Question

设函数f(x)=ax2+bx+c (a>0)且f(1)=-a/2

Answer 1

(1)

x=1,f(x)=-a/2代入函数方程：

a+b+c=-a/2

b=-3a/2-c

对于方程ax^2+bx+c=0,由韦达定理,得

x1+x2=-b/a

x1x2=c/a

(x1-x2)^2

=(x1+x2)^2-4x1x2

=(-b/a)^2-4c/a

=(b^2-4ac)/a^2

=9/4-c/a+(c/a)^2

=[(c/a)-1/2]^2+2≥2

|x1-x2|≥√2

(2)

a>0

f(0)=c

f(1)=-a/20,f(1)