Algebra 2 Problems of the Day (Jan 2020)

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January 2020, Part III

All Questions in Part III are worth 4 credits. Work must be shown. Partial credit is given.


33. Sonja is cutting wire to construct a mobile. She cuts 100 inches for the first piece, 80 inches for the second piece, and 64 inches for the third piece. Assuming this pattern continues, write an explicit equation for an, the length in inches of the nth piece.

Sonja only has 40 feet of wire to use for the project and wants to cut 20 pieces total for the mobile using her pattern. Will she have enough wire? Justify your answer.

Answer:
80/100 = .8 or 4/5. 64/80 = .8 or 4/5. r = .8
The formula is a_n = 100(.8)^{n - 1}

Remember that the exponent is n - 1 because the first number has to be 100, so the exponent needs to be zero.

To find out whether she has enough for 20 pieces of wire, for the sum of the first 20 lengths using the formula:
S_n = (a₁ - a₁(r)ⁿ) / (1 - r)
S₂₀ = (100 - 100(.8)²⁰) / (1 - .8) = 494.235... inches

40 feet of wire = 40 * 12 = 480 inches.
She will not have enough wire. She will need approximately another 14.24 inches of wire.

34. Graph the following function on the axes below.

f(x) = log₃(2 - x)
State the domain of f.
State the equation of the asymptote.

Answer:
Plug the function into your graphing calculator. You will see the "easy" points are (-7, 2), (-1, 1) and (1, 0). Also, f(0) is approximately 0.6. The asymptote is x = 2.
Graph it as follows:

The domain of f is x < 2. (Not x < 2.)
The equation of the asymptote is x = 2. (It says equation, so don't just says that "it's 2".)

Comments and questions welcome.

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