Question 1016425
{{{f(x)=-2x^3+bx^2+cx+k}}}.


{{{-2*0^3+b*0^2+c*0+k=2}}} meaning that {{{k=2}}}.
Revising f gives  {{{f(x)=-2x^3+bx^2+cx+2}}}, and you can form your other two equations to solve them as a system for b and c.


Hopefully you can do that.