Question 1159778
No.  The parabola has to come below the x-axis to have a circular base.
So lets bring it down below the x-axis to (0,-2) for the vertex.
Then the vertex equation is 
y = a(x-h)^2 + k
(h,k) = (0,-2)
and a point the parabola goes through above the x-axis is (6,8)
so 8 = a(6-0)^2 -2
   10=36a
    a = 10/36 = 5/18
The equation would be:  y = (5/18)(x)^2 -2