# 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?

Jun 2, 2015

I would build a system of three equations in the three unknowns $a , b , c$ and solve it using Cramer's Method:

Jun 2, 2015

The question gives us 3 simultaneous equations to solve:

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

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

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

Subtracting a couple of these we find:

$30 = - 16 + 46 = f \left(3\right) - f \left(- 3\right)$

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

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

Then

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

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

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

Add $10$ to both ends to get:

$- 48 = 16 a$

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

Then

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

Subtract $2$ from both ends to get:

$c = - 4$

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

Check:

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

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

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