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

##### 1 Answer

#### 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.

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

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

We have a system of equations now:

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

Multiply first row by

Add

Add

Multiply second row by

Add

Add

Doing this series of operations to the solution vector gives:

So reading off the solutions we have

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