已知二次函數(shù)y=x2-(2m+2)x+(m2+4m-3)中,m為不小于0的整數(shù),它的圖象與x軸交于點A和點B,點A在原點左邊,點B在原點右邊.
(1)求這個二次函數(shù)的解析式;
(2)點C是拋物線與y軸的交點,已知AD=AC(D在線段AB上),有一動點P從點A出發(fā),沿線段AB以每秒1個單位長度的速度移動,同時,另一動點Q從點C出發(fā),以某一速度沿線段CB移動,經(jīng)過t秒的移動,線段PQ被CD垂直平分,求t的值;
(3)在(2)的情況下,求四邊形ACQD的面積.
【答案】
分析:(1)根據(jù)二次函數(shù)的圖象與x軸有兩個交點,得到△>0,求出m的取值范圍,結合m為不小于0的整數(shù),
求出m的整數(shù)解;再將整數(shù)解代入二次函數(shù)解析式,找到符合題意的二次函數(shù);
(2)根據(jù)題意畫出圖象,證出DQ∥AC,從而得到△BDQ∽△BAC,然后利用相似三角形的性質求出t的值;
(3)由于△BDQ∽△BAC,求出S
△BAC=6,利用相似三角形的面積比等于相似比的平方,求出S
△DQB,二者相減,即可得到S
四邊形ACQD.
解答:解:(1)∵二次函數(shù)的圖象與x軸有兩個交點,
∴△=[-(2m+2)]
2-4(m
2+4m-3)=-8m+16>0,
∴m<2.
∵m為不小于0的整數(shù),
∴m取0、1.
當m=1時,y=x
2-4x+2,圖象與x軸的兩個交點在原點的同側,不合題意,舍去;
當m=0時,y=x
2-2x-3,符合題意.
∴二次函數(shù)的解析式為:y=x
2-2x-3;
(2)∵AC=AD,
∴∠ADC=∠ACD
∵CD垂直平分PQ,
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∴DP=DQ,
∴∠ADC=∠CDQ.
∴∠ACD=∠CDQ,
∴DQ∥AC
∴△BDQ∽△BAC,
∴
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=
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,
∵AC=
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,BD=4-
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,AB=4.
∴DQ=
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-
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,
∴PD=
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-
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.
∴AP=AD-PD=
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,
∴t=
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÷1=
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,
(3)∵△BDQ∽△BAC,
∵AB=4,AD=AC=
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=
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,
∴BD=4-
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,
∴
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=(
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)
2=(
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),
∵S
△BAC=6,
∴S
△BOQ=
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,
∴S
四邊形ACQD=6-
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=
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.
點評:本題考查了二次函數(shù)圖象與x軸的交點與判別式的關系,相似三角形的性質,坐標與圖形的面積等知識,綜合性很強,需要從各角度進行分析解答.