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求证:当a>0时,函数y=ax^2+bx+c的最小值是(4ac-b^2)/4a; 当a
人气:289 ℃ 时间:2020-04-08 13:58:33
解答
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
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