Question

抛物线y=x^2+2mx+3m的顶点的纵坐标最大值为

Answer 1

9/4

y=x^2+2mx+3m=(x+m)^2+3m-m^2所以顶点纵坐标为3m-m^2,

令t=3m-m^2=-(m-3/2)^2+9/4则t最大值为9/4,即顶点的纵坐标最大值