Question 1179625
max height will be at vertex


 {{{y=-16x^2+228x+71}}}...rewrite in vertex form by completing square


 {{{y=(-16x^2+228x)+71}}}...factor out {{{-16}}}


{{{y=-16(x^2-(57/4)x+b^2)-16b^2+71}}}.......{{{b=(57/4)/2=(57/8)}}}


{{{y=-16(x^2-(57/4)x+(57/8))-(-16(57/8)^2)+71}}}


{{{y=-16(x-57/8)^2+3249/4+71}}}


{{{y=-16(x-57/8)^2+3533/4}}}=> {{{h=57/8}}} and {{{k=3533/4}}}


where {{{k=3533/4}}} is {{{y}}} coordinate of the vertex 

so, max height is {{{y=3533/4}}} at {{{x = 57/8}}}

to the nearest tenth of a foot:  {{{y=883.3}}}