Question

高一数学题（课本例题）原文：由二次函数的知识,对于函数h(t)=-4.9t^2+14.7t+18,我们有：当t=-14.

Answer 1

h(t)=-4.9t^2+14.7t+18

=-4.9(t^2-14.7t/4.9-18/4.9)

=-4.9{[t-14.7/(2*4.9)]^2-18/14.9-[14.7/(2*4.9)]^2}

=-4.9[t-14.7/(2*4.9)]^2+18+4.9*[14.7/(2*4.9)]^2

=-4.9[t-14.7/(2*4.9)]^2+29

显然-4.9[t-14.7/(2*4.9)]^2是非正数,当-4.9[t-14.7/(2*4.9)]^2=0时,函数有最大值29,

即当t=-14.7/2*(-4.9)=1.5时,4.9[t-14.7/(2*4.9)]^2=0,函数的最大值是29