How do you find a quadratic function f(x)=ax^2+bx+c for which f(1)=-2, f(-3)=-46, and f(3)=-16?

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Answer 1 · 2015-06-02T18:16:40

I would build a system of three equations in the three unknowns $a, b, c$ $a, b, c$ and solve it using Cramer's Method:
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Answer 2 · 2015-06-02T18:32:57

The question gives us 3 simultaneous equations to solve:

$- 2 = f (1) = a + b + c$ $-2 = f(1) = a+b+c$

$- 46 = f (- 3) = 9 a - 3 b + c$ $-46 = f(-3) = 9a-3b+c$

$- 16 = f (3) = 9 a + 3 b + c$ $-16 = f(3) = 9a+3b+c$

Subtracting a couple of these we find:

$30 = - 16 + 46 = f (3) - f (- 3)$ $30 = -16+46 = f(3) - f(-3)$

$= (9 a + 3 b + c) - (9 a - 3 b + c) = 6 b$ $= (9a+3b+c)-(9a-3b+c) = 6b$

So $b = \frac{30}{6} = 5$ $b = 30/6 = 5$

Then

$- 58 = - 46 - 16 + 4 = f (- 3) + f (3) - 2 f (1)$ $-58 = -46-16+4 = f(-3)+f(3)-2f(1)$

$= (9 a - 3 b + c) + (9 a + 3 b + c) - 2 (a + b + c)$ $=(9a-3b+c)+(9a+3b+c)-2(a+b+c)$

$= 16 a - 2 b = 16 a - 10$ $=16a-2b = 16a-10$

Add $10$ $10$ to both ends to get:

$- 48 = 16 a$ $-48 = 16a$

So $a = - \frac{48}{16} = - 3$ $a = -48/16 = -3$

Then

$- 2 = f (1) = a + b + c = - 3 + 5 + c$ $-2 = f(1) = a+b+c = -3 + 5 + c$ = 2 + c#

Subtract $2$ $2$ from both ends to get:

$c = - 4$ $c = -4$

So $f (x) = - 3 x^{2} + 5 x - 4$ $f(x) = -3x^2+5x-4$

Check:

$f (1) = - 3 + 5 - 4 = 5 - 7 = - 2$ $f(1) = -3+5-4 = 5-7 = -2$

$f (- 3) = - 3 \cdot 9 - 3 \cdot 5 - 4 = - 27 - 15 - 4 = - 46$ $f(-3) = -3*9-3*5-4 = -27-15-4 = -46$

$f (3) = - 3 \cdot 9 + 3 \cdot 5 - 4 = - 27 + 15 - 4 = - 16$ $f(3) = -3*9+3*5-4 = -27+15-4 = -16$