Question 318797
The equation is in vertex form, {{{y=a(x-h)^2+k}}}, where (h,k) is the vertex.
Comparing,
(h,k)=({{{-1}}},{{{9}}})
The vertex lies on the axis of symmetry,
{{{x=-1}}}
The coefficient of the {{{x^2}}} term is positive, so the parabola opens upwards and the minimum for the parabola occurs at the vertex.
{{{ymin=9}}}
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{{{drawing(300,300,-8,8,-2,14,grid(1),circle(-1,9,.3),blue(line(-1,-10,-1,20)),graph(300,300,-8,8,-2,14,(1/4)(x+1)^2+9))}}}