Question 1166634
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If you graph the parabola so that the vertex is at the origin, the center of the roadway would be at the point (0,-30).  The towers are 400 feet apart, so they are 200 feet either side of center.  Considering the 30 foot offset, the points of attachment of the cables to the tower are at (-200,120) and (200,120).


The equation of a parabola in vertex form with the vertex at the point (h,k) is *[tex \Large y\ =\ a(x\,-\,h)^2\ +\ k].  Since we chose the origin for the vertex, the equation simplifies to *[tex \Large y\ =\ ax^2]


Using one of the points of attachment of the cable:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(200)^2\ =\ 120]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ \frac{120}{40000}\ =\ 0.003]


Making the desired function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(x)\ =\ 0.003x^2]


The height of the cable <i>above the *[tex \Large x]-axis</i> at 50 feet in from one of the towers is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(150)\ =\ 0.003(150)^2]


Then the height <i>above the roadway</i> is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(150)\ +\ 30\ =\ 0.003(150)^2\ +\ 30]


You get to do your own arithmetic. 

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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