There is a simple formula that I like to use to find the #x#-coordinate of the vertex of parabolas in the form #f(x) = ax^2 + bx + c#:
#x = -b/(2a)#.
Using this formula, plug in #b# and #a# from your original function.
#x = -b/(2a)#
#x = - (-3)/(2 * 2)#
#x = 3/4#
Therefore, the #x#-coordinate of the vertex is #3/4#, and the axis of symmetry is also #3/4#. Now, plug in your value of #x# (which you have found to be the #x#-coordinate of the vertex of the parabola) to find the #y#-coordinate of the vertex.
#y = 2x^2 - 3x + 2#
#y = 2(3/4)^2 - 3 (3/4) + 2#
#y = 0.875 or 7/8#
Now you have found both the #x#- and #y#-coordinates of the vertex as well as the axis-of-symmetry, so write your answers:
Vertex = #(3/4, 7/8)#
Axis of Symmetry = #3/4#
I hope that helps!
