Question 1198215
the parabola is open downward and its origin is at the point  ({{{ 0}}},{{{ 0}}}),

vertex is  at  ({{{ 0}}},{{{ 29}}}) 

 the {{{ x}}}-intercepts are  {{{ 348/2=174}}} feet away from the origin at  ({{{ -174}}},{{{ 0}}})  ({{{ 174}}},{{{ 0}}})


use three points above


 {{{ y = ax^2 +bx+c}}}

since  vertex  at  ({{{0}}},{{{29}}}) 

{{{29 = a*0^2 +b*0+c}}}

{{{c=29}}}


so far your equation is

{{{y = ax^2 +bx+29}}}


use   ({{{-174}}},{{{0}}}) 


{{{0= a(-174) ^2 +b(-174) +29}}}

{{{0=30276a -174b +29}}}

{{{a=(174b -29)/30276}}}..........eq1


use   ({{{174}}},{{{0}}}) 

{{{0= a(174) ^2 +b(-174) +29}}}

{{{0=30276a +174b +29}}}

{{{a=(-174b -29)/30276}}}..........eq2


then


{{{(174b -29)/30276=(-174b -29)/30276}}}

{{{174b -29=-174b -29}}}

{{{348 b = 0}}}

 {{{b = 0}}}


go to


{{{a=(174*0 -29)/30276}}}..........eq1 plug in {{{b = 0}}}

{{{a= -29/30276}}}

{{{a= -1/1044}}}



your   equation is:


{{{y = -(1/1044)x^2 +29}}}




then  find   {{{x}}} when {{{y=11}}} 

{{{11 = -(1/1044)x^2 +29}}}

{{{x^2 = 18792}}}

{{{x = sqrt(18792)}}}

{{{x = 137.1ft}}} 


  the distance from the center at which the height is {{{11}}}  feet is {{{ 137.1ft}}}