Question 1203474
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Bridge shaped like a parabolic arch has a horizontal distance of 20 feet. 
The height of a point 1 foot from the center is 8 feet. 
What is the maximum height of the bridge if it is located at the center?
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<pre>
Let x-axis be horizontal at the ground level, with the origin under the upper 
point of the bridge; y-axis vertical.


Then the parabola has x-intercepts at x= -10 ft and x= 10 ft.


So, the parabola has the form  y = a*(x-(-10))*(x-10) = a*(x+10)*(x-10) = a*(x^2-100).


Coefficient "a" is some real negative number (since the parabola is opened downward).
We do not know this number. It is the only unknown in this problem,
and our goal is to find this single unknown.


To find "a", use the fact that at x= 1 ft we have y= 8 ft, from the problem.
It gives you this equation

    8 = a*(1^2 - 100),  or  8 = -99a,  which implies  a = {{{-8/99}}} = -0.08080808...


Thus the quadratic function is y = {{{(-8/99)*(x^2-100)}}},

and it has the maximum at x= 0  (at the axis of symmetry).


Thus the maximum height of the bridge is  {{{(-8/99)*(-100)}}} = {{{800/99}}} = 8.08080808... ft, 
or, after rounding, about 8.08 ft.    <U>ANSWER</U>
</pre>

Solved (with the minimum necessary calculations).