點的坐標(biāo)特征等知識點的理解和掌握,此題是一個拔高的題目,有一定的難度,用的數(shù)學(xué)思想是分類討論思想.