Question 1009400
A simple sketch and reference to STANDARD FORM instead of the format that you show, will allow you to identify the vertex, a maximum point, and the zeros of the equation.  You can then find the factor "a", again in reference to {{{y=a(x-h)^2+k}}}.


Assuming that discussion makes sense for you,
{{{highlight_green(y=a(x-100)^2+500)}}}.
Vertex is  (100, 500).
Zeros are x at 0 and at 200.


What about the factor a ?
{{{a(x-100)^2=y-500}}}
{{{a=(y-500)/(x-100)^2}}}
Use coordinates of either of the zeros....
{{{a=(0-500)/(200-100)^2}}}
{{{a=-500/10000}}}
{{{a=-5/100}}}
or simpler
{{{a=-1/20}}}
-
The resulting STANDARD FORM equation is  {{{highlight(y=-(1/20)(x-100)^2+500)}}}.


The factor of 4 shown in the model you give, does not seem correct.  In fact, the factor shown there,  "4x", does not fit a quadratic equation, nor parabola.



ANSWERS FROM THIS:
h=100
k=500
What is c supposed to be? What is this supposed to mean?



---
If you meant to write the form,  {{{(x-h)^2=4p(y-k)}}}, this is the typical result when deriving the equation of a parabola using a given vertex  (h,k) and directrix  p units from the vertex and focus p units from the vertex but on the other side from the vertex.  If your book does not have this derivation, then you can find a video showing it:  <a href="https://www.youtube.com/watch?v=Wworlx39KfQ">vertex of parabola not at origin - derivation of equation using focus and directrix</a>