已知函数f(x)=x³+ax²+bx+c在x=-2处取得极值,且它的图像与直线y=-3x+3在点（1

f'(x) = 3x² + 2ax + b
在x=-2处取得极值,f'(-2) = 12 - 4a + b = 0 (1)
图像过(1,0):f(1) = 1 + a + b + c = 0 (2)
在点（1,0）处斜率=-3:f'(1) = 3 + 2a + b = -3 (3)
(3)-(1):-9 + 6a = -3,a = 1
带入(3):b = -8
带入(2):c = 6
f(x) = x³+x²-8x+6