设x1＜x2≤-b/2a
则f(x1)-f(x2)
=ax1^2-ax2^2+bx1-bx2
=a(x1-x2)(x1+x2)+b(x1-x2)
=(x1-x2)(ax1-ax2+b)
=a(x1-x2)(x1+x2+b/a)
因为x1＜x2≤-b/2a
所以x1-x2＜0 x1+x2+b/a＜0
又因为a＜0
所以f(x1)-f(x2)＜0
所以f(x)在（-∞,-b/2a]上是增函数
设x1＜x2≤-b/2a
则f(x1)-f(x2)
=ax1^2-ax2^2+bx1-bx2
=a(x1-x2)(x1+x2)+b(x1-x2)
=(x1-x2)(ax1-ax2+b)
=a(x1-x2)(x1+x2+b/a)
因为x1＜x2≤-b/2a
所以x1-x2＜0 x1+x2+b/a＜0
又因为a＜0
所以f(x1)-f(x2)＜0
所以f(x)在（-∞,-b/2a]上是增函数