初中二次函数已知抛物线y=x^2
y=x^2-(2m+4)x+m^2-10
=[x-(2m+4)/2]^2-(2m+4)^2/4+m^2-10
对称轴 x=m+2
C的值=-(2m+4)^2/4+m^2-10=-4m-14 即C点坐标（m+2,-4m-14)
x1+x2=2m+4 x1*x2=m^2-10
AB间距离 =√( x1-x2)^2=√[x1+x2)^2-4x1*x2] =√[(2m+4)^2-4(m^2-10)]
=√(16m+56)
=4√(m+4)
ABC为等边三角形 则 ( CA)^2 -(AB/2)^2=(-4m-14)^2
[4√(m+4)]^2-[2√(m...全部
y=x^2-(2m+4)x+m^2-10
=[x-(2m+4)/2]^2-(2m+4)^2/4+m^2-10
对称轴 x=m+2
C的值=-(2m+4)^2/4+m^2-10=-4m-14 即C点坐标（m+2,-4m-14)
x1+x2=2m+4 x1*x2=m^2-10
AB间距离 =√( x1-x2)^2=√[x1+x2)^2-4x1*x2] =√[(2m+4)^2-4(m^2-10)]
=√(16m+56)
=4√(m+4)
ABC为等边三角形 则 ( CA)^2 -(AB/2)^2=(-4m-14)^2
[4√(m+4)]^2-[2√(m+4)]^2=(-4m-14)^2
化简 4m^2+25m+37=0
m=(-25+√33)/8 或m=（-25-√33）/8
将m代入y=x^2-(2m+4)x+m^2-10得该抛物线解析式。
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