Question 324289
The equation is already in vertex form, {{{y=a(x-h)^2+k}}} where ({{{h}}},{{{k}}}) is the vertex.
Comparing, 
({{{h}}},{{{k}}})=({{{2}}},{{{5}}})
The vertex lies on the axis of symmetry, {{{x=2}}}.
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{{{drawing(300,300,-8,8,-2,14,grid(1),circle(2,5,0.3),blue(line(2,-20,2,20)),graph(300,300,-8,8,-2,14,(x-2)^2+5))}}}