Question 397259
   <pre><font size = 3 color = "indigo"><b>
Hi

Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
{{{f(x)=(1/4)(x+1)^2+5}}} |1/4 = a > 0, parabola opens upward
 Vertex is Pt(-1,5)is minimum point and the Line of symmetry is x = -1
f(x) = 5 is the minimum value for the function
{{{drawing(300,300, -6, 6, -6, 6,blue(line(-1,6,-1,-6))    grid(1),
circle(-1, 5,0.3),
graph( 300, 300, -6, 6, -6, 6,.25(x+1)^2+5))}}}