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

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Answer 1 · 2016-07-09T12:47:49

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

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]}