设f(x)=ax^2+(b+3)x+b的图像关于y轴对称,定义域为[a-1,2a],求f(x)的值域
图象对称则定义域对称:
a-1==-2a,
a=1/3,
f(x)=ax^2+(b+3)x+b
= 1/3 x^2+(b+3)x+b
= b - 3/4 (3 + b)^2 + 1/3 ((3 (3 + b))/2 + x)^2
对称轴 x = 3 (3 + b))/2 = 0,
b= -3,
f(x)=1/3 x^2+(-3+3)x -3
= -3 + x^2/3
≥-3
值域[-3,+∞).