Kaelyn S.
asked • 04/02/20The amount of trout y (in tons) caught in a lake from 1995 to 2014 can be modeled by the equation y=-0.08x2 + 1.6x + 10, where x is the number of years since 1995.
Part a.) When were about 15 tons of trout caught in the lake?
Part b.) Do you think this model can be used to determine the amounts of trout caught in future years?
Dianna J. answered • 04/02/20
Math Teacher with 10 + years of education experience.
Kaelyn S.
Thank you so much04/02/20
x = number of years since 1995. y = number of tons of trout caught in the lake between 1995 and 2014
part a).
To find x for y=15 need to solve equation 15 = 0.08x**2 + 1.6x + 10
Subtracting 15 from both sides:
0.08x**2 + 1.6x -5 = 0
Using the quadratic formula with a = 0.08, b = 1.6 and c = -5
x=2.74 or -22.74. 2.74 is the only answer that makes sense, which if we round off to the nearest year (3), puts us at 1998.
part b) If the model was generated based on past data then it could only used to model future years if the current trends continue.
Kaelyn S.
i’m not understanding how you got 2.74 and -22.74. So if you round -22.74 to the nearest year, what would you get?04/02/20
Dianna J.
the a value is incorrect. it should be -.08, according to your provided problem; which explains the incorrect x values given in this solution04/02/20
Dianna J.
I show that there are two solutions for part a and explain why you should not use this quadratic for future predictions (answering part b) in the below video. I apologize for the video's length and my less than thrilling presentation but it should do the trick! I also apologize for the doc cam imaging not being perfect. By the time I realized the doc cam was not being super cooperative, I had spent 20 minutes recording already. . . it should still be readable and helpful though! EDIT: It looks like the video is still posting so I will summarize my findings from the video here.... part a) 15 tons of fish are caught in 1999 (or the end of 1998 depending on how you round) and 15 tons of fish are caught in 2011 per the quadratic formula. (a=-.08 b=1.6 c=-5) (you get x=3.8 and x=16.12) part b) if you find the zeros of your graph using your quadratic you will find that one of your x intercepts occur at 2020. (x=-5 and x=25 are the zeros of the overall function when you plug in a=-.08, b=1.6 and c=10) Meaning all y values after 2020 are negative since the parabola opens down. You cannot use this to predict fish being caught after 2020 because you cannot catch a negative amount of fish. If you do use this model, you would have to assume the lake just has no more trout, or people are actively pouring tons of trout into the lake . . . neither of which seems likely.04/02/20
Kaelyn S.
thank you so much04/02/20
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Dianna J.
Oh my goodness! I had no idea they would make this a youtube video. I thought you would be able to full screen the video like in the classroom application. I am so sorry. See my summary in the comments on Jon S.'s answer part a) 15 tons of fish are caught in 1999 (or the end of 1998 depending on how you round) and 15 tons of fish are caught in 2011 per the quadratic formula. (a=-.08 b=1.6 c=-5) (you get x=3.8 and x=16.12) part b) if you find the zeros of your graph using your quadratic you will find that one of your x intercepts occur at 2020. (x=-5 and x=25 are the zeros of the overall function when you plug in a=-.08, b=1.6 and c=10) Meaning all y values after 2020 are negative since the parabola opens down. You cannot use this to predict fish being caught after 2020 because you cannot catch a negative amount of fish. If you do use this model, you would have to assume the lake just has no more trout, or people are actively pouring tons of trout into the lake . . . neither of which seems likely.04/02/20