equation of a specific parabola

eliminator · Aug 7th 2007, 10:59 AM

I'm stumped. I need the equation of a parabola that opens down, passes through the origin, passes through the point (n,0), and has vertex (n/2,h). Thanks in advance.

**topsquark** · Aug 7th 2007, 11:24 AM

Originally Posted by eliminator

I'm stumped. I need the equation of a parabola that opens down, passes through the origin, passes through the point (n,0), and has vertex (n/2,h). Thanks in advance.

Well, the form of a vertically opening (either up or down) parabola is
$y = a(x - p)^2 + q$
where (p, q) is the vertex. (We usually use (h, k) as the vertex, but you already have an "h" in the problem.)

It opens down, so a is negative. So let's write this as
$y = -b(x - p)^2 + q$
and let's make "b" positive, just to remind us.

It has a vertex $\left ( \frac{n}{2}, h \right )$ so your parabola is
$y = -b \left ( x - \frac{n}{2} \right ) ^2 + h$

Now, the parabola passes through (0, 0) so...
$0 = -b \left ( 0 - \frac{n}{2} \right ) ^2 + h$
and it passes through (n, 0), so....
$0 = -b \left ( n - \frac{n}{2} \right ) ^2 + h$

So we need to solve the simultaneous equations for h and b in terms of n:
$0 = -\frac{bn^2}{4} + h$

$0 = -\frac{bn^2}{4} + h$

So
$b = \frac{4h}{n^2}$

Thus the parabola is:
$y = - \frac{4h}{n^2} \cdot \left ( x - \frac{n}{2} \right ) ^2 + h$

-Dan

**topsquark** · Aug 7th 2007, 11:28 AM

Originally Posted by eliminator

I'm stumped. I need the equation of a parabola that opens down, passes through the origin, passes through the point (n,0), and has vertex (n/2,h). Thanks in advance.

Another way to do this is to note that two of the points we have on this parabola are the x-intercepts: (0, 0) and (n, 0), so the parabola must have the form:
$y = -a(x - n)(x - 0) = -ax(x - n)$
(The outside coefficient must be negative for it to open downward.)

We know it must pass through the vertex $\left ( \frac{n}{2}, h \right )$ :
$h = -a \frac{n}{2} \cdot \left ( \frac{n}{2} - n \right )$

$h = \frac{an^2}{4}$

Solve this for a:
$a = \frac{4h}{n^2}$

Thus the parabola is
$y = - \frac{4h}{n^2} \cdot x(x - n)$

You can verify that $\left ( \frac{n}{2}, h \right )$ is indeed the vertex of this parabola.

-Dan

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