若二次函数fx=ax^2+bx+2013满足fx1=fx2(其中x1≠x2),则f(x1+x2)=
问题描述:
若二次函数fx=ax^2+bx+2013满足fx1=fx2(其中x1≠x2),则f(x1+x2)=
答
x1+x2=-b/a,
f(x1+x2)=2013
答
∵f(x1)=f(x2),x1≠x2
∴(x1+x2)/2=-b/2a
则x1+x2=-b/a
f(x1+x2)=a(x1+x2)²+b(x1+x2)+2013
=a·(-b/a)²+b·(-b/a)+2013
=b²/a-b²/a+2013
=2013