题目

求证:当a>0时,函数y=ax^2+bx+c的最小值是(4ac-b^2)/4a; 当a

提问时间：2021-02-02

答案

y = ax^2 + bx + c
= a[x^2 + (b/a)x] + c
= a(x + b/2a)^2 - a*(b/2a)^2 + c
= a(x + b/2a)^2 -b^2/4a + c
= a(x + b/2a)^2 -(b^2 - 4ac)/4a
= a(x + b/2a)^2 + (4ac - b^2)/4a
a>0时,抛物线开口向上,
最小值为(4ac-b^2)/4a
a