如图,二次函数y=x2+bx-3b+3的图象与x轴交于A、B两点(点A在点B的左

如图,二次函数y=x2+bx3b+3的图象与x轴交于AB两点(点A在点B的左边),交y轴于点C,且经过点(b22b25b1.

1)求这条抛物线的解析式;

2)⊙MABC三点,交y轴于另一点D,求点M的坐标;

3)连接AMDM,将∠AMD绕点M顺时针旋转,两边MAMDx轴、y轴分别交于点EF,若△DMF为等腰三角形,求点E的坐标.

 


   

答案

解析:(1)把点(b22b25b1)代入解析式,得

2b25b1=b22+bb2)-3b+3  

解得b=2.

∴抛物线的解析式为y=x2+2x3.          

2)由x2+2x3=0,得x=3x=1.

A(-30)、B10)、C0,-3.

抛物线的对称轴是直线x=1,圆心M在直线x=1.

∴设M(-1n),作MGx轴于GMHy轴于H

连接MCMB.

MH=1BG=2.                                      

MB=MC,∴BG2+MG2=MH2+CH2

4+n2=1+3+n2,解得n=1,∴点M(-1,-1   

3)如图,由M(-1,-1),得MG=MH.

MA=MD,∴RtAMGRtDMH,∴∠1=2.

由旋转可知∠3=4. ∴△AME≌△DMF.

若△DMF为等腰三角形,则△AME为等腰三角形.         

Ex0),△AME为等腰三角形,分三种情况:

AE=AM=,则x=3,∴E30);

②∵MAB的垂直平分线上,

MA=ME=MB,∴E10                           

③点EAM的垂直平分线上,则AE=ME.

AE=x+3ME2=MG2+EG2=1+(-1x2,∴(x+32=1+(-1x2,解得x=,∴E0.∴所求点E的坐标为(30),(10),(0

 


    

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