Hi

y = -x^2+2x+3 | y-intercept Pt(0,3)

Using the vertex form of a parabola,  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



 

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