# How do you find a quadratic function f(x) = ax² + bx + c given minimum value -4 when x=3; one zero is 6?

Jul 9, 2016

$f \left(x\right) = \frac{4}{9} {x}^{2} - \frac{8}{3} x$

#### Explanation:

Quadratic functions are symmetric about their vertex line, ie at x = 3 so this implies the other zero will be at x = 0.

We know the vertex occurs at x = 3 so the first derivative of the function evaluated at x = 3 will be zero.

$f ' \left(x\right) = 2 a x + b$

$f ' \left(3\right) = 6 a + b = 0$

We also know the value of the function itself at x=3,

$f \left(3\right) = 9 a + 3 b + c = - 4$

We have two equations but three unknowns, so we'll need another equation. Look at the known zero:

$f \left(6\right) = 0 = 36 a + 6 b + c$

We have a system of equations now:

$\left(\begin{matrix}6 & 1 & 0 \\ 9 & 3 & 1 \\ 36 & 6 & 1\end{matrix}\right) \left(\begin{matrix}a \\ b \\ c\end{matrix}\right) = \left(\begin{matrix}0 \\ - 4 \\ 0\end{matrix}\right)$

To read off the solutions we want to reduce our coefficient matrix to reduced echelon form using elementary row operations.

Multiply first row by $\frac{1}{6}$

$\left(\begin{matrix}1 & \frac{1}{6} & 0 \\ 9 & 3 & 1 \\ 36 & 6 & 1\end{matrix}\right)$

Add $- 9$ times the first row to the second row:

$\left(\begin{matrix}1 & \frac{1}{6} & 0 \\ 0 & \frac{3}{2} & 1 \\ 36 & 6 & 1\end{matrix}\right)$

Add $- 36$ times the first row to the third:

$\left(\begin{matrix}1 & \frac{1}{6} & 0 \\ 0 & \frac{3}{2} & 1 \\ 0 & 0 & 1\end{matrix}\right)$

Multiply second row by $\frac{2}{3}$

$\left(\begin{matrix}1 & \frac{1}{6} & 0 \\ 0 & 1 & \frac{2}{3} \\ 0 & 0 & 1\end{matrix}\right)$

Add $- \frac{2}{3}$ times the third row to the second row:

$\left(\begin{matrix}1 & \frac{1}{6} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$

Add $- \frac{1}{6}$ times the second to the first

$\left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$

Doing this series of operations to the solution vector gives:

$\left(\begin{matrix}\frac{4}{9} \\ - \frac{8}{3} \\ 0\end{matrix}\right)$

So reading off the solutions we have $a = \frac{4}{9} \mathmr{and} b = - \frac{8}{3}$

$f \left(x\right) = \frac{4}{9} {x}^{2} - \frac{8}{3} x$

graph{4/9x^2 - 8/3x [-7.205, 12.795, -5.2, 4.8]}