- 31 Oct '11 08:22The graph of f(x) = ax^2 + bx + c is a parabola if a does not equal zero. The effect of a and c on the graph are fairly simple, a can flip it upside down and make it tighter or wider, c moves the apex in the direction of the y-axis.

b is less obvious. It would seem to move the apex of the parabola along a curve, but what curve? - 31 Oct '11 12:17

Hint:*Originally posted by talzamir***The graph of f(x) = ax^2 + bx + c is a parabola if a does not equal zero. The effect of a and c on the graph are fairly simple, a can flip it upside down and make it tighter or wider, c moves the apex in the direction of the y-axis.**

b is less obvious. It would seem to move the apex of the parabola along a curve, but what curve?It\'s not, strictly speaking, a curve.

Richard - 31 Oct '11 22:23 / 2 editsy = ax^2 + bx + c

dy/dx = 2ax + b

At the apex this is zero

2ax + b = 0

x = -b/(2a) at the apex

y = ax^2 + bx + c

at the apex:

y = b^2/(4a) - b^2/(2a) + c

y = -b^2/(4a) + c

Since the intercept y scales with b^2 and the intercept x scales with b, b moves the intercept along a parabola - 01 Nov '11 06:38 / 1 editIndeed it does - nicely done.

Putting it into a more familiar format of y(x) instead of the paired equations of y(b) and x(b).

Coordinates of the apex being, as you solved it,

x = -b/(2a)

y = -b^2/(4a) + c

Squaring the first equation gives x^2 = b^2/(4a^2).

Tweaking the second equation and then substituting x into it gives

y = -b^2/(4a) + c = -a * b^2 / (4a^2) + c = -ax^2 + c

So it seems not just any parabola, but identical in shape with the original, upside down, and has the apex at (0,c) which is where all the parabolas y = ax^2 + bx + c pass through the y-axis.