Quadratic Functions Problem

Question

Question: The main section of a certain bridge has cables in the shape of a parabola. Suppose that the points on the tops of the towers where the cables are attached are 168m apart and 24 vertically above the minimum height of the cables.

• Choose two other locations for the origin. Write the corresponding quadratic function for the shape of the cables for each.

So far I have found that the vertex form that represents the shape of the cables is 1/294x^2. When I try to attempt the question I listed above I get a completely wrong answer and don't know where I went wrong, so I'm assuming I must be using a wrong origin? Anyways, thanks to anyone who can help.

Answer 1

You got the right answer, namely $y = \frac 1 {294} x^2$, if, and only if, the origin is at the lowest point of the cable and in the middle point between the two towers. In this case, the vertex of the parabola coincides with the origin, so we have $V = (0, 0)$.

If you choose another point as the origin, your vertex will not have coordinates $(0, 0)$. This time, the parabola will have general form $y = a(x - x_v)^2 + y_v$, where $x_v$ and $y_v$ are the coordinates of the vertex. Put in another way, we have $V = (x_v, y_v)$. Let's say we choose as origin the point which lies below the leftmost tower, at the same level as the lowest point of the cable:

Now, how do we determine the equation of the parabola with respect to the new system of axes? Well, the coefficient $a$ is obviously the same, because the parabola is the same. What's changed? The vertex position! The vertex is now $V = (84, 0)$, yielding the equation $y = \frac 1 {294} (x - 84)^2 = \frac 1 {294} x^2 - \frac 47 x + 24$.

If you have to do it with another origin, the easiest case is to take the rightmost tower. I'm sure you can do it. If you have any doubts, do not hesitate to ask.

Answer 2

vertex form is $ y = a(x-h)^2 + k$ where (h,k) is the vertex. The easiest vertex would obviously be (0,0):

$y = a(x-0)^2 + 0$
$ y = ax^2$
$ 24 = a(84)^2$
$a = \frac{1}{294}$

final form of $y = \frac{1}{294}x^2$

Another easy vertex would be (0,1), in which the equation would merely have to be shifted 1 up, which could be accomplished by:
$y = \frac{1}{294}x^2 + 1$

or
$y = a(x-0)^2 + 1$
$ y = ax^2+1$
$ 24+1 = a(84)^2+1$ (24 + 1 because ends are 24 above middle, which I set at 1)
$a = \frac{1}{294}$ so with a vertex at (0,1), the equation would be:
$y = \frac{1}{294}x^2 + 1$

This is my first post, so any comments for improvements are appreciated.