For an equation in the form #y=a(x-h)^2+k#, while axis of symmetry is #x-h=0#, vertex is #(h,k)#.

As we can write #y=x^2-3# as #y=(x-0)^2+(-3)#, axis of symmetry is #x=0# i.e. #y#-axis and vertex is #(0,-3)#.

For #x#-intercept, we should put #y=0# i.e. #x^2=3# and hence #x#-intercepts are at #(sqrt3,0)# and #(-sqrt3,0)#. For #y#-intercepts we should put #x=0# and only #y#-inercept is at #(0,-3)#.

While there are no limitations on #x# and hence domain for #y=x^2-3#, is #(-oo,oo)#. But as #x^2>=0#, #y# cannot be less than #-3# and hence range is #[-3,oo)#.

graph{ x^2-3 [-10, 10, -5, 5]}