【答案】
分析:(1)把x=0和x=2代入得出關(guān)于t的方程,求出t即可;
(2)把A的坐標(biāo)代入拋物線,即可求出m,把A的坐標(biāo)代入直線,即可求出k;
(3)求出點(diǎn)B、C間的部分圖象的解析式是y=-
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(x-3)(x+1),得出拋物線平移后得出的圖象G的解析式是y=-
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(x-3+n)(x+1+n),-n-1≤x≤3-n,直線平移后的解析式是y=4x+6+n,若兩圖象有一個(gè)交點(diǎn)時(shí),得出方程4x+6+n=-
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(x-3+n)(x+1+n)有兩個(gè)相等的實(shí)數(shù)解,求出判別式△=6n=0,求出的n的值與已知n>0相矛盾,得出平移后的直線與拋物線有兩個(gè)公共點(diǎn),
設(shè)兩個(gè)臨界的交點(diǎn)為(-n-1,0),(3-n,0),代入直線的解析式,求出n的值,即可得出答案.
解答:(1)解:∵二次函數(shù)y=(t+1)x
2+2(t+2)x+
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在x=0和x=2時(shí)的函數(shù)值相等,
∴代入得:0+0+
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=4(t+1)+4(t+2)+
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,
解得:t=-
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,
∴y=(-
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+1)x
2+2(-
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+2)x+
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=-
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x
2+x+
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,
∴二次函數(shù)的解析式是y=-
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x
2+x+
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.
(2)解:把A(-3,m)代入y=-
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x
2+x+
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得:m=-
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×(-3)
2-3+
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=-6,
即A(-3,-6),
代入y=kx+6得:-6=-3k+6,
解得:k=4,
即m=-6,k=4.
(3)解:由題意可知,點(diǎn)B、C間的部分圖象的解析式是y=-
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x
2+x+
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=-
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(x
2-2x-3)=-
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(x-3)(x+1),-1≤x≤3,
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則拋物線平移后得出的圖象G的解析式是y=-
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(x-3+n)(x+1+n),-n-1≤x≤3-n,
此時(shí)直線平移后的解析式是y=4x+6+n,
如果平移后的直線與平移后的二次函數(shù)相切,
則方程4x+6+n=-
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(x-3+n)(x+1+n)有兩個(gè)相等的實(shí)數(shù)解,
即-
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x
2-(n+3)x-
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n
2-
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=0有兩個(gè)相等的實(shí)數(shù)解,
判別式△=[-(n+3)]
2-4×(-
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)×(-
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n
2-
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)=6n=0,
即n=0,
∵與已知n>0相矛盾,
∴平移后的直線與平移后的拋物線不相切,
∴結(jié)合圖象可知,如果平移后的直線與拋物線有公共點(diǎn),
則兩個(gè)臨界的交點(diǎn)為(-n-1,0),(3-n,0),
則0=4(-n-1)+6+n,
n=
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,
0=4(3-n)+6+n,
n=6,
即n的取值范圍是:
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≤n≤6.
點(diǎn)評(píng):本題考查了二次函數(shù)和一次函數(shù)的性質(zhì),平移的性質(zhì),根的判別式等知識(shí)點(diǎn)的應(yīng)用,通過做此題培養(yǎng)了學(xué)生的分析問題和解決問題的能力,題目綜合性比較強(qiáng),有一定的難度.