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25 April, 13:15

The national catch of Atlantic Cod has been declining in the past few decades. We can model this decline with the equation y = -.003x2 +.120x + 1.909 where x is the number of years since 1950 and y is the catch in millions of metric tons. In which year did the catch reach its maximum? In which year will the catch reach 0? For what years was the catch increasing? Decreasing? When was the catch greater than 2.5 million metric tons? Show all of your work or explain how you came up with your solutions.

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  1. 25 April, 15:07
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    Y = -.003x² +.120x + 1.909

    Standard form of a parabola is y = ax² + bx + c. The x coordinate of the vertex can be found with - b/2a.

    -.120 / (2 * -.003) + 20

    So the max is at 1950 + 20 because x is years after 1950. Which is 1970.

    To find when y will be 0, use the quadratic equation x = [-b + / - √ (a² - 4ac) ]/2a. + / - means plus or minus because I can't type the actual symbol.

    x = [-.120 + / - √ (-.003² - 4 (-.003) (1.909)) ]/2 (-.003) Plug in info

    x = [-.120 + / - √ (.000009 - 4 (-.005727)) ]/-.006

    x = [-.120 + / - √ (.000009 +.022908) ]/-.006

    x = [-.120 + / - √ (.022917) ]/-.006

    x=-.120/-.006 + / - √ (.022917) / -.006

    x = 20 + / -.15138361866/-.006

    x = 20 + / - - 25.2306031108

    Because it is plus or minus you split it and do both.

    x = 20 - - 25.2306031108 x = 20 + - 25.2306031108

    x = 45.2306031108 x = - 5.23060311077

    so y is 0 when x is 45.2306031108 because there is no such thing as negative time so it can't be the other one. So the answer is 1995 because you add 45 to 1950.

    Since the max is at 1970, it is increasing from 1950 to 1975 and decreasing from 1975 to 1995.

    Sorry I took so long, I had to look up how to do it because I couldn't remember.
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