the answer:
the vertex formula is V( -b/2a, f((-b/2a) )
so if The graph of y=ax^2+bx+c is a parabola that opens up and has a vertex at (0,5), then -b/2a = 0, implies b = 0, and when b =0 f((-b/2a = 0/2a=0))=f(0)
in our case, f(x)=ax^2+bx+c, and it implies f(0) = c ≠ 5
so there is no value of a, b and c which verifies the equation of vertex, the set solution is {∅} (the empty set)
