Question 624238
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Hi,
Note; vertex form of a Parabola opening up(a>0) or down(a<0), {{{y=a(x-h)^2 +k}}} 
where *[tex \large\ (h,k)] is the vertex  and {{{ x = h }}} is the Line of Symmetry
f(x)=1/5 (x+5)^(2)+6  || V(-5,6)  1/5> 0, parabola opens Up, vertex in a min point
The line of symmetry is? x = -5
the minimum value of f(x)  is 6
the value, f(-5) = 6, is a minimum 
{{{drawing(300,300,   -10, 6, -6, 20,  blue(line(-5,20,-5,-6))  , grid(1),
circle(-5, 6,0.3),
graph( 300, 300, -10, 6, -6, 20,0,.2(x+5)^(2)+6  ))}}}