Question 1027710
{{{f(x) = x^3+ax^2+bx+c }}}

==> f'(x) = {{{3x^2+2ax + b}}} ==> f"(x) = 6x + 2a

Setting f" to zero, we get {{{x = -a/3}}}.
To the right of -a/3, f" >0, while to its left f" < 0.  (a > 0!)

==> To the right of -a/3, f is concave up, while to its left f is concave down.

Hence {{{x = -a/3}}} is an inflection point of f.

If (0,-2) is an inflection point then {{{-a/3 = 0}}}, and so a = 0.

==> {{{f(x) = x^3+bx+c }}} ==> {{{f(0) = 0^3+b*0+c  = -2}}} ==> c = -2

==> {{{highlight(f(x) = x^3+bx-2) }}}.
==> f'(x) = {{{3x^2+ b}}}.

Since f' > 0 for all x because b > 0, f will not have any critical points, and the problem is solved.