Answer
$$\eqalign{
& {\text{Vertex }}\left( {h,k} \right) \to {\text{Vertex }}\left( { - 1,1} \right) \cr
& \,\,\,\,\,\,\,{\text{Horizontal axis of symmetry}} \cr
& \,\,\,\,\,\,\,{\text{Domain: }}\left[ {h,\infty } \right):\left[ { - 1,\infty } \right) \cr
& \,\,\,\,\,\,\,{\text{Range:}}\left( { - \infty ,\infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& x = 2{y^2} - 4y + 1 \cr
& x - 1 = 2{y^2} - 4y \cr
& x - 1 = 2\left( {{y^2} - 2y} \right) \cr
& x - 1 + 2\left( 1 \right) = 2\left( {{y^2} - 2y + 1} \right) \cr
& x + 1 = 2{\left( {y - 1} \right)^2} \cr
& {\left( {y - 1} \right)^2} = \frac{1}{2}\left( {x - \left( { - 1} \right)} \right) \cr
& {\text{The equation is written in the form }}{\left( {y - k} \right)^2} = 4p\left( {x - h} \right) \cr
& {\left( {y - 1} \right)^2} = \frac{1}{2}\left( {x - \left( { - 1} \right)} \right) \to k = 1,\,\,\,h = - 1 \cr
& \cr
& {\text{With: }} \cr
& \,\,\,\,\,\,\,\,{\text{Vertex }}\left( {h,k} \right) \to {\text{Vertex }}\left( { - 1,1} \right) \cr
& \,\,\,\,\,\,\,{\text{Horizontal axis of symmetry}} \cr
& \,\,\,\,\,\,\,{\text{Domain: }}\left[ {h,\infty } \right):\left[ { - 1,\infty } \right) \cr
& \,\,\,\,\,\,\,{\text{Range:}}\left( { - \infty ,\infty } \right) \cr} $$
