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Satellite dish antennas have the shape of a parabola. Consider a satellite dish that is 80 cm across. Its cross-sectional shape can be described by the function d(x) = 0.0125x2 - x, where d is the depth, in centimetres, of the dish at a horizontal distance of x centimetres from one edge of the dish.

 

a) What is the domain of this function?

b) Graph the function to show the cross-sectional shape of the satellite dish.

c) What is the maximum depth of the dish? Does this correspond to the maximum value of the function? Explain.

d) What is the range of the function?

e) How deep is the dish at a point 25 cm from the edge of the dish?

Guest Oct 26, 2014

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 #2
avatar+80865 
+8

Thanks for that analysis, geno......very nice......here's the graph......https://www.desmos.com/calculator/0eueskjaft

Graphing this one first is definitely an easier (lazier) route !!!

 

CPhill  Oct 27, 2014
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2+0 Answers

 #1
avatar+17705 
+8

How you approach this depend upon whether or not you use a calculator.

1)  The domain of the function is:  -∞ < x < ∞

2)  If you have to do this by hand, you may want to complete the square:

       d(x)  =  0.0125(x² - 80x + 1600) - 20

        d(x)  =  0.0125(x - 40)² - 20  

      This shows that the vertex occurs at (40, -20).

      Substituting 0 for x gives you the point (0, 0).

      Substituting 80 for x gives you the point (80, 0).

      The parabola passes through these points.

3)  The maximum depth of the dish occurs at the vertex; the depth will be 20 cm.

      This is the minimum of the function, not the maximum.

4)  Range:  -20 ≤ y < ∞

5)  Substituting 25 for x, d(x)  =  -17.1875.  

     Distance from vertex:  20 - 17.1875  =  2.8125; distance from top of dish:  17.1875 cm.

geno3141  Oct 27, 2014
 #2
avatar+80865 
+8
Best Answer

Thanks for that analysis, geno......very nice......here's the graph......https://www.desmos.com/calculator/0eueskjaft

Graphing this one first is definitely an easier (lazier) route !!!

 

CPhill  Oct 27, 2014

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