Question 389589
<pre><font size = 3 color = "indigo"><b>
Hi,  
the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
vertex (-1,-108)
y =a(x+1)^2 -108
using y-intercept pt(0,-105) to find a
-105 = a(1)^2 -108
3 = a
y = 3(x+1)^2 -108  
finding x-intercepts when y = 0
0 = 3(x+1)^2 -108 
108 = 3(x+1)^2
 36 = (x+1)^2
  6 = (x+1)  x = 5
 -6 = (x+1)  x -7
 x-intercepts are Pt(5,0) and Pt(-7,0)