(1)y = x^2+(2m-1)x+m^2+2根据二次游梁函数的性质没磨运,x=(1-2m)/2 的时候取最小值y_min = [(1-2m)/2]^2 + (2m-1)*(1-2m)/2 + m^2 + 2 = m + 7/4m + 7/4 = 2 故有 m = 1/4.(2)将 m 代回二次函数枯梁之中y = x^2 - 1/2 x + 33/16顶点座标即为 ( 1/4 , 2 )
m=0(0,0)y轴
将2带入对称轴-2a/b带入x