【名师支招】一道二次函数经典30问答案解析,全网最全!
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01
特别说明
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02
针对变式题目
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03
形定问题
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1、分析:二次函数有三种表达形式:
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由A、C两点坐标可以求出表达式里b、c两个参数的值,由顶点公式可求出顶点D的坐标。(也可以采用设一般式或交点式方法求解)
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2、分析:由点A、C、D的坐标可以求出AD、AC、CD的长度,从而判断出ΔACD的形状。
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04
线段问题
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3、分析:因为BE=CE,则点E为BC的垂直平分线与y轴的交点。根据勾股定理找出直角ΔOBE三边的关系从而求出点E的坐标。
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4、分析:设出点P的坐标,然后再求出直线AC的表达式,从而表示出交点N的坐标,从而用x表示PM和MN的长度,根据PM和MN的关系求出点P的坐标。
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05
线段最值问题
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5、分析:求线段PH可以转化为求PF的最值,用含有x的表达式表示出PF的长度,根据二次函数求最值的方法从而求出PF的最大值。根据PH和PF的数量关系求出PH的最大值和P的坐标。(或者根据面积法求出高PH的函数表达式,同理可求)
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06
线段最值问题
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6、分析:由(5)知PH=GH,矩形PEGH为正方形。C矩形PEGH=4PH,当PH最大时成立。
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7、分析:△BCP的周长为:BC+BP+PC,BC长度是定值,当PB+PC最小时,△BCP的周长最小。
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07
面积问题
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10、分析:四边形ABCD为不规则图形,可以采用隔或者补的方法转化为规则的图形。
解:过点D作DE垂直于x轴于点E。
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08
特殊图形
1
直角三角形
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2
等腰三角形
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09
平行四边形存在性
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10
相似三角形
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11
角度问题
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27、分析:若使直线AC与BM的夹角等于∠ACB的2倍,则∠MCB=∠MBC,则MC=MB,利用勾股定理用点M的坐标表示出MC和MB的长度,从而求出点M的坐标。
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28、分析:若使∠BCO+∠BNO=∠BAC,可以在∠BAC上截取∠OAE=∠BCO,过点E作EF垂直AC于点F,则∠BNO=∠EAF,根据∠EAF的正切值求出点ON的长度即可。
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12
旋转问题
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29、分析:如图所示,分两种情况,根据旋转前后图形全等,设出抛物线一个点的坐标,根据数量关系表示出另一个抛物线的点的坐标,代入抛物线解析式,从而求出点O’坐标。
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30、分析:如图所示,构造两个直角三角形,求出点E的坐标,从而求出直线AE的表达式,与二次函数联立,从而求出交点M的坐标。
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