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Classroom activity Grade 8 Math

Wolverton Commons Fixture Payback: A Linear Function Station Rotation

A grade 8 station rotation where groups build net-savings functions for water-saving fixtures at a fictional apartment and find each fixture's payback month.

Objectives

  1. Students will build a net-savings linear function from an upfront cost and a monthly savings.
  2. Students will find and compare the payback month for several water-saving fixtures.

Vocabulary

Key words: linear function, rate of change, initial value, payback

Teacher overview

This is a four-station rotation for small groups of three or four students. Each station describes one water-saving fixture with an upfront cost and a monthly savings: a fixture's net savings after m months is a linear function whose rate of change is the monthly savings and whose initial value is the negative upfront cost. Groups build that function, then find the payback month, the point where net savings reaches zero, so the fixture has paid for itself.

Activity Directions

  1. Before class, print the four station cards (A, B, C, D) and place them at four spots around the room. Each card lists the fixture, its upfront cost, and its monthly savings, matching the handout. Print one handout per student.
  2. Divide the class into four groups and send each group to a different station to start. Give each group a calculator and each student the handout.
  3. At each station (about 8 minutes), the group reads the fixture's upfront cost and monthly savings, writes the net-savings function in the form N = (monthly savings) times m minus (upfront cost), and finds the payback month by setting N equal to zero and solving for m.
  4. When you call time, groups rotate clockwise to the next station and repeat, until every group has completed all four stations.
  5. After the rotations (about 8 minutes), groups return to their seats and complete handout items 6, 7, and 8, which compare the four fixtures.
  6. Close with a debrief (about 5 minutes): ask "Which fixture pays back fastest, and which saves the most each month?" and "Are those the same fixture?" Use the answers to point out that the fastest payback and the steepest savings line are not always the same choice.
Wolverton Commons Fixture Payback: A Linear Function Station Rotation — Grade 8 Math
Name: ____________________________ Date: ______________

Student Handout

Wolverton Commons is a fictional apartment community comparing four water-saving fixtures. For each one, m stands for the number of months since it was installed.

  • Station A, WaterSense showerhead: upfront cost $24, saves $4 each month.
  • Station B, faucet aerators: upfront cost $21, saves $3 each month.
  • Station C, high-efficiency toilet: upfront cost $180, saves $9 each month.
  • Station D, drip-repair kit: upfront cost $8, saves $2 each month.
  1. For Station A, write the net-savings function N = 4m - 24. State its rate of change and its initial value, and explain what each one means.
  2. For Station A, solve 4m - 24 = 0 to find the payback month.
  3. For Station B, find the payback month.
  4. For Station C, find the payback month.
  5. For Station D, find the payback month.
  6. Rank the four fixtures from the fastest payback to the slowest.
  7. Which fixture's net-savings line is the steepest, and why?
  8. Explain why a fixture with a high upfront cost, like the toilet, can still be a good choice, using the ideas of rate of change and initial value.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: rate of change 4, initial value -24.
    In N = 4m - 24, the rate of change is $4, the savings added each month, and the initial value is -$24, the cost the community is behind before any savings, on the day the fixture is installed.
  2. Answer: 6 months.
    Set 4m - 24 = 0, so 4m = 24 and m = 6 months.
  3. Answer: 7 months.
    Net savings N = 3m - 21. Set 3m - 21 = 0, so 3m = 21 and m = 7 months.
  4. Answer: 20 months.
    Net savings N = 9m - 180. Set 9m - 180 = 0, so 9m = 180 and m = 20 months.
  5. Answer: 4 months.
    Net savings N = 2m - 8. Set 2m - 8 = 0, so 2m = 8 and m = 4 months.
  6. Answer: D, A, B, C.
    The payback months are D at 4, A at 6, B at 7, and C at 20, so fastest to slowest is D, A, B, C.
  7. Answer: Station C.
    Station C saves $9 each month, the greatest monthly savings, so its net-savings line has the steepest slope.
  8. Answer: A high upfront cost can still pay off if the monthly savings is large.
    The toilet has the largest initial cost ($180) but also the steepest savings rate ($9 per month), so its line climbs fastest and, given enough months, it saves the most overall even though its payback takes longer.

Differentiation

Support

Give students the net-savings function already written at each station so they only need to solve for the payback month. Provide a worked example (a $30 fixture saving $5 per month, payback at 6 months) on the handout for reference.

Extension

Ask students to find, for any two stations, the month at which the two fixtures have saved the same total amount, by setting their net-savings functions equal and solving, then explain which fixture is ahead before and after that month.

ELL supports

Pre-teach "upfront cost," "savings," and "payback" with a simple picture of a cost being slowly repaid month by month. Provide the sentence frame "Fixture ___ pays back in ___ months because it saves $___ each month."