y=ax^2-2x+1 =a(x -1/a)^2 + 1-1/a^2(1)当a ≤0时， y=ax2-2x+1在区间[0,1]上的最小值，ymin =y(0) = 1(2)当0<1/a<1即a>1时， y=ax^2-2x+1在区间[0,1]上的最小值，ymin =y(1/a)= 1-1/a^2(3)当1/a≥1,即 0 y=ax^2-2x+1在区间[0,1]上的最小值，ymin =y(1) =a-1
