Question 948258
The equation is already in vertex form.
The vertex occurs at ({{{-5}}},{{{2}}}).
The axis of symmetry is {{{x=-5}}}.
The parabola opens upwards since {{{(1/4)>0}}}.
Since it opens upwards then the minimum occurs at the vertex {{{y[min]=2}}}.
{{{drawing(300,300,-10,6,-8,8,grid(1),circle(-5,2,0.35),blue(line(-5,-10,-5,10)),graph(300,300,-10,6,-8,8,(1/4)(x+5)^2+2))}}}