Question 1204076
a. Find the equation of the parabola.


a span of  {{{256}}}  meters ,  {{{x}}}  coordinate of the vertex is  {{{256/2=128}}} meters  and a maximum height of  {{{32}}} 

Vertex at ( {{{128}}} , {{{32}}} )


vertex form of the equation is

 {{{y=a(x-h)^2+k}}} , ( {{{ h}}} , {{{ k}}} ) is the vertex

 {{{y=a(x-128)^2+32}}} 

the parabola starts at ( {{{0}}} , {{{0}}} ) that is a point ( {{{ x}}} , {{{ y}}} )

 {{{0=a(0-128)^2+32}}} 
 {{{a(-128)^2 = -32}}} 
 {{{16384a=-32}}} 
 {{{a=-32/16384}}} 
 {{{a=-1/512}}} 

so, equation is

 {{{y=(-1/512)(x-128)^2+32}}} 


b. Determine the distance from the center at which the height is  {{{12}}}  meters.

the height of the arch  {{{12}}}  feet from the center
 center is at  {{{128}}}, {{{ 12}}}  meters from the center so  {{{x=128+12=140}}} 


 {{{y=(-1/512)(140-128)^2+32}}} 

 {{{y=1015/32}}} 

 {{{y=31.7}}}  approximately


{{{ drawing( 600, 600, -10, 260, -10, 50, 
circle(128,32,1.3), locate(128,34,V(128,32)),
circle(140,31.7,1.3), locate(140,31.7,p(140,31.7)),
graph( 600, 600, -10, 260, -10, 50, (-1/512)(x-128)^2+32)) }}}