Question 1132760
An arch in the shape of a parabola has height {{{630ft}}} and width {{{630ft}}}. How wide is the arch {{{300ft}}} up? 

The width of the bridge is {{{630}}} feet so the parabola crosses the x-axis with x-coordinates ± {{{630/2}}} = ± {{{315}}}

=> x-intercepts are at ({{{-315}}},{{{0}}}) and ({{{315}}},{{{0}}})

Assume that it stands on the x axis, that it is symmetric with respect to the y-axis, and that it is {{{630}}} feet wide at the base and {{{630}}} feet high at the {{{center}}}.

 Then the maximum height occurs at {{{x = 0}}}; so, the vertex of the parabola is  =>({{{0}}},{{{630}}})


Since the curve is a parabola which opens downward its equation can be written 

{{{f(x) = a(x-h)^2 +k}}}

since {{{h=0}}} and {{{k=630}}}

{{{f(x) = a(x-0)^2 +630}}}

{{{f(x) = ax^2 +630}}}

x-intercepts are at ({{{-315}}},{{{0}}}), use x-intercept to find {{{a}}}

{{{0 = a(-315)^2 +630}}}

{{{0 =99225*a +630}}}

{{{99225 a =-630}}}

{{{ a =-630/99225}}}

{{{a =-2/315}}}


and, equation is

{{{f(x) = -(2/315)x^2 +630}}}


to find how wide is the arch {{{300ft}}} up, substitute {{{f(x)= 300ft}}}

{{{300 = -(2/315)x^2 +630}}}

{{{(2/315)x^2 =630-300}}}

{{{(2/315)x^2 =330}}}

{{{x^2 =330/(2/315)}}}

{{{x^2 =(330*315)/2}}}

{{{x^2 =165*315}}}

{{{x^2 =51975}}}

{{{x =sqrt(51975)}}}

{{{x}}} = ± {{{227.98}}}

the arch {{{300}}} ft have intercepts at ({{{-227.98}}},{{{300}}}) and ({{{227.98}}},{{{300}}}), so width will be the distance from {{{x=-227.98}}} to {{{x=227.98}}} which is {{{ 2*227.98=455.96}}} ft

answer: the arch {{{300}}} ft up is {{{ 455.96}}} ft wide