Math 1 · Unit 7 · Lesson 6 · part 3 of 5
Using the model: forward and backward
Once you have the model, two kinds of question.
Using the model: forward and backward
Once you have the model, two kinds of question. Forward: given x, find y. Put x in and evaluate, power first. Backward: given y, find x. There is no equation-solving tool for that yet (Unit 13), so you use a table or the graph.
Forward: the phone after 3 years
y = 800 · 0.803 = 800 · 0.512 = $409.60. Power first: 0.8 · 0.8 · 0.8 = 0.512, then times 800.
Backward, by table: when is it under $200?
| x | y |
|---|---|
| 0 | 800 |
| 1 | 640 |
| 2 | 512 |
| 3 | 410 |
| 4 | 328 |
| 5 | 262 |
| 6 | 210 |
| 7 | 168 |
Year 6 is still $210; year 7 is $168. It first drops under $200 during year 7.
Backward, by graph
Draw the target as a horizontal line, y = 200, and find where the curve crosses it. That crossing is the answer, the same way a system of two equations was solved by its crossing point in Unit 6, Lesson 1. The curve and the line meet a little after x = 6.
Discrete or continuous?The phone loses value once a year, so the model really only means something at x = 0, 1, 2, 3: whole years. That is a discrete situation, like the discrete stories in Unit 6, Lesson 6, and the honest picture is a row of dots. We draw the smooth curve anyway because it shows the shape and lets us read the crossing, but the answer to when is it under $200 is a whole year: year 7. Bacteria growing all the time, or a temperature cooling, would be continuous: any x makes sense, and the curve is the honest picture.