Answer
$$\eqalign{
& {\text{Vertex }}\left( {h,k} \right) \to {\text{Vertex }}\left( {2,5} \right) \cr
& \,\,\,\,\,\,\,{\text{Horizontal axis of symmetry}} \cr
& \,\,\,\,\,\,\,{\text{Domain: }}\left[ {h,\infty } \right):\left[ {2,\infty } \right) \cr
& \,\,\,\,\,\,\,{\text{Range:}}\left( { - \infty ,\infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& x = 4{\left( {y - 5} \right)^2} + 2 \cr
& x - 2 = 4{\left( {y - 5} \right)^2} \cr
& 4{\left( {y - 5} \right)^2} = x - 2 \cr
& {\left( {y - 5} \right)^2} = \frac{1}{4}\left( {x - 2} \right) \cr
& {\text{The equation is written in the form }}{\left( {y - k} \right)^2} = 4p\left( {x - h} \right) \cr
& {\left( {y - 5} \right)^2} = \frac{1}{4}\left( {x - 2} \right) \to k = 5,\,\,\,h = 2 \cr
& \cr
& {\text{With: }} \cr
& \,\,\,\,\,\,{\text{Vertex }}\left( {h,k} \right) \to {\text{Vertex }}\left( {2,5} \right) \cr
& \,\,\,\,\,\,\,{\text{Horizontal axis of symmetry}} \cr
& \,\,\,\,\,\,\,{\text{Domain: }}\left[ {h,\infty } \right):\left[ {2,\infty } \right) \cr
& \,\,\,\,\,\,\,{\text{Range:}}\left( { - \infty ,\infty } \right) \cr} $$
