How can the vertex be found in a parabolic curve equation?

[karmatic]

Parabolic Curve Question!

Homework Statement

The parabolic curve described by the cable of a suspension bridge, as shown in the diagram below, is

y=12(x/20-1)^2+3

where x is the distance in metres measured from one end of the
bridge. The deck, or load-bearing portion, is at y = 0 and is hung
below the suspension cable on vertical suspenders. At either end
of the span shown the suspension cable is anchored to identical
supporting pillars. What distance does the bridge span between supporting pillars? How high are the supporting pillars at either end of the span? How long is the vertical suspender in the centre of the span?

Homework Equations

y=ax^2+bx+c

The Attempt at a Solution

Should the co-ordinates of the vertex be 1,3 or 20,3?. The value for a is the 12, but I have no idea how to find the values for b or c! The equation given is the vertex form, I found a formula to convert it back to the standard form which is just to expand and simplify the binomial, so I'm wanting to take (x/20-1)^2 and expand it but I'm not sure if I should be substituting an arbitrary value for x or not! I'm also not sure of how to incorperate the /20 into it, I know if that weren't there it would simply be (x-1)(x-1)=x^2-2x+1.

Can someone shed a little light on this one for me please!?

 

[Unit]

I suggest drawing a picture. As for finding the vertex, consider the following:

$$y = 12 (\frac{x}{20}-1)^2 + 3$$

Factor 1/20:
$$y = 12 (\frac{1}{20} (x-20))^2 + 3$$

Square 1/20:
$$y = \frac{12}{400} (x-20)^2 + 3$$

Reduce fractions:
$$y = \frac{3}{100} (x-20)^2 + 3$$

This tells you that the vertex is located at (20, 3).