Question 707573
{{{p(x)=-x^3+bx^2+cx+d}}}


derivative of p(0) must be zero because of tangent at the origin.  Taking derivative, setting it equal zero, seems c=0.


p(2)=8, so carrying this through did not seem to help.  8=-8+4b+d, not too useful.


Point (0,0) must satisfy our p(x) polynomial, so {{{0=-0^3+b*0^2+d}}},
apparantly d=0.


So far we seem to have {{{p(x)=-x^3+bx^2}}}.
Let's try point (2,8) again.  
{{{p(x)=-(x^2)(x-b)}}}, also.
Using the point, {{{8=-(2)^2*(2-b)}}}
.
b=4


THIS should work well:  {{{p(x)=-(x^2)(x-4)}}}