SOLUTION: 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 horizontial distance from the bank of

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 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 horizontial distance from the bank of      Log On


   

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 horizontial distance from the bank of the river(in feet). Assume that the surface of the river is modeled by y=0. 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.) b. Your boat is directly above the deepedst 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 botton.Explain. 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.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
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.
:
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.)
:
Plot the above equation from x= 0 to x=45 ; your graph should look like this
+graph%28+300%2C+200%2C+-10%2C+50%2C+-30%2C+10%2C+.0581x%5E2-2.3238x%29+
:
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.
:
Find axis of symmetry (the distance your boat is from the shore)
x = %28-b%29%2F%282a%29 is formula; in this equation: a=.0581; b = -.23238
:
x = %28-%28-2.3238%29%29%2F%282%2A.0581%29
x = 2.3238%2F.1162
x = 19.998 ~ 20 ft from the shore
:
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)
:
y = .0581(400)-2.3238(20)
:
y = 23.24 - 46.476
:
y = -23.2 ft is the depth at that point (and the deepest point as you can see on the graph)
:
We have to say that a sinker on the end of 25 ft line would rest on the bottom
:
:
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.
:
This question amounts to finding the value of y when x = 8
:
y = .0581(8^2)-2.3238(8)
:
I'll let you determine whether not the fish detector can be right, if you have any question about this you can email me