Question 1044048
{{{dy/dx=3ax^2+4x+a^2=0}}}----Must be for {{{x=-1}}}.


Let x=-1;
{{{3a-4+a^2=0}}}, from the derivative being 0 when x is -1;



ALSO from the original equation,
{{{-a+2-a^2+b=0}}}
so the problem gives the system of equations,



{{{system(a^2+3a-4=0,a^2+a-2-b=0)}}}
Solve this system (first, for "a", and then for b).


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First equation of the system is factorable.
{{{(a-4)(a+1)=0}}}
{{{system(a=-1,or,a=4)}}}
and the description gave,  ",... has a minimum point at (-1,0)".


Either the equation becomes {{{y=-x^3+2x^2+x+b}}}  or  {{{y=4x^3+2x^2+16x+b}}}.  Do  you believe that the second-derivative might give further information about x at -1 being minimum or maximum?