二次函數的值域及極值

求以下各二次函數的值域及極值：
1. f(x) = x^2+3x+5
Sol
x^2+3x+5
=(x^2+3x+9/4)－9/4+5
=(x+3/2)^2+11/4
(x+3/2)^2>=0
(x+3/2)^2+11/4>=11/4
f(x)>=11/4

2. f(x) = x^2+5x+3
Sol
x^2+5x+3
=(x^2+5x+25/4)－25/4+3
=(x+5/2)^2－13/4
(x+5/2)^2>=0
(x+5/2)^2-13/4>=－13/4
f(x)>=－13/4

3. f(x) = 2x^2－3x－6 (x >=－1)
Sol
2x^2－3x－6
=2(x^2－3x/2)－6
=2(x^2－3x/2+9/16)－9/8－6
=2(x－3/4)^2－57/8
x>=－1
x－3/4>=－7/4
(x－3/4)^2>=0
2(x－3/4)^2>=0
2(x-3/4)^2－57/8>=－57/8
f(x)>=－57/8

4. f(x) = 3x^2+7x+3 (－3<= x<=－2 或 －1<=x<=0)
Sol
3x^2+7x+3
=3(x^2+7x/3)+3
=3[x^2+7x/3+(7/6)^2]－3*49/36+3
=3(x+7/6)^2－39/36
=3(x+7/6)^2－13/12
－3<=x<=－2
－3+7/6<=x+7/6<=－2+7/6
－11/6<=x+7/6<=－5/6
25/36<=(x+7/6)^2<=121/36
25/12<=3(x+7/6)^2<=121/12
1<=3(x+7/6)^2－13/12<=9
1<=f(x)<=9………………….(1)
－1<=x<=0
1/6<=x+7/6<=7/6
1/36<=(x+7/6)^2<=49/36
1/12<=3(x+7/6)^2<=49/12
－1<=3(x+7/6)^2－13/12<=3
－1<=f(x)<=3………….…(2)
綜合(1)(2)
－1<=f(x)<=9