Question 465710
  <pre><font face = "Tohoma" size = 4 color = "indigo"><b> 
Hi
y = -x^2+2x+3 | y-intercept Pt(0,3)
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
y = -x^2+2x+3    |completing the square
y = -[(x-1)^2 -1]+3
y = -(x-1)^2 +4   Vertex (1,4) a = -1 <0  Parbola opens downward, vertex is max pt
Line of symmetry is x = 1  and when y = 0 (x-1)^2 = 4  x = 3
{{{drawing(300,300,   -6, 6, -6, 6,  blue(line(1,6,1,-6))  , grid(1),
circle(0, 3,0.3),
circle(3, 0,0.3),
graph( 300, 300, -6, 6, -6, 6,0,-x^2+2x+3 ))}}}