Question 1044081
Turning points of a curve are points where the derivative {{{dy/dx = 0}}}.

===> the derivative {{{dy/dx=a(x-p)(x-q)}}} is actually {{{ dy/dx=a(x-2)(x-1)}}},

and wouldn't matter whether p=2 and q=1, or p=1 and q=2.

===> {{{ dy/dx=a(x-2)(x-1) = a(x^2-3x+2)}}},

===> {{{y = a(x^3/3-3x^2/2+2x)+c}}}, after getting the integral.

Plugging in the respective coordinates of the point (2,0) into the last equation gives 

{{{0 = 2a/3+c}}}.   (Verify!)

Similarly, for the point (1,1), we get 

{{{1 = 5a/6 + c}}}.   (Again verify!)

===> {{{highlight(a = 6)}}}, c = -4, and the curve itself is 

{{{y = 2x^3 - 9x^2 + 12x -4}}}.