# 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 while now, I can never get the same answer twice...

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 while now, I can never get the same answer twice and keep getting strange fraction answers! I understand that you should plug in the numbers and then eliminate the variables but I can not get it to work! Help please!

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Substitute the three points into the equation

# (-1,9)(-3,1)(2,-9)

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

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8a - 2b = -8 Which we can reduce to 4a - b = -4 (4)

(3) - (1)

4a + 2b + c = -9

-a + b - c = -9

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3a + 3b = -18 Which we can reduce to a + b = -6 (5)

Now compute (4) + (5)

4a - b = -4

a +b = -6

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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