Question

已知f(x)=x^3+bx^2+cx+d图像过点p(1,-3),且在点(0,f(0))的切线方程为9x+y-2=0,求常

Answer 1

f(x)=x^3+bx^2+cx+d

f(0)=d,即切点坐标为(0,d)

f′(x)=3x^2+2bx+c

f(x)图像过点p(1,-3)

1+b+c+d=-3 ①

f(x)在点(0,f(0))的切线方程为9x+y-2=0

f′(0)=-9且(0,d)在直线9x+y-2=0上

c=-9 ②

d-2=0 ③

因此c=-9,d=2,b=3