If the graph of y=ax^2+bx+c passes through the points(-1,12),(0,5) and (2,-3) find the value of a+b+c

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- Feb 25th 2007, 03:45 PMGodfathergraphing problem
If the graph of y=ax^2+bx+c passes through the points(-1,12),(0,5) and (2,-3) find the value of a+b+c

- Feb 25th 2007, 04:41 PMJhevon
Here's how we go about it. We know that the y-intercept occurs when x is 0. And that in a polynomial of this form, the constant is the y-intercept. We see that one point is (0,5), that means when x is 0, y is 5, this implies c = 5. Now we use the other two points to obtain simultaneous equations.

using (x,y) = (-1,12) we get

a - b + 5 = 12..................................(1)

using (x,y) = (2,-3) we get

4a + 2b + 5 = -3...............................(2)

Thus our equations are:

a - b = 7...........................(1)

4a + 2b = -8......................(2)

So

a - b = 7............................(1)

2a + b = -4.........................(3) which we get by equation(2)/2

Adding these equations we obtain

3a = 3................................(1)+(3)

so a = 1

since a - b = 7

=> b = -6

so the numbers are a = 1, b = -6 and c = 5