Question

用配方法求一元二次方程的最大值与最小值

Answer 1

2x²-7x+2

=2(x²-7x/2)+2

=2(x²-7x/2+49/16-49/16)+2

=2(x²-7x/2+49/16)-49/8+2

=2(x-7/4)²-33/8

所以x=7/4,最小值=-33/8

-3x²+5x+1

=-3(X^2-5X/3)+1

=-3(X^2-5X/3+25/36-25/36)+1

=-3(X-5/6)^2+25/12+1

=-3(X-5/6)^2+37/12

当X=5/6时,函数的最大值为133/25

不懂的欢迎追问,