例已知0a1,且s0,t0,试用平面区域表示点(as+at,a2s+a2t)存在的范围.解显然0as1,0at1,故关于u的二次方程u2-(as+at)u+asat=0的两根均在(0,1)之间,令x=as+at,y=a2s+a2t,则x2-2asat=y,即asat=21(x2-y).因此函数f(u)=u2-xu+21(x2-y),其对称轴u=2x=as+2 at,因两根在(0,1)内, Example is known as 0a1, and s0,t0, the trial plane area represents the range in which the point (as+at, a2s+a2t) exists. The solution is obviously 0as1, 0at1, so the quadratic equation u2-(as+at)u+ about u Two of asat=0 are between (0,1), and let x=as+at,y=a2s+a2t, then x2-2asat=y, that is, asat=21(x2-y). So the function f( u)=u2-xu+21(x2-y), whose axis of symmetry u=2x=as+2at, since two roots are within (0,1),