Question 395575
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Hi
f(x) = -x^2 +2x+15
Finding x-intercepts when f(x)=0
 -x^2 +2x+15 = 0
{{{x = (-2 +- sqrt( 64 ))/(-2)= (-2 +- 8)/-2 }}}  |Using Calculator
x = -3, 5  |x-intercepts P(-3,0) and Pt(5,0)
Finding vertex
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} 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)
{{{drawing(300,300, -20,20,-20,20, grid(1),
circle(-3, 0,0.8),
circle(5, 0,0.8),
circle(1, 16,0.8),
graph( 300, 300, -20,20,-20,20, -x^2 +2x+15))}}}