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1.3 — Growing, Growing Dots
Human review completed .
1. In class: A Develop Understanding Task
Multiplying by the same factor
At time 0 there are 3 dots. Each minute, every group doubles in size: the counts are 3, 6, 12, 24, 48. This is a geometric sequence: consecutive values are multiplied by the same number, called the common ratio. Here that ratio is 2. In 1.2, we added 4 each minute; here we multiply by 2.
| Time (minutes) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Dots | 3 | 6 | 12 | 24 | 48 | 96 |
Recursive first: use the previous value
A recursive rule says how to find the next term from an earlier term. It needs a starting value. Let \(D(t)\) be the number of dots after \(t\) minutes:
Read this as “start with 3 dots; the next dot count is twice the current count.” \(D(t+1)\) is the output one minute later, not \(D(t)+1\).
Explicit: go directly to a time
An explicit rule calculates an output directly from its input. After \(t\) minutes, there have been \(t\) doublings. Repeated multiplication by 2 is \(2^t\):
The coefficient 3 is the starting dot count. The base 2 is the growth factor per minute. At time 0, \(2^0=1\), so the rule gives the correct starting value 3. A nonzero number to the power 0 is 1.
Worked example — Main 2
Original problem: Find the number of dots at 5 minutes.
Worked solution: Double 48 once to get 96, or calculate \(3\cdot2^5=3\cdot32=96\). Answer: 96 dots.
2. Ready
Work independently first. Use the answers below to check your work after trying the problems, or for help when you are stuck.
Use the two rows of the packet’s table together. If you are given an output, find it in the bottom row and read its input from the top row. Then perform any requested change to that input and look up the new output.
Worked example — Ready 1
Original problem: A: when \(f(n)=12\), find \(n\). B: find \(f(n-1)\).
Worked solution: The table puts 12 below \(n=5\). Therefore \(n-1=4\), and \(f(4)=7\). Do not subtract 1 from the output 12; \(f(n-1)\) changes the input first.
Check your work — Ready 1–6
| Problem | A — n | B — Output |
|---|---|---|
| 1 | 5 | 7 |
| 2 | 6 | 12 |
| 3 | 9 | 37 |
| 4 | 3 | 17 |
| 5 | 8 | -3 |
| 6 | 1 | 37 |
3. Set
Work independently first. Use the answers below to check your work after trying the problems, or for help when you are stuck.
If the requested term is just one step after a known term, the recursive rule is usually quicker. If it is far away, the explicit rule avoids calculating every term between them. Other correct choices are possible if justified.
Check your work — Set 7–12
| Problem | Value | A convenient method and reason |
|---|---|---|
| 7 | 19 | Recursive: 16+3, one step. |
| 8 | 157 | Explicit: 3(50)+7. |
| 9 | 198 | Explicit: 10(20)−2. |
| 10 | 88 | Recursive: 78+10. |
| 11 | 160 | Recursive: 80×2. |
| 12 | 640 | Explicit: 5×2⁷=640; three doublings from 80 also work. |
4. Go
Work independently first. Use the answers below to check your work after trying the problems, or for help when you are stuck.
Evaluate the power before multiplying by any outside coefficient or negative sign. Parentheses change the base: \((-3)^2=9\), but \(-3^2=-(3^2)=-9\). Thus \((-3)^x\) alternates signs at whole-number inputs, while \(-3^x\) is always negative.
Check your work — Go 13–17
| Problem / rule | Outputs at x=1,2,3,4,5 |
|---|---|
| 13 | 4, 16, 64, 256, 1024 |
| 14 | −3, 9, −27, 81, −243 |
| 15 | −3, −9, −27, −81, −243 |
| 16 | 10, 100, 1000, 10000, 100000 |
17: \(f(4)=5^4=625\).
Additional learning resources
| Resource | What it helps with and scope |
|---|---|
| Khan Academy — Evaluating functions given their formula | Review function inputs and outputs; supports substituting into explicit rules. Not specific to geometric dot patterns. |
| Khan Academy — Intro to geometric sequences | Close match for the initial term and common ratio. The video may use subscript notation; this lesson uses function notation for the same term values. |
Teacher/parent review of the source packet
No substantial mathematical issues were found in this section.
Prompts and data adapted from Mathematics Vision Project, Secondary Mathematics I, Module 1 student packet, Section 1.3, pages 11–14, licensed under CC BY 4.0. Explanations and diagrams prepared for this supplement.
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