Question 356175
{{{f(x)= (1/5)(x+6)^2 +4}}}
The equation is already in vertex form, {{{y=a(x-h)^2+k}}} where ({{{h}}},{{{k}}}) is the vertex.
Comparing,
({{{h}}},{{{k}}})=({{{-6}}},{{{4}}})
The vertex lies on the line of symmetry, {{{x=-6}}}
Since {{{a>0}}}, the parabola opens upwards and the value at the vertex is the minimum.
{{{y[min]=4}}}
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{{{drawing(300,300,-12,2,-2,12,circle(-6,4,0.2),grid(1),blue(line(-6,12,-6,-12)),graph(300,300,-12,2,-2,12,(1/5)(x+6)^2+4))}}}