Question 389425
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Hi,  
the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
vertex (4,-1)
y =a(x-4)^2 -1
using y-intercept pt(0,15) to find a
15 = a(-4)^2 - 1
16 = a*16
 1 = a
y = (x-4)^2 - 1  OR y = x^2 - 8x + 15
x^2 - 8x + 15 = 0
factoring
(x-3)(x-5) = 0
(x-3) = 0   x = 3
(x-5) = 0   x = 5
x-intercepts are Pt(3,0) and pt(5,0)
{{{drawing(300,300, -20,20,-20,20,
 grid(1),
circle(4, -1,0.6),
circle(0, 15,0.6),
circle(3, 0,0.6),
circle(5, 0,0.6),

graph( 300, 300, -20,20,-20,20,(x-4)^2 - 1))}}}