Convert to standard form, which is #y = ax^2 + bx + c, a !=0#.

#y = 2(x - 3)^2 - x + 3#

#y = 2(x^2- 6x + 9) - x + 3#

#y = 2x^2 - 12x + 18 - x + 3#

#y = 2x^2 - 13x + 21#

Now, to determine the vertex, convert to vertex form, which is #y = a(x - p)^2 + q, a !=0#

#y = 2(x^2 - 13/2x + m - m)^2 + 21#

The goal here is to convert to a perfect square. #m# is given by #(b/2)^2#, where #b = (ax^2 + bx + ...) inside the parentheses.

#m = ((-13/2)/2)^2 = 169/16#

#y = 2(x^2 - 13/2x + 169/16 - 169/16) + 21#

#y = 2(x^2 - 13/2x+ 169/16) - 169/8 + 21#

#y = 2(x- 13/4)^2 - 1/8#

In vertex form, #y = a(x - p)^2 + q, a !=0#, the vertex is located at #(p, q)#. Hence, the vertex is at the coordinates #(13/4, -1/8)#.

Hopefully this helps!