Question 389427
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Hi,  
the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
vertex (-5,80)
y =a(x+5)^2 + 80
using y-intercept pt(0,45) to find a
45 = a(5)^2 + 80
-35 = a*25
 -35/25 = a
y = -7/5(x+5)^2 +80  OR y = -7/5[x^2 +10x + 25] + 80
finding x-intercepts  (y = 0)
-7/5[x^2 +10x + 25] + 80 = 0
 -7/5x^2 - 14x -35 + 80 = 0
 -7/5x^2 - 14x +45 = 0
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (14 +- sqrt( 448 ))/(2.8) }}}