Question

若二次函数f(x)=ax^2+bx+c,当x=3/2时取得最小值-3/4、又f(x)=0的两根x1、

Answer 1

由最小值,可知f(x)=a(x-3/2)^2-3/4,且a>0

f(x)=a(x^2-3x+9/4)-3/4=ax^2-3ax+(9a-3)/4

x1+x2=3a/a=3

x1x2=(9a-3)/(4a)

9=x1^3+x2^3=(x1+x2)(x1^2-x1x2+x2^3)=(x1+x2)[(x1+x2)^2-3x1x2]=3[9-3(9a-3)/(4a)]

1=3-(9a-3)/(4a)

解得a=3

因此f(x)=3x^2-9x+6