Question 413884
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Hi
Identify the maximum or minimum (vertex), zeros, and axis of symmetry lin
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
 y = (-1/4)x^2+2x           |Completing the square
 y = -.25(x-4)^2 +4       |Vertex is Pt(4,4)& a Maximum as Parbola opens downward (a< 0)    x = 4, the line of symmetry
 0 = -.25(x-4)^2 +4      x = 4 ħsqrt(16)  x = 4 ħ4  {0,8} 
{{{drawing(300,300,   -10,10,-10,10, blue(line(4,10,4,-10)),   grid(1),
circle(4, 4,0.4),
circle(8,0,0.4),
circle(0,0,0.4),
graph( 300, 300, -10,10,-10,10,0, (-1/4)x^2+2x  ))}}}