Question

已知二次函数f(x)=ax^2+bx+c(a≠0)满足f(-2+x)=f(-2-x)且f92）=4

Answer 1

f(-2+x)=a(x-2)^2+(x-2)b+c

f(-2-x)=a(x+2)^2-(x+2)b+c

f(-2+x)=f(-x-2)

则,(x-2+x+2)b=a[(x+2)^2-(x-2)^2]=a*4*2x=2xb

2x(4a-b)=0

4a=b.①

又,f(2)=4

即,4a+8a+c=4

12a+c=4.②

f(x)=ax^2+4ax+4-12a

对于任意实数,

f(x)-0.5(x^2+4)=ax^2+4ax+4-12a-2-0.5x^2

=a(x+6)(x-2)+0.5*(2-x)(2+x)

=(2-x)(-ax-6a+0.5x+1)

=(2-x)[(-a+0.5)x+(1-6a)]