f'(x)=12x^2+2ax+b
f'(1)=12+2a+b
f(1)=9+a+b
于是函数在点（1,f（1））处的切线为
y-f(1)=f'(1)(x-1)
y=f'(1)(x-1)+f(1)
=f'(1)x-f'(1)+f(1)
=(12+2a+b)x-(a+3)
与y=-12x 对比得
12+2a+b=-12,a+3=0
于是 a=-3,b=-18
故 f（x）=4x³-3x²-18x＋5