An equation of the form #y=a(x-h)^2+k# has #x-h=0# as axis of symmetry and #(h,k)# as vertex. Such an equation is a vertical parabola and if #a# is positive, it open up and if #a# is negative it opens down.
Hence in #y=2/3(x-2)^2-5/3#, we have a parabola whose vertex is #(2,5/3)#, axis of symmetry is #x-2=0# and it opens up. As it opens up, we have the minima appearing at vertex i.e. at #(2,-5/3)#.
To draw the parabola select a few points around #x=2# i.e. on both sides of axis of symmetry. Here I select values of #x# as #{-7,-4,-1,2,5,8,11}# and find the corresponding value of #y# given by
#y=2/3(x-2)^2-5/3# and I get the points through which parabola passes as #(-7,52 1/3),(-4,22 1/3),(-1,4 1/3),(2,-1 2/3),(5,4 1/3),(8,22 1/3),(11,52 1/3)#. Observe that I have selected points at an interval of #3# so that they get cancelled out as we have #a=2/3#.
Joining these points gives us the following graph and we have marked axis of symmetry and vertex too.
graph{(2/3(x-2)^2-5/3-y)(x-2)((x-2)^2+(y+5/3)^2-0.05)=0 [-18.3, 21.7, -3.04, 16.96]}
