已知二次函数y=(2m-1)x^2-(5m+3)x+3m+5

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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