<PRE><B>Question 237556</B>
{{{graph(600,600,-100,100,-100,100,.005*x^2+10,60,10)}}}


The graph shows the cable between the posts at x = -100 and x = 100


The lines are at y = 10 and y = 60 to show you that the cable meets the specifications at the tower and at the minimum point.


The equation of the graph is y = .005x^2 + 10


The minimum point is at x = 0


The x value extends from x = -100 to x = +100


The road bed is at y = 0.


y = f(x) is equal to the height of the cable from the roadbed.


D(x) = f(x), so:


D(x) = .005x^2 + 10


The domain of D(x) is the value of x from x = -100 to x = +100


The range of D(x) is from 10 to 60.


Values of x and values of y are expressed in feet.


7 equally spaced cables supporting the roadbed would extend from the suspension cable to the roadbed.


These would be equally spaced 25 feet apart.


They would be at:


x = -75
x = -50
x = -25
x = 0
x = 25
x = 50
x = 75


To find the height at these points from the roadbed, solve for D(x) at each of these points.


D(-75) = (.005)*(-75)^2 + 10 = 38.125 feet from the roadbed.
D(-50) = (.005)*(-50)^2 + 10 = 22.5 feet from the roadbed.
D(-25) = (.005)*(-25)^2 + 10 = 13.125 feet from the roadbed.
D(0) = (.005)*(0)^2 + 10 = 10 feet from the roadbed.
D(25) = (.005)*(25)^2 + 10 = 13.125 feet from the roadbed.
D(50) = (.005)*(50)^2 + 10 = 22.5 feet from the roadbed.
D(75) = (.005)*(75)^2 + 10 = 38.125 feet from the roadbed.


Each Tower will be at x = -100 and at x = 100 feet.


D(-100) = (.005)*(-100)^2 + 10 = 60 feet from the roadbed.
D(100) = (.005)*(100)^2 + 10 = 60 feet from the roadbed.


The formula was created as follows:


Standard formula for a quadratic equation is ax^2 + bx + c = 0


The minimum point of a quadratic equation is at x = -b/2a.


Since that had to be at x = 0, that meant that -b/2a was equal to 0 which meant that b had to be equal to 0.


When x = 0, the quadratic equation becomes c = 0 because the a factor drops out since it is being multiplied by x = 0.  The b factor was already canceled out because it was equal to 0.


Since the minimum point of the cable had to be 10, that meant that c = 10.


The equation became:


ax^2 + 10 = 0


At x = -100, the height of the cable had to be 60.


That meant that ax^2 + 10 = 60 when x = -100.


equation became:


a*(-100)^2 + 10 = 60


that became:


10,000*a + 10 = 60


Solve for a to get a = 50/10000 = .005


This same value was also applicable at x = 100.


Equation became:


y = .005x^2 + 10


Let D(x) = y and equation became:


D(x) = .005x^2 + 10
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