Question 173547
{{{-15=A(-2)^2+B(-2)+C}}}
{{{-3=A(0)^2+B(0)+C}}}
{{{-27=A(4)^2+B(4)+C}}}
:
-15=4A-2B+C...eq 1
{{{highlight(-3 =C)}}}.........eq 2
-27=16A+4B+C..eq 3
:
lets take C's value in eq 2 and plug it into eq 1 and 3
:
4A-2B+(-3)=-15....eq (4)---->4A-2B=-12----mult by 2---->8A-4B=-24
16A+4B+(-3)=-27...eq (5)---->16A+4B=-24
:
8A-4B=-24.....eq (4) revised
16A+4B=-24....eq (5) revised
:
now lets solve by elimination after we multiplied eq 4 by 2 .   Add eq 4 and 5 together and as you can see the B terms will be eliminated(-4B+4B=0) and we are left with
:8A+16A=-24-24
:
24A=-48
:
{{{highlight(A=-2)}}}
:
plug A's found value back into eq 4.--> 8(-2)-4B=-24
:
-4b=-8
:
{{{highlight(B=2)}}}
so 
since {{{y=Ax^2+Bx+C}}}...plug in the found values
{{{highlight(y=-2x^2+2x-3)}}}