Question

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

Answer 1

`f(x) = 4/9x^2 - 8/3x`

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'(x) = 2ax + b`

`f'(3) = 6a + b = 0`

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

`f(3) = 9a + 3b + c = -4`

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

`f(6) = 0 = 36a + 6b + c`

We have a system of equations now:

`((6, 1, 0),(9,3,1),(36,6,1))((a),(b),(c)) = ((0),(-4),(0))`

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

Multiply first row by `1/6`

`((1, 1/6, 0),(9,3,1),(36,6,1))`

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

`((1, 1/6, 0),(0,3/2,1),(36,6,1))`

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

`((1, 1/6, 0),(0,3/2,1),(0,0,1))`

Multiply second row by `2/3`

`((1, 1/6, 0),(0,1,2/3),(0,0,1))`

Add `-2/3` times the third row to the second row:

`((1, 1/6, 0),(0,1,0),(0,0,1))`

Add `-1/6` times the second to the first

`((1, 0, 0),(0,1,0),(0,0,1))`

Doing this series of operations to the solution vector gives:

`((4/9),(-8/3),(0))`

So reading off the solutions we have `a=4/9 and b = -8/3`

`f(x) = 4/9x^2 - 8/3x`

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