已知f（m）＝㎡+（lga+2）x+lgb,f（-1）＝-2.当x∈R,f（x）≥2x恒成立,求实数a的值与f（x）的最小值.

f(x)=x^2+(lga+2)x+lgb,f(-1)=-2
f(-1)=1-(lga+2)+lgb=-2
①lga-lgb=1.lgb=lga-1
f(x)=x^2+(lga+2)x+lgb≥2x
x^2+(lga+2)x-2x+lgb≥0
x^2+(lga+2-2)x+lgb≥0
②lga^2-4lgb≥0
lga^2-4lga+4≥0
(lga-2)^2≥0.lga≥2=lg10(2)=lg100
当lga=2时最小.lgb=2-1=1
f(x)=x^2+(lga+2)x+lgb
=x^2+4x+1
=x^2+4x+4-4+1=(x+2)^2-3
此时f(x)的最小值为-3