题目
抛物线 y=x^2+mx+16的顶点在x轴上,则m的值为__________
提问时间:2021-03-13
答案
顶点在x轴上,即最值等于0
此处开口向上,所以是最小值等于0
所以可以写成y=(x-h)^2+0=x^2-2hx+h^2
所以m=-2h
h^2=16
h=4,-4
所以m=-8,m=8
此处开口向上,所以是最小值等于0
所以可以写成y=(x-h)^2+0=x^2-2hx+h^2
所以m=-2h
h^2=16
h=4,-4
所以m=-8,m=8
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