(1)
f(-4)=f(0)=3
16-4b+c=3
f(-2)=-1
4-2b+c=-1
b=4
c=x²+4x+3
(2)
图像略
定义域x≥-4
-4≤x<0
f(x)≤3
x≥0
f(x)≤3
综上所述
f(x)≤3
(3)
x>0
-x+3<0
x>3
-4≤x<0
x²+4x+3>0
x>-1或x<-3
综合上述情况
x>3或-1
对称轴x=-a
-a≤1即a≥-1
f(x)max=f(3)
f(x)min=f(1)
-a≥3即a≤-3
f(x)max=f(1)
f(x)min=f(3)
(1+3)/2≤-a≤3即-3≤a≤-2
f(x)max=f(1)
f(x)min=f(-a)
1≤-a≤(1+3)/2即-2≤a≤-1
f(x)max=f(3)
f(x)min=f(-a)
3
对称轴x=1
t+2≤1即t≤-1
f(x)max=f(t)
f(x)min=f(t+2)
t≥1
f(x)max=f(t+2)
f(x)min=f(t)
[t+(t+2)]/2<1,t<1,t+2>1
f(x)max=f(t)
f(x)min=f(1)
[t+(t+2)]/2>1,t<1,t+2>1
f(x)max=f(t+2)
f(x)min=f(1)那第四题呢？嘿嘿~第四题?????