Question 1205668
The standard form of a parabola:

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

given:

height is {{{h=535ft}}} (which is {{{y}}} value of the vertex)

width is {{{4100ft}}} 

if we place the origin in the midpoint, then we have {{{x=4100ft/2=2050ft}}} from each side of the origin

than, the vertex is at ({{{0}}},{{{535}}}) and {{{x}}}-intercepts are at ({{{-2050}}},{{{0}}}) and ({{{2050}}},{{{0}}})


{{{y=ax^2 + bx + c}}}....plug in {{{x=0}}}, {{{y=535}}}

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

{{{c=535}}}



plug in {{{x=2050}}}, {{{y=0}}}

{{{0=a*2050^2 + b*2050 + 535}}}

{{{4202500 a + 2050 b + 535=0}}}....solve for {{{a}}}

{{{a =- 2050b /4202500- 535/4202500}}}

{{{a =-b/2050-107/840500}}}....eq.1



plug in {{{x=-2050}}}, {{{y=0}}}

{{{0=a*(-2050)^2 + b*(-2050 )+ 535}}}

{{{-4202500 a + 2050 b - 535 = 0}}}....solve for {{{a}}}

 {{{2050 b - 535 = 4202500 a}}}

{{{a =2050b /4202500- 535/4202500}}}

{{{a =b/2050-107/840500}}}....eq.2


from eq.1 and eq.2 we have

{{{-b/2050-107/840500=b/2050-107/840500}}}

{{{-b/2050=b/2050}}}

{{{b=0}}}



go to

{{{a =b/2050-107/840500}}}....eq.2, substitute{{{ b}}}

{{{a =0/2050-107/840500}}}

{{{a =-107/840500}}}


your equation is:


{{{y=-(107/840500)x^2 + 535}}}



The factored form of a parabola:

{{{y = a(x - p)(x - q)}}}, where {{{p}}} and {{{q }}}are the x-intercepts of the parabola

substitute x-intercepts


{{{y = -(107/840500)(x - 2050)(x - (-2050))}}}

{{{y = -(107/840500)(x - 2050)(x +2050)}}}