Question 378095
The vertex form is {{{y = a(x - h)^2+k}}}.  Since the line of symmetry is x = 3, then the x-ccordinate of the vertex is h = 3.  Thus the equation now is {{{y = a(x-3)^2 +k}}}. 
We have to find a and k.  Use the coordinates of the given point.
From (5, -3):  {{{-3 = a(5-3)^2 + k}}}, or -3 = 4a + k, or k = -4a-3.
From (-1, 9):  {{{9 = a(-1-3)^2 + k}}}, or 9 = 16a + k, or k = 9 - 16a.

Hence -4a - 3 = 9 - 16a.
12a = 12, or a = 1.

Then k = -4*1-3 = -7.  Therefore the equation of the parabola is 

{{{y = (x - 3)^2 - 7}}}.