Question 12588
You are correct.  Remember that you are trying to find a family of parabolas, which means that there are perhaps an infinite number of such parabolas.  This is what happens when you have three unknowns and only two conditions (i.e., equations) to satisfy.  


Both equations come down to A + C = -1.  So, let B=1, and solve for either C or A in terms of the other.  Say, C= -A -1.


The equation is the family of parabolas:
{{{y= Ax^2 + Bx + C}}}


{{{y = Ax^2 + x -A -1}}}, where A represents any non-zero number.  Of course, if A = 0, then it is NOT a parabola.


Any other value of A should give you points that include (1,0) and (-1,-2).


For examples, let A = 1:  {{{y = x^2 + x -2}}} Do both points satisfy this?  If x =1, then y= 0, and if x=-1, then y =-2.


Let A = 2:  (((y = 2x^2 +x -3}}}


Let A = -1:  {{{y = -x^2 + x }}} 


R^2 at SCC