Question 680660
Find vertex, axis, domain and range of  
{{{2x=y^2-4y+6}}}
complete the square
2x=(y^2-4y+4)-4+6
2x=(y-2)^2+2
2x-2=(y-2)^2
(y-2)^2=2(x-1)
This is an equation of a parabola that opens rightwards.
Its standard form: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given equation: {{{2x=y^2-4y+6}}}
vertex:(1,2)
axis of symmetry: y=2
domain: [1,∞)
range: (-∞,∞)
see graph below:

y=(2(x-1))^.5+2
{{{ graph( 300, 300, -10, 10, -10, 10,(2(x-1))^.5+2,-(2(x-1))^.5+2) }}}