1.
函数f(x)=x^2-4ax+2a+6求函数的值域为【0,+无穷大)
可知判别式一定等于0（方程）
16a^2-4(2a+6)=0
2a^2-a-3=0
(a+1)(2a-3)=0
a=-1,a=3/2
2.
题意可知x^2-4ax+2a+6≥0恒成立
则△=16a²-4(2a+6)≤0
2a²-a-3≤0
=> -1≤a≤3/2
g(a)=2-a|a+3|
则因-1≤a≤3/2,则a+3>0
g(a)=2-a(a+3)
=-(a+3/2)²+17/4 ( -1≤a≤3/2)
g(a)min=g(3/2)=-19/4,g(a)max=g(-1)=4
g(a)∈[-19/4,4]