Question 134638
a.This equation stands for the river's cross section at its deepest. y=0.0581x^2-2.3238x, where y is the depth of the river(in feet) and x the horizontal distance from the bank of the river(in feet). Assume that the surface of the river is modeled by y=0.
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Graph the river's cross section labeling the vertex, axis of symmetry.(use a modified coordinate plane;the left bank will be one of your solutions at the origin.)
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Plot the above equation from x= 0 to x=45 ; your graph should look like this
{{{ graph( 300, 200, -10, 50, -30, 10, .0581x^2-2.3238x) }}}
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 b. Your boat is directly above the deepest point of the river. You drop your line with a sinker at the end. Your line is 25 ft. long. Will your sinker fouch the bottom.Explain.
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Find axis of symmetry (the distance your boat is from the shore)
x = {{{(-b)/(2a)}}} is formula; in this equation: a=.0581; b = -.23238
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x = {{{(-(-2.3238))/(2*.0581)}}} 
x = {{{2.3238/.1162}}}
x = 19.998 ~ 20 ft from the shore
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The depth (y) at that point can be found by finding the vertex:
Substitute 20 for x in the original equation
y = .0581(20^2)-2.3238(20)
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y = .0581(400)-2.3238(20)
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y = 23.24 - 46.476
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y = -23.2 ft is the depth at that point (and the deepest point as you can see on the graph)
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We have to say that a sinker on the end of 25 ft line would rest on the bottom
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c. There is a fish detector on the boat. The detector shows that there is a school of fish 18 ft. directly below you. You are 8 ft. from the bank of the river. Is the fish detector accurate? Explain.
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This question amounts to finding the value of y when x = 8
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y = .0581(8^2)-2.3238(8)
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 I'll let you determine whether not the fish detector can be right, if you have any question about this you can email me