Tuesday, April 24, 2018

Algebra 2 Problems of the Day (open-ended)

Continuing with daily Algebra 2 questions and answers.

More Algebra 2 problems.

January 2018, Part II

Questions in Part II are worth 2 credits. All work must be shown or explained for full credit. A correct numerical answer without work is only worth 1 credit.

29. Researchers in a local area found that the population of rabbits with an initial population of 20 grew continuously at the rate of 5% per month. The fox population had an initial value of 30 and grew continuously at the rate of 3% per month.
Find, to the nearest tenth of a month, how long it takes for these populations to be equal.

The equation for rabbit population is y = 20e.05x.
The equation for fox population is y = 30e.03x.
Set them equal to each other 20e.05x = 30e.03x
Divide by 30: (20/30)e.05x = e.03x
Divide by the e term: (2/3) = (e.03x)/(e.05x)
Simplify the exponent: 2/3 = e-.02x
ln (2/3) / -.02 = -.02x / -.02
20.273... = x
20.3 months.

You also could have graphed the two equations, found the point of intersection, record that point of intersection on your exam paper as an ordered pair, and then state the answer of 20.3 months based on the x coordinate.

30. Consider the function h(x) = 2sin(3x) + 1 and the function q represented in the table below.

Determine which function has the smaller minimum value for the domain [-2,2]. Justify your answer.

The minimum for q(x) is -8, according to the table. You don't need to work out what the equation is.
The minimum for h(x) is -1. You don't need to explain how you know this as it would likely be assumed that you got it from the calculator, or you just knew that y = 2 sin(3x) has a minimum of -2 and a maximum of 2, so if you add 1 to those numbers, the minimum is -1.
So q(x) has the smaller minimum. Don't forget to mention this. It isn't enough to just state the two minimums. But you DO need to state the two minimums.

Comments and questions welcome.

More Algebra 2 problems.

No comments: