The three points should be sufficient to find the three unknowns: a, b and c.
Putting the given points into y=ax^2+bx+c we have:
A. -1=4a+2b+c
B. 2=16a+4b+c
C. -1=36a+6b+c
Equating A and C (because they both equal -1):
D. 0=32a+4b; 8a+b=0, so b=-8a
-4A-B: E. -6=4b+3c, so c=(-6-4b)/3=(-6+32a)/3 or (32a-6)/3
We now have b and c in terms of a. Using equation A we can substitute for b and c:
-1=4a-16a+(32a-6)/3; -3=-36a+32a-6; 3=-4a so a=-3/4. Now we can find b and c:
b=-8(-3/4)=6; c=(32a-6)/3=(-24-6)/3=-10, and y=-(3/4)x^2+6x-10 or 4y=-3x^2+24x-40.