# Parabola Equation

• November 9th 2011, 11:49 PM
squelia
Parabola Equation
So here's my question: (by the way, please don't penalise me for putting it in the wrong forum - i didn't see one for graphs!)

The cross section of a hayshed is in the shape of an inverted parabola. It can be
modelled by the equation:

h = 0.3w(8 - w)

where w is the width of the hayshed, and h is the height. Solve w (width).

Please explain how you worked it out?
• November 10th 2011, 12:09 AM
Deveno
Re: Parabola Equation
multiply the right side out, and then "complete the square":

$ax^2 + bx = ax^2 + 2a(\frac{b}{2a})x$

$= a(x^2 + 2(\frac{b}{2a})x) = a(x^2 + 2(\frac{b}{2a})x + (\frac{b}{2a})^2 - (\frac{b}{2a})^2)$

$= a(x + \frac{b}{2a})^2 - \frac{b^2}{4a}$

(except you're using "w" instead of "x". can you figure out what your "a" and "b" are?)

get the square on one side, and everything else on the other, take the square root, and then subtract to "isolate" your variable (in this case, w).