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About LinkBacksI'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 isOriginally Posted by eliminator
$\displaystyle y = a(x - p)^2 + q$
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
$\displaystyle y = -b(x - p)^2 + q$
and let's make "b" positive, just to remind us.
It has a vertex $\displaystyle \left ( \frac{n}{2}, h \right ) $ so your parabola is
$\displaystyle y = -b \left ( x - \frac{n}{2} \right ) ^2 + h$
Now, the parabola passes through (0, 0) so...
$\displaystyle 0 = -b \left ( 0 - \frac{n}{2} \right ) ^2 + h$
and it passes through (n, 0), so....
$\displaystyle 0 = -b \left ( n - \frac{n}{2} \right ) ^2 + h$
So we need to solve the simultaneous equations for h and b in terms of n:
$\displaystyle 0 = -\frac{bn^2}{4} + h$
$\displaystyle 0 = -\frac{bn^2}{4} + h$
So
$\displaystyle b = \frac{4h}{n^2}$
Thus the parabola is:
$\displaystyle y = - \frac{4h}{n^2} \cdot \left ( x - \frac{n}{2} \right ) ^2 + h$
-Dan
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:
$\displaystyle y = -a(x - n)(x - 0) = -ax(x - n)$
(The outside coefficient must be negative for it to open downward.)
We know it must pass through the vertex $\displaystyle \left ( \frac{n}{2}, h \right )$:
$\displaystyle h = -a \frac{n}{2} \cdot \left ( \frac{n}{2} - n \right )$
$\displaystyle h = \frac{an^2}{4}$
Solve this for a:
$\displaystyle a = \frac{4h}{n^2}$
Thus the parabola is
$\displaystyle y = - \frac{4h}{n^2} \cdot x(x - n)$
You can verify that $\displaystyle \left ( \frac{n}{2}, h \right )$ is indeed the vertex of this parabola.
-Dan
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