Find the quadratic function y=ax^2 + bx + c whose graph passes through the points (-1,9)(-3,1)(2,-9)I have been working on this problem for quite a...

Substitute the three points into the
equation

x = -1, y = 9
9 = a(-1)^2 +
b(-1) + c  which simplifies to   9 = a - b + c

x = -3, y =
1
1 = a(-3)^2 + b(-3) + c  which simplifies to 1 = 9a -3b +
c

x = 2, y = -9
-9 = a(2)^2 + b(2) + c    which
simplifies to -9 = 4a + 2b + c

So now we have a set of 3
equations which we can solve using elimination

a -   b + c
=  9  (1)
9a - 3b + c =  1  (2)
4a+ 2b + c = -9 
(3)

(2) - (1)
9a - 3b + c = 1
-a +  b
-  c = -9
----------------------
8a - 2b = -8  Which we can reduce
to 4a - b = -4  (4)

(3) - (1)
4a + 2b + c =
-9
-a +   b -  c = -9
----------------------
3a + 3b =
-18  Which we can reduce to a + b = -6 (5)

Now compute (4)
+ (5)

4a - b = -4
a +b =
-6
------------------
5a  = -10  solving for a we get a =
-2
Using (5) we get  -2 + b = -6, or b = -4
Using (1) we get -2 - 
-4 + c =  9, c = 7

So our function is
f(x) =
-2x^2 - 4x + 7
Checking:
f(2) = -2(2)^2 - 4(2) + 7 = -8 - 8 + 7 =
-9  gives point (2,-9)
f(-1) = -2(-1)^2 - 4(-1) + 7 = -2 + 4 + 7 = 9 gives
point (-1,9)
f(-3) = -2(-3)^2 - 4(-3) + 7 = -18 + 12 + 7 = 1 gives point
(-3,1)
Which are the points we were given.
Again our solution is
f(x) = -2x^2 - 4x + 7