Answer:
Step-by-step explanation:
Not sure what form you need this in, but it really doesn't matter, as you'll see in the final equation. I used the vertex form and solved for a:
[tex]y=a(x-h)^2+k[/tex]
We are given the vertex (h, k) as the origin (0, 0), and we have a point that the graph goes through as (4, -64). That's our x and y. Plugging in what we have:
[tex]-64=a(4-0)^2+0[/tex] gives us
-64 = 16a and
a = -4. That means that the quadratic equation is
[tex]y=-4x^2[/tex] which is both vertex form and standard form here, no difference.