Hi

f(x) = -x^2 +2x+15

Finding x-intercepts when f(x)=0

 -x^2 +2x+15 = 0

  |Using Calculator

x = -3, 5  |x-intercepts P(-3,0) and Pt(5,0)

Finding vertex

Using the vertex form of a parabola,  where(h,k) is the vertex

f(x) = -x^2 +2x+15   |completing square to put into vertex form

f(x) = -[(x-1)^2 -1] + 15

f(x) = -(x-1)^2 +1 + 15

f(x) = -(x-1)^2 + 16  Vertex is Pt(1,16)



 

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