Question 32369
If you have such a parabola, I would immediately call "Ripley's Believe it or not"
It is not possible to graph a parabola whose vertx is at (0, 16) with x-intercepts at (0, -16) and (0, 16).

I expect that you have inadvertently interchanged the x- and y-coordinates of your x-intercept points.
I think that the x-intercepts really should be: (-16, 0) and (16, 0), so we'll proceed on this assumption.

Since you are given the location of the vertex, let's use the "vertex" form of the equation for a parabola:{{{y = a(x-h)^2+k}}} for which we know that the vertex of the parabola is located at (h, k), so h = 0 and k = 16, thus wre can write:
{{{y = a(x-0)^2 + 16}}} But now we need to find the value of a.
Now, the value of the y-coordinate of the vertex is positive and the parabola crosses the x-axis, it must therefore open downward and sign of a must be negative.
{{{y = -ax^2+16}}} so let's find the value of a by substituting one of the x-intercept point into this equation. Lets use (16, 0)
{{{0 = a(16^2)+16}}} solve for a. Subtract 16 from both sides.
{{{-16 = a(16^2)}}} Divide both sides by {{{16^2}}}
{{{-16/16^2 = a}}} Simplify.
{{{a = -1/16}}} Now you can write the equation of the parabola.

{{{y = -(1/16)x^2 + 16}}}
Here's the graph:
{{{graph(300,200,-20,20,-10,20,-(1/16)x^2+16)}}}