SOLUTION: find a quadratic model for the set of values (-2,-15),(0,-3),(4,-27)

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Question 173547: find a quadratic model for the set of values (-2,-15),(0,-3),(4,-27)
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
-15=A%28-2%29%5E2%2BB%28-2%29%2BC
-3=A%280%29%5E2%2BB%280%29%2BC
-27=A%284%29%5E2%2BB%284%29%2BC
:
-15=4A-2B+C...eq 1
highlight%28-3+=C%29.........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%28A=-2%29
:
plug A's found value back into eq 4.--> 8(-2)-4B=-24
:
-4b=-8
:
highlight%28B=2%29
so
since y=Ax%5E2%2BBx%2BC...plug in the found values
highlight%28y=-2x%5E2%2B2x-3%29