Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
5. Solving Quadratic Equations Using the Quadratic Formula
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Exercise 24 Page 521

Practice makes perfect
a In this exercise, we are given an equation with two variables and definitions for those variables. For Part A, we need to find the year when 15 tons of trout were caught. Let's substitute y=15 into the equation, then solve for x.
y=-0.08x^2+1.6x+10
15=-0.08x^2+1.6x+10
15-15=-0.08x^2+1.6x+10-15
0 = -0.08x^2+1.6x-5
-0.08x^2+1.6x-5=0
Now we have our quadratic in standard form, we can apply the Quadratic Formula and solve for x.
-0.08x^2+1.6x-5=0
x=-( 1.6) ± sqrt(( 1.6)^2-4( -0.08)( - 5))/2 ( -0.08)
Simplify right-hand side
x=-1.6 ± sqrt(2.56-1.6)/-0.16
x = -1.6 ± sqrt(0.96)/-0.16
x ≈ -1.6 ± 0.98/-0.16
The solutions for this equation are x ≈ -1.6 ± 0.98-0.16. Let's separate the two cases.
x ≈ -1.6 ± 0.98/-0.16
x_1 ≈ x_2 ≈
-1.6 + 0.98/-0.16 -1.6 - 0.98/-0.16
-0.62/-0.16 -2.58/-0.16
3.88 16.12

Since the exercise states that the model works from 1995 to 2014, both solutions are viable. The first solutions x ≈ 3.88, says that about 3.88 years since 1995, in 1998, there were 15 tons of trout caught. The second solution, x_2 ≈ 16.12, tells us that 15 tons of trout were caught in 2011.

b When answering a question about how a model works, we need to talk about both the characteristics of the equation and the physical situation of the lake before we tie them together.

The Model

The equation, y=-0.08x^2 +1.6x+10, is a downward facing quadratic and therefore will rise to a certain point, then go back down.

The Lake

The lake's ecology is a limiting factor in how many fish can be caught. Among the factors that will limit the amount of fish in a lake are: size, availability of food, oxygen (DO), predators, death rate, and the trout's reproduction rate. Furthermore the number of fisher people and their ability to catch trout can limit the number caught.

Conclusion

Weather or not the quadratic model given took into account these other factors will determine if it is a viable model. Let's have a look at the model again, but add in some elements from the lake situation.

It is possible that it is effective for a few more years, but eventually the quadratic will have negative solutions and negative years, which limit it's ability to model real world for an extensive period of time. After 2005 the tons of trout caught will start to decline and after 2021 the function gives us negative values for x.