Question 411041
  <pre><font size = 3 color = "indigo"><b>
Hi
f(x)=1/4(x+2)^2+1
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
the vertex: Pt(-2,1)
the line of symmetry: x = -2 
and the maximum or minimum value of f(x): 
f(-2) = 1 is a minimun  |parabola opens upward as a = (1/4)>0
{{{drawing(300,300,   -6, 6, -6, 6, blue(line(-2,6,-2,-6))  , grid(1),
circle(-2, 1,0.3),
circle(0, 2,0.3),
graph( 300, 300, -6, 6, -6, 6,0,.25(x+2)^2+1))}}}

f(x)= -22x^2+2x+6 OR f(x) = -22(x- 1/11)^2 + 64/11 |completing the square
{{{drawing(300,300,   -10,10,-10,10,  grid(1),
graph( 300, 300, -10,10,-10,10,0,-22x^2+2x+6))}}}