Question 1192447
<br>
One form of the equation for a parabola with vertex (h,k) that opens upward is<br>
{{{y=(1/(4p))(x-h)^2+k}}}<br>
With the equation in that form, p is the directed distance (i.e., might be negative) from the vertex to the focus.<br>
Your graph shows the vertex at (0,2); and it shows that the directed distance from the vertex to the focus is +1.  So p=1 and the equation is<br>
{{{y=(1/(4))(x)^2+2}}}<br>
You can use the graph to verify the equation by using x=2 or x=-2 and seeing that for both of those x values the value of y is 3.<br>
Outside the scope of your question, note that 4p=4 is the length of the latus rectum -- the segment that is perpendicular to the axis of the parabola through the focus, with endpoints on the parabola.<br>