1.2 Representing a Growing Pattern
SLO F-IF
Interpreting Functions. Understand what a function is and use function notation;
Learning Objectives
By the end of this section, you will be able to:
- turn a growing pattern of figures into a table of values, and read the pattern of change off the table;
- decide from a table whether a pattern grows by adding the same amount each step or by multiplying by the same amount each step;
- write a recursive rule and an explicit rule for a pattern, and choose whichever one makes a given question easier;
- sketch and read a graph of a situation, and say what its flat, rising, and falling parts mean;
- move in any direction among picture, table, graph, and rule for the same pattern.
1.2.1 The Border, Now as a Table
Section 1.1 left the cafeteria floor with four expressions on it. The floor is an \(N\) by \(N\) block of center tiles inside a border exactly two tiles thick on every side — that thickness is where every number below comes from, and it is why the whole floor measures \(N+4\) tiles on a side. Marisol, Jamal, Priya and Wen each cut that border into different pieces, and each wrote down what they saw:
| Student | How they cut the border | Expression |
|---|---|---|
| Marisol | whole floor minus the center hole | \((N+4)^2 - N^2\) |
| Jamal | four pinwheel strips, each 2 thick and \(N+2\) long | \(4 \cdot 2(N+2)\) |
| Priya | four 2-by-\(N\) sides plus four 2-by-2 corners | \(4(2N) + 4(4)\) |
| Wen | two strips across the top and bottom, two up the sides | \(2 \cdot 2(N+4) + 2 \cdot 2N\) |
Those are the total border tiles, colored and white together — the left-hand column of §1.1's Table 1.1.3. Section 1.1 then went one step further and halved them, because the contractor only wanted the colored ones; that is why §1.1 finished on \(4N + 8\) and this table starts from \(8N + 16\). The halving is settled and does not come back. From here on, "border tiles" means all of them.
Section 1.1 answered one question about those four: are they the same? Multiply each one out and they all collapse to the same thing.
Simplify each of the four students' expressions. What do all four turn out to be?
Solution
Take them in the order they appear, starting with the one that looks least like the others.
Marisol — whole floor minus the center hole. The whole floor is \((N+4)\) tiles on a side, so its area is \((N+4)^2\). Removing the center leaves
$$ (N+4)^2 - N^2 = N^2 + 8N + 16 - N^2 = 8N + 16 $$The \(N^2\) terms cancel, which is why a quadratic-looking expression describes something that grows steadily.
Jamal — four pinwheel strips.
$$ 4 \cdot 2(N+2) = 8(N+2) = 8N + 16 $$Priya — four sides plus four corners.
$$ 4(2N) + 4(4) = 8N + 16 $$Already there, with no work needed.
Wen — two strips across, two strips up.
$$ 2 \cdot 2(N+4) + 2 \cdot 2N = 4(N+4) + 4N = 4N + 16 + 4N = 8N + 16 $$Answer: All four give \(8N + 16\). They were four correct descriptions of one border the whole time.
That question is closed. Here is the one §1.1 never asked: what does the border do as \(N\) grows?
Nobody can answer that from a single floor. You need several, side by side — which is a table.
| Center size, \(N\) | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Border tiles, \(B\) | 24 | 32 | 40 | 48 | 56 | 64 |
The \(N = 5\) column is the floor from §1.1, and it says 56, which is what all four students counted. That agreement is the point of building the table at all: the new representation has to reproduce what you already knew before you trust it with what you did not.
Before a table tells you about \(N = 6\), it has to agree with §1.1 about \(N = 5\). That is not busywork — it is the only reason to believe the new column.
Now look along the bottom row instead of down a column. Each entry is 8 more than the one before it. Every time the center square grows by one tile on a side, the border needs exactly 8 more tiles — and the expression says so out loud, because the \(8\) in \(8N + 16\) is the number multiplying \(N\).
This is the move the whole section is built on. Four expressions became one expression in §1.1 by simplifying. One expression became a visible pattern here by tabulating. Each representation makes a different fact obvious, and none of them makes every fact obvious.
Summary of the four representations. The table below collects what each one is good at, for quick reference.
| Representation | What it looks like | What it is good at |
|---|---|---|
| Picture / diagram | a sequence of figures, one per step | showing why the rule works — you can point at the new pieces |
| Table | inputs in one column, outputs in the next | spotting how the output changes from one step to the next |
| Graph | points or a curve on a coordinate plane | showing the overall shape of the growth at a glance |
| Rule (equation) | symbols such as \(B = 8N + 16\) | jumping straight to a far-away step without listing everything before it |
Read that last column again, because it is a working strategy and not just a summary. Each representation has a question it answers almost instantly and a question it answers badly. "What happens next?" is a table question. "Is this speeding up or slowing down?" is a graph question. "What is the value at step 500?" is a rule question. "Why is the rule that and not something else?" is usually a picture question.
The skill, then, is not picking a favorite. It is being able to get from any one of the four to any other.
One more thing carries forward from §1.1, and it is worth saying before the new patterns arrive. Jamal's \(4 \cdot 2(N+2)\) and Priya's \(4(2N) + 4(4)\) are the same rule wearing different clothes, and the shorter form, \(8N + 16\), is not automatically the better one. It is the fastest to compute with and the worst at explaining itself: nothing in \(8N + 16\) points at a corner or a side. When the goal is to explain a pattern rather than compute with it, the messier expression is frequently the more useful one. That will be true of every rule in this section.
Use the border table to answer both parts.
a) What is \(B\) when \(N = 7\)? Answer it twice — once by continuing the table's pattern, once from the rule \(B = 8N + 16\).
b) A border took 104 tiles. How big was the center square?
Solution
a) From the table. The table ends at \(N = 6\) with 64 tiles. Each step adds 8, so \(N = 7\) gives
$$ 64 + 8 = 72 $$From the rule.
$$ B = 8(7) + 16 = 56 + 16 = 72 $$Both give 72. Two representations agreeing is the check described above.
b) Working backwards. We want \(8N + 16 = 104\).
$$ 8N = 104 - 16 = 88 $$ $$ N = \frac{88}{8} = 11 $$Answer: a) 72 tiles. b) The center square was 11 by 11.
Notice that part (b) is a rule question, not a table question — continuing the table to \(N = 11\) would work, but it is five more rows of arithmetic to get the same answer.
1.2.2 A Pattern That Grows by Adding
Here is the first of the two anchor patterns. It is called Growing Dots.
A single dot sits in the middle of the page. That is the picture at 0 minutes. After one minute, four new dots have appeared, one out along each diagonal from the center — upper left, upper right, lower left, lower right — so the figure now looks like a small letter X. After two minutes, four more dots have appeared, one at the end of each of the four arms, so each arm is now two dots long. The pattern continues: every minute, each of the four arms grows by exactly one dot.
Step 1: from the picture to a table
Counting the dots in each figure gives a table. The input is the number of minutes; the output is the number of dots.
| Minutes, \(t\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Dots, \(D\) | 1 | 5 | 9 | 13 | 17 | 21 |
Even without the pictures, this table answers the task's second question: at 3 minutes there are 13 dots.
Step 2: from the table to the pattern of change
Look at what happens between neighboring columns. Write the differences underneath:
$$5 - 1 = 4 \qquad 9 - 5 = 4 \qquad 13 - 9 = 4 \qquad 17 - 13 = 4$$The same number every time — exactly as the border's row of differences was 8 every time. That is the signature of the first kind of growth, and it is worth a name.
A pattern has a constant difference when you get from each output to the next by adding the same number. That fixed number is the difference.
The outputs tell you where the pattern is; the gaps tell you what it is doing. Almost every question in this section is answered by looking at the gaps.
Definition 1.2.1 — Constant Difference: four new dots arrive every minute, and that is where the 4 comes from.
Here the constant difference is \(4\), and the picture explains where the 4 comes from: four arms, one new dot on each, every minute. You can point at the four new dots. That is the picture doing a job the table cannot do.
Step 3: a recursive rule
A recursive rule writes the constant difference down formally. It has two parts: where the pattern starts, and how to get from one step to the next.
A recursive rule is the staircase. Each step is easy, and you have to take every one of them. An explicit rule, coming next, is the elevator: press the floor number and arrive. Both reach floor 100. Only one of them is worth doing by hand.
A recursive rule gives the first value of a pattern and then tells you how to build each new value from the value before it.
Definition 1.2.2 — Recursive Rule: one hop gets you the next step, and the hundredth costs a hundred hops.
For Growing Dots the recursive rule is:
$$D(0) = 1, \qquad D(t) = D(t-1) + 4$$Read aloud, that is: "Start with 1 dot. Every minute, the number of dots is whatever it was last minute, plus 4." The notation \(D(t-1)\) just means the value one step back. Some books write this as now = previous + 4, which says the same thing in words.
A recursive rule is exactly what you want for the dots at 3 minutes: take the count at 2 minutes and add 4. Done. It is exactly what you do not want for the dots at 100 minutes.
Step 4: an explicit rule
Finding the dots at 100 minutes recursively means adding 4 to itself a hundred times, and the recursive rule offers no shortcut. An explicit rule repairs that.
An explicit rule computes the value at any step directly from the step number, without needing the value of the step before it.
Definition 1.2.3 — Explicit Rule: four arms of t dots plus one center dot is exactly what 4t + 1 says.
Build it from the picture rather than by guessing, the way §1.1's four students built theirs from the floor. At \(t\) minutes:
- there is 1 dot in the center, which was there from the start and never changes;
- there are 4 arms, and each arm has grown by one dot per minute for \(t\) minutes, so each arm holds \(t\) dots;
- the four arms together hold \(4t\) dots.
Every piece of that expression points at something you can see. The \(4\) is the number of arms. The \(t\) is how long they have been growing. The \(+1\) is the center dot. That is the same test §1.1 applied to \(4N + 8\) — point at a term and name what it counts — and it is the difference between understanding a rule and having memorized one.
Verify \(D(t) = 4t + 1\) against the table, then find the number of dots at 100 minutes.
Solution
Step 1 — check every value you already know. Do not trust a rule you have not checked.
| \(t\) | \(4t + 1\) | Table says |
|---|---|---|
| 0 | \(4(0) + 1 = 1\) | 1 |
| 1 | \(4(1) + 1 = 5\) | 5 |
| 2 | \(4(2) + 1 = 9\) | 9 |
| 3 | \(4(3) + 1 = 13\) | 13 |
Four agreements. The rule reproduces everything the table already knew.
Step 2 — now ask it something the table cannot answer.
$$D(100) = 4(100) + 1 = 400 + 1 = 401$$Answer: 401 dots at 100 minutes. Recursively this would have taken a hundred additions; explicitly it takes one multiplication and one addition.
And the general question is already answered — \(D(t) = 4t + 1\) is the count at \(t\) minutes, for every \(t\) at once.
Step 5: the graph
Plot the pairs from the table as points: \((0, 1)\), \((1, 5)\), \((2, 9)\), \((3, 13)\), \((4, 17)\). Put minutes on the horizontal axis and dots on the vertical axis.
Those five points lie on a straight line. They have to. Every time you move one unit to the right, you move exactly 4 units up, because the difference is constant — and "the same rise for every run" is what makes a line straight. The line meets the vertical axis at \((0, 1)\), which is the single starting dot.
So the constant difference of 4 is visible in all four representations at once:
| Representation | Where the 4 shows up |
|---|---|
| Picture | four new dots appear each minute |
| Table | each output is 4 more than the one before |
| Graph | the line climbs 4 units for every 1 unit across |
| Rule | the coefficient of \(t\) in \(D(t) = 4t + 1\) |
The border from §1.2.1 would do the same thing with an 8 in place of the 4. Two different situations, one shape of growth.
A different dot pattern starts with 3 dots and gains 6 dots every minute.
a) Write the recursive rule and the explicit rule.
b) How many dots at 12 minutes?
c) A figure in this pattern has 87 dots. How many minutes have passed?
Solution
a) Recursive. Start at 3, add 6 each minute:
$$P(0) = 3, \qquad P(t) = P(t-1) + 6$$Explicit. The 3 never changes; the 6 arrives \(t\) times:
$$P(t) = 6t + 3$$b) At 12 minutes.
$$P(12) = 6(12) + 3 = 72 + 3 = 75$$c) Working backwards. We want \(6t + 3 = 87\).
$$6t = 84 \qquad t = 14$$Answer: a) \(P(0) = 3,\; P(t) = P(t-1) + 6\) and \(P(t) = 6t + 3\). b) 75 dots. c) 14 minutes.
Check part (c) forwards: \(6(14) + 3 = 84 + 3 = 87\). Agrees.
1.2.3 A Pattern That Grows by Multiplying
Now the companion task, Growing, Growing Dots. It is set up to look like the first one and behave completely differently.
Again there is a single dot at 0 minutes. But this time, at the end of every minute, every dot in the picture is replaced by two dots. Nothing is added to the outside of the figure; instead the whole figure doubles.
The table
| Minutes, \(t\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Dots, \(N\) | 1 | 2 | 4 | 8 | 16 | 32 |
The task asks for the count at 5 minutes: 32 dots.
The pattern of change
The doubling table does not have a constant difference. Show that, then find what it does have.
Solution
Step 1 — try the first test. Subtract each output from the next one:
$$2 - 1 = 1 \qquad 4 - 2 = 2 \qquad 8 - 4 = 4 \qquad 16 - 8 = 8$$The differences are \(1, 2, 4, 8\). They are not constant, so this pattern does not grow by adding a fixed amount.
Step 2 — a failed test is information. It does not mean "no pattern." It means try the other test.
$$\dfrac{2}{1} = 2 \qquad \dfrac{4}{2} = 2 \qquad \dfrac{8}{4} = 2 \qquad \dfrac{16}{8} = 2 \qquad \dfrac{32}{16} = 2$$Answer: The differences are not constant, but the ratios are all \(2\). This is the multiplying kind of growth, with ratio 2.
Constant. That is the signature of the second kind of growth.
When the differences are not constant, you have not run out of options — you have eliminated one of two, which is exactly what a test is for.
A pattern has a constant ratio when you get from each output to the next by multiplying by the same number. That fixed number is the ratio.
Definition 1.2.4 — Constant Ratio: every dot becomes two dots, so the whole count is multiplied by 2.
Here the constant ratio is \(2\), and again the picture says why: each dot becomes two dots, so the whole count is multiplied by 2.
The two rules
The recursive rule swaps the "\(+\)" for a "\(\times\)":
$$N(0) = 1, \qquad N(t) = 2 \cdot N(t-1)$$In words: start with 1 dot; every minute, double whatever you had. In the "now = previous" form used in some of the practice sets, that is now = previous \(\times\) 2.
The explicit rule needs a way to say "doubled \(t\) times." That is what exponent notation is for:
$$\underbrace{2 \cdot 2 \cdot 2 \cdots 2}_{t \text{ factors}} = 2^{t}$$so
$$N(t) = 2^{t}$$Check it:
| \(t\) | \(2^{t}\) | Expanded | Table says |
|---|---|---|---|
| 0 | \(2^{0}\) | — | 1 |
| 1 | \(2^{1}\) | \(2\) | 2 |
| 2 | \(2^{2}\) | \(2 \cdot 2\) | 4 |
| 3 | \(2^{3}\) | \(2 \cdot 2 \cdot 2\) | 8 |
| 5 | \(2^{5}\) | \(2 \cdot 2 \cdot 2 \cdot 2 \cdot 2\) | 32 |
You are only using exponent notation here as shorthand for repeated multiplication. The rules for working with exponents — what happens when you multiply two powers, what a negative exponent means, why \(2^0 = 1\) — are developed in §1.5.
The graph, and the real difference
Plot \((0,1)\), \((1,2)\), \((2,4)\), \((3,8)\), \((4,16)\), \((5,32)\).
These points do not lie on a straight line. From \(t = 0\) to \(t = 1\) the climb is 1 unit. From \(t = 4\) to \(t = 5\) the climb is 16 units. The points start out nearly flat and then sweep upward, getting steeper the further right you go. A constant ratio produces a curve that bends upward, and it bends harder and harder.
Set the two anchor patterns side by side. Both start at 1 dot. Both are perfectly regular.
| Minutes | Growing Dots \((4t+1)\) | Growing, Growing Dots \((2^t)\) |
|---|---|---|
| 0 | 1 | 1 |
| 1 | 5 | 2 |
| 2 | 9 | 4 |
| 3 | 13 | 8 |
| 4 | 17 | 16 |
| 5 | 21 | 32 |
| 10 | 41 | 1,024 |
| 20 | 81 | 1,048,576 |
For the first four minutes the adding pattern is comfortably ahead. It never recovers. By 20 minutes the multiplying pattern is more than ten thousand times larger, and the gap keeps widening forever. This is not a quirk of these particular numbers; multiplying repeatedly eventually overwhelms adding repeatedly, no matter how big the amount being added is.
Keep \(N(t) = 2^{t}\) where you can find it. It comes back twice: §1.3 evaluates it in function notation, and §1.5 takes the exponent apart.
A colony starts with 5 bacteria and triples every hour.
a) Write the recursive rule and the explicit rule.
b) How many bacteria after 4 hours?
c) Both this colony and a second one that starts at 5 and adds 30 every hour have 5 bacteria at hour 0. Which is larger at hour 2, and which at hour 5?
Solution
a) Recursive.
$$C(0) = 5, \qquad C(t) = 3 \cdot C(t-1)$$Explicit. Starting at 5 and tripling \(t\) times:
$$C(t) = 5 \cdot 3^{t}$$b) After 4 hours.
$$C(4) = 5 \cdot 3^{4} = 5 \cdot 81 = 405$$c) The two colonies. The adding colony is \(A(t) = 30t + 5\).
| Hour | Tripling \(5 \cdot 3^{t}\) | Adding \(30t + 5\) |
|---|---|---|
| 2 | \(5 \cdot 9 = 45\) | \(60 + 5 = 65\) |
| 5 | \(5 \cdot 243 = 1{,}215\) | \(150 + 5 = 155\) |
Answer: a) \(C(0) = 5,\; C(t) = 3 \cdot C(t-1)\) and \(C(t) = 5 \cdot 3^{t}\). b) 405 bacteria. c) The adding colony is larger at hour 2 (65 against 45); the tripling colony is far larger at hour 5 (1,215 against 155).
This is the same crossover the comparison table shows: adding wins early, multiplying wins permanently.
1.2.4 Telling the Two Apart from a Table
Given a bare table with no picture and no story, there is a short procedure that settles which kind of growth you are looking at — or tells you it is neither.
1. Subtract each output from the next. If those differences are all the same, the pattern grows by adding. Record the difference.
2. If not, divide each output by the one before it. If those ratios are all the same, the pattern grows by multiplying. Record the ratio.
3. If neither test comes out constant, the pattern is neither — it is still a pattern, just not one of these two.
Both tests need to be run on every neighboring pair, not just the first one. Two matching differences prove nothing; the sequence \(1, 3, 5, 11, 21\) starts out looking like it adds 2.
Run the procedure on four short tables.
| Term | 1st | 2nd | 3rd | 4th | 5th | Test | Verdict |
|---|---|---|---|---|---|---|---|
| Value | 2 | 4 | 8 | 16 | 32 | ratios all \(2\) | multiplying |
| Value | 66 | 50 | 34 | 18 | 2 | differences all \(-16\) | adding (a negative amount) |
| Value | 160 | 80 | 40 | 20 | 10 | ratios all \(\tfrac{1}{2}\) | multiplying (shrinking) |
| Value | \(-9\) | \(-2\) | 5 | 12 | 19 | differences all \(+7\) | adding |
Two of those rows deserve comment.
The second row goes down by 16 each time. It is still the adding kind of growth — the amount being added is just negative. "Growth" is the traditional word, but the pattern is shrinking, and nothing about the test changes.
The third row also shrinks, and it is the multiplying kind: each value is half the one before it. A ratio between 0 and 1 makes the values shrink toward zero without ever quite reaching it. On a graph, this is the upward-sweeping curve of §1.2.3 running in reverse — steep at first, then flattening out.
Naming comes later
These two kinds of growth do have names, and Chapter 4 gives them. A name is what you call the answer after the test has given it to you; it is not a shortcut to the answer. Someone who remembers only "the fast one grows by multiplying" will say that about every fast-growing table, and will be wrong about half the time. For now, run the test and report what it found.
A pattern that passes neither test is neither. That is a real verdict, not a failure to classify.
Worked classifications
For each table below: describe how to get the next term, write a recursive rule, describe how to jump to the \(n^{\text{th}}\) term, write an explicit rule, and say which kind of growth the tests found.
The table pairs \(x = 0, 1, 2, 3, 4\) with \(y = 5, 8, 11, 14, \; ?\). Classify it and write both rules.
Solution
Step 1 — run test one. Differences: \(8 - 5 = 3\), \(11 - 8 = 3\), \(14 - 11 = 3\). Constant, so the pattern grows by adding, and the first test settles it.
Step 2 — next term. Add 3 to the previous term: \(14 + 3 = 17\).
Step 3 — recursive rule.
$$f(0) = 5, \qquad f(n) = f(n-1) + 3$$Step 4 — explicit rule. To find the \(n^{\text{th}}\) term, start with 5 and add 3 a total of \(n\) times:
$$f(n) = 5 + 3n$$Check. \(f(3) = 5 + 9 = 14\). Agrees with the table.
Answer: Grows by adding, constant difference 3. Recursive \(f(0) = 5,\; f(n) = f(n-1) + 3\); explicit \(f(n) = 5 + 3n\).
Table A. \(x = 1, 2, 3, 4\) pairs with \(y = 5, 10, 20, 40\). Classify it and write both rules.
Solution
Step 1 — test one. Differences are \(5, 10, 20\) — not constant.
Step 2 — test two. Ratios are \(\tfrac{10}{5} = 2\), \(\tfrac{20}{10} = 2\), \(\tfrac{40}{20} = 2\) — constant.
Step 3 — the rules. Next term: multiply the previous term by 2, so the 5th term is \(80\).
$$f(1) = 5, \qquad f(n) = 2 \cdot f(n-1)$$For the explicit rule, start at 5 and double it \(n - 1\) times, because the first term has not been doubled at all:
$$f(n) = 5 \cdot 2^{\,n-1}$$Check at \(n = 4\): \(5 \cdot 2^{3} = 5 \cdot 8 = 40\). Agrees.
Answer: Grows by multiplying, constant ratio 2. Recursive \(f(1) = 5,\; f(n) = 2 \cdot f(n-1)\); explicit \(f(n) = 5 \cdot 2^{\,n-1}\).
The \(n - 1\) in the exponent is the part worth slowing down for. The table starts at \(n = 1\), not \(n = 0\), so by the time you reach term \(n\) you have doubled one fewer time than the term number suggests.
Table B. \(x = 1\) through \(6\) pairs with \(y = -8, -17, -26, -35, -44, -53\).
- Next term: subtract 9 from the previous term.
- Recursive rule: \(f(1) = -8\), \(\;f(n) = f(n-1) - 9\).
- To find the \(n^{\text{th}}\) term: start at \(-8\) and subtract 9 a total of \(n - 1\) times.
- Explicit rule: \(f(n) = -8 - 9(n-1)\), which simplifies to \(f(n) = -9n + 1\).
- Grows by adding.
Check at \(n = 4\): \(-9(4) + 1 = -36 + 1 = -35\). Agrees.
Table C. \(x = 1\) through \(6\) pairs with \(y = 2, 6, 18, 54, 162, 486\).
Ratios: \(\tfrac{6}{2} = 3\), \(\tfrac{18}{6} = 3\), \(\tfrac{54}{18} = 3\), \(\tfrac{162}{54} = 3\), \(\tfrac{486}{162} = 3\).
- Next term: multiply the previous term by 3.
- Recursive rule: \(f(1) = 2\), \(\;f(n) = 3 \cdot f(n-1)\).
- Explicit rule: \(f(n) = 2 \cdot 3^{\,n-1}\).
- Grows by multiplying.
Table D. \(x = 1\) through \(6\) pairs with \(y = 3, 15, 27, 39, 51, \; ?\).
- Next term: add 12. The 6th term is \(63\).
- Recursive rule: \(f(1) = 3\), \(\;f(n) = f(n-1) + 12\).
- Explicit rule: \(f(n) = 3 + 12(n-1) = 12n - 9\).
- Grows by adding.
Table E. \(x = 0\) through \(5\) pairs with \(y = 3, 4, 7, 12, 19, \; ?\). Classify it.
Solution
Step 1 — test one. Differences: \(1, 3, 5, 7\). Not constant, so it does not grow by adding.
Step 2 — test two. Ratios: \(\tfrac{4}{3}\), \(\tfrac{7}{4}\), \(\tfrac{12}{7}\) — nowhere near constant, so it does not grow by multiplying either.
Step 3 — the verdict is "neither," and it is still a pattern. The differences \(1, 3, 5, 7\) grow by 2 each time, so the next difference is 9 and the next term is \(19 + 9 = 28\).
Step 4 — an explicit rule anyway.
$$f(n) = n^{2} + 3$$Check: \(f(0) = 0 + 3 = 3\); \(f(3) = 9 + 3 = 12\); \(f(5) = 25 + 3 = 28\). Agrees throughout.
Answer: Neither kind, with explicit rule \(f(n) = n^{2} + 3\).
Table E is the reason step 3 of the procedure exists. Plenty of perfectly orderly patterns are neither, and calling one of them the multiplying kind because it grows quickly is a guess, not a verdict.
Table F. \(x = 1\) through \(6\) pairs with \(y = -1,\; 0.2,\; -0.04,\; 0.008,\; -0.0016,\; 0.00032\). Classify it.
Solution
Step 1 — the signs are alarming; the test does not care. Divide as usual:
$$\dfrac{0.2}{-1} = -0.2 \qquad \dfrac{-0.04}{0.2} = -0.2 \qquad \dfrac{0.008}{-0.04} = -0.2$$Constant, so the pattern grows by multiplying, with constant ratio \(-0.2\).
Step 2 — the rules.
$$f(1) = -1, \qquad f(n) = -0.2 \cdot f(n-1)$$ $$f(n) = -1 \cdot (-0.2)^{\,n-1}$$Answer: Grows by multiplying, constant ratio \(-0.2\).
Two features fall straight out of that one number. A negative ratio is what makes the signs flip back and forth. A ratio whose size is less than 1 is what makes the values shrink toward zero. Neither requires a new idea — both are read off the ratio you already computed.
Choosing the rule that makes the question easy
Both rules describe the same pattern, so both give the same answers. They do not cost the same amount of work. Given a sequence starting \(10, 13, 16, \ldots\) with explicit rule \(y = 3x + 7\) and recursive rule now = previous \(+ 3\):
| Question | Better tool | Why | Answer |
|---|---|---|---|
| Find the 4th term | recursive | the 3rd term (16) is already in front of you; add 3 | \(19\) |
| Find the 50th term | explicit | recursion would need 47 more additions | \(3(50) + 7 = 157\) |
The same trade-off with a different sequence. The 8th term is 78, the sequence increases by 10 each step, the explicit rule is \(y = 10x - 2\):
| Question | Better tool | Work | Answer |
|---|---|---|---|
| Find the 9th term | recursive | \(78 + 10\) | \(88\) |
| Find the 20th term | explicit | \(10(20) - 2\) | \(198\) |
And once more with a multiplying sequence. The 4th term is 80, the sequence doubles each step, the explicit rule is \(y = 5(2^{x})\):
| Question | Better tool | Work | Answer |
|---|---|---|---|
| Find the 5th term | recursive | \(80 \times 2\) | \(160\) |
| Find the 7th term | explicit | \(5(2^{7}) = 5(128)\) | \(640\) |
The pattern in these three tables is worth stating outright: a recursive rule is fastest for the step right next to one you already know; an explicit rule is fastest for a step far away. Neither is the better rule in general. Reach for the one that matches the question.
Classify each table, and write an explicit rule for it.
a) \(9, 16, 23, 30, 37\)
b) \(3, 12, 48, 192\)
c) \(1, 4, 9, 16, 25\)
Solution
a) Differences: \(7, 7, 7, 7\). Constant, so it grows by adding, with constant difference 7. Starting at 9 with \(n = 1\):
$$f(n) = 9 + 7(n-1) = 7n + 2$$Check at \(n = 3\): \(21 + 2 = 23\). Agrees.
b) Differences: \(9, 36, 144\) — not constant. Ratios: \(\tfrac{12}{3} = 4\), \(\tfrac{48}{12} = 4\), \(\tfrac{192}{48} = 4\). Constant, so it grows by multiplying, with constant ratio 4:
$$f(n) = 3 \cdot 4^{\,n-1}$$Check at \(n = 3\): \(3 \cdot 16 = 48\). Agrees.
c) Differences: \(3, 5, 7, 9\) — not constant. Ratios: \(4, \tfrac{9}{4}, \tfrac{16}{9}, \tfrac{25}{16}\) — not constant. Neither. These are the square numbers:
$$f(n) = n^{2}$$Answer: a) grows by adding, \(f(n) = 7n + 2\). b) grows by multiplying, \(f(n) = 3 \cdot 4^{\,n-1}\). c) neither, \(f(n) = n^{2}\).
Part (c) grows quickly, and growing quickly is not the same as multiplying by a constant. Only the ratio test decides that.
1.2.5 Graphs of Situations
Not every relationship arrives as a picture of dots. Often it arrives as a story, and the representation that captures a story best is a graph — because the shape of a graph carries meaning that a table of numbers hides.
The basic vocabulary is short.
Summary of graph shapes. The table below collects what each shape says about the situation.
| What the graph does | What it means about the situation |
|---|---|
| rises left to right | the output quantity is increasing |
| falls left to right | the output quantity is decreasing |
| stays flat (horizontal) | the output quantity is not changing |
| rises steeply | it is increasing quickly |
| rises gently | it is increasing slowly |
| a straight segment | the change is happening at a constant rate |
| a bending curve | the rate of change is itself changing |
One situation, two graphs
Go back to the cafeteria floor. Marisol and Jamal are laying the 56 border tiles of the \(N = 5\) design, and they lay 8 tiles an hour between them. Three hours in they stop for an hour, then finish the job.
The next example graphs that afternoon twice, changing nothing about the afternoon.
Graph the tiles laid so far against hours elapsed. Then graph the tiles still to lay against hours elapsed.
Solution
Step 1 — tiles laid so far. At 8 tiles an hour, three hours of work lays \(3 \times 8 = 24\) tiles. The break adds none. The remaining \(56 - 24 = 32\) tiles take \(32 \div 8 = 4\) more hours, finishing at hour 8.
| Hours | What is happening | Tiles laid | Shape of the graph |
|---|---|---|---|
| 0 to 3 | laying at a steady 8 an hour | 0 up to 24 | a straight segment rising |
| 3 to 4 | the break | stays at 24 | a horizontal segment |
| 4 to 8 | back to a steady 8 an hour | 24 up to 56 | a straight segment rising, at the same steepness as the first |
Step 2 — tiles still to lay. Same afternoon, different quantity on the vertical axis.
| Hours | Tiles remaining | Shape of the graph |
|---|---|---|
| 0 to 3 | 56 down to 32 | a straight segment falling |
| 3 to 4 | stays at 32 | a horizontal segment |
| 4 to 8 | 32 down to 0 | a straight segment falling |
Answer: One graph climbs from 0 to 56; the other drops from 56 to 0. Both have a flat piece from hour 3 to hour 4, and both are made of straight pieces because the work happened at a constant rate.
The flat stretch is the part people get wrong. An hour of standing around still feels like something happening, and the instinct is to draw something busy. But the graph is tracking tiles laid, and that quantity does absolutely nothing for that hour. A flat graph does not mean nothing is happening. It means the quantity being graphed is not changing.
Same floor, same workers, same break, same eight hours — and one graph climbs while the other drops. Nothing about the situation changed. What changed is which quantity you decided to put on the vertical axis, and that choice does more to the shape of a graph than anything the workers did. Section 1.4 comes back to this hard, because a graph you have misread is usually a graph whose axis you did not check.
Reading a point off a graph
A point on a graph is always written \((\text{input}, \text{output})\) — the horizontal coordinate first. On the tiles-laid graph, the point \((2, 16)\) means two hours in, 16 tiles had been laid. It does not mean 16 hours and 2 tiles, and it does not mean 2 tiles per hour. Reading a point correctly means reading the axis labels, every time.
That rule is also what lets a graph act as a table you can read continuously. The table above lists five moments; the graph fills in every moment between them, which is why you can ask it about hour \(2\tfrac{1}{2}\) and get an answer.
Where the graph meets the axes
Two places on any graph get asked about so often that they have names, and you have already met both without being told:
- Where the tiles-remaining graph reaches the horizontal axis, the count has hit zero and the job is finished.
- Where either graph meets the vertical axis, you are reading the value at the moment the clock started — 0 tiles laid, or 56 still to lay.
Those two places are the intercepts, and the highest and lowest values a graph reaches have names too. All four are defined properly in §1.4, where they are the section's actual subject and where reading them off a graph is the work. For now it is enough to notice that they are worth asking about, and that each one answers a question about the situation rather than about the picture.
Priya fills a bucket from a tap for 2 minutes, carries it across the yard for 1 minute, then pours it out over 30 seconds.
a) Sketch the graph of the water in the bucket against time, and say what each piece means.
b) One of the three pieces is flat. Explain why "flat" is the right shape even though Priya is working hardest during it.
Solution
a) Three pieces.
| Time | What is happening | Water in the bucket | Shape |
|---|---|---|---|
| 0 to 2 min | filling from the tap | rising from empty to full | a straight segment rising |
| 2 to 3 min | carrying it across the yard | stays full | a horizontal segment |
| 3 to 3.5 min | pouring it out | falling from full to empty | a steep straight segment falling |
The pouring piece is steeper than the filling piece because the same amount of water leaves in a quarter of the time.
b) Why flat is right. The graph tracks water in the bucket, and carrying a bucket changes where the water is, not how much of it there is. Priya's effort is not the quantity on the vertical axis. This is the same trap as the tile-layers' break, arriving from the opposite direction: there, nothing was happening and the graph was flat; here, plenty is happening and the graph is still flat.
Answer: Rising, then flat, then steeply falling — and the flat piece is flat because the graphed quantity is unchanged, not because Priya is idle.
1.2.6 Moving Between the Representations
Everything above is one skill practiced in different directions. Collected in one place:
Summary of the eight moves. The table below collects every route between the four representations.
| Starting from | Getting to | How |
|---|---|---|
| Picture | Table | count the objects in each figure; record step number and count |
| Table | Rule | test differences, then ratios; use whichever is constant |
| Rule | Table | substitute step numbers into the rule and evaluate |
| Table | Graph | plot each (input, output) pair as a point |
| Graph | Table | read coordinates off labeled points |
| Story | Graph | translate "increasing / decreasing / unchanged" into rising / falling / flat |
| Graph | Story | describe each straight or curved piece in the language of the situation |
| Picture | Rule | name what each visible part of the figure contributes, then add the parts |
Two habits make all of this reliable.
Check across representations, not within one. Once you have a rule, put a table value into it. Once you have a graph, confirm two of its points against the table. A mistake almost never shows up in the representation where it was made; it shows up when a second representation disagrees. That is exactly what the \(N = 5\) column did back in §1.2.1: the table had to reproduce §1.1's 56 before it was worth trusting for \(N = 6\).
Say what each symbol means in the situation. In \(D(t) = 4t + 1\), the 4 is dots-per-minute and the 1 is the starting dot. In \(B = 8N + 16\), the 8 is tiles-per-step and the 16 is the four corners. A rule you can narrate is a rule you can check, extend, and fix. A rule you cannot narrate is a string of characters you are hoping is right.
A rule is given as \(y = 5x - 1\).
a) Build a table for \(x = 0\) through \(4\).
b) Describe the graph of those points without drawing it.
c) Classify the pattern, and say which of the four representations you used to decide.
Solution
a) Rule to table. Substitute each \(x\):
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(y\) | \(-1\) | 4 | 9 | 14 | 19 |
b) Table to graph. The differences are \(5, 5, 5, 5\), so the points climb 5 units for every 1 unit across — a straight line. It crosses the vertical axis at \((0, -1)\), which is the value the rule gives at \(x = 0\).
c) Classify. Constant difference of 5, so the pattern grows by adding. The table is what decided it: the rule alone would have told you if you already knew that a rule of the form \(y = mx + b\) always adds a fixed amount, but the differences are the direct evidence.
Answer: a) \(-1, 4, 9, 14, 19\). b) A straight line rising 5 for every 1, meeting the vertical axis at \((0, -1)\). c) Grows by adding, decided from the table's constant difference.
Here is what §1.2 quietly assumed, and what §1.3 will make honest.
Throughout this section, phrases like "the number of dots at \(t\) minutes" and "the tiles laid after 2 hours" were used as though each named a single, definite number — as though feeding a value in always produced exactly one value out. For the dot patterns and the tile count that assumption is plainly safe.
It is not always safe. A bagel shop posts this sign:
| Number of bagels | Price |
|---|---|
| 1 | $1.25 |
| 6 | $6.00 |
| 9 | $8.00 |
| 12 | $10.00 |
A customer asks for 13 bagels. The worker charges 13 singles, $16.25. But two 6-packs and a single is $13.25; a 9-pack and four singles is $13.00; a 12-pack and one single is $11.25. Every one of those readings uses the sign correctly. The input "13 bagels" has produced four different outputs, and the shop has to choose one before the sign means anything definite.
That difference — between a relationship where each input has exactly one output and a relationship where it does not — is the subject of §1.3, and it is the idea the entire rest of this chapter rests on. Section 1.3 gives it a name, gives it a notation, and gives you tests for spotting it in each of the four representations you have just learned to move between.
Problem Set 1.2
Review — reaching back to §1.1.
Problem 1. In §1.1, Priya cut the border into four 2-by-\(N\) sides plus four 2-by-2 corners and wrote \(4(2N) + 4(4)\). Multiply it out, and say which term counts the corners.
Solution
Step 1 — Distribute: Multiply each term in \(4(2N) + 4(4)\) through:
$$ 4(2N) + 4(4) = 8N + 16 $$Step 2 — Identify the corner term: Each corner is a 2-by-2 block, so one corner is \(2 \cdot 2 = 4\) tiles, and there are four of them: \(4(4) = 16\). That is where the corners live in the expanded form.
Answer: \(4(2N) + 4(4) = 8N + 16\); the constant term \(16\) counts the four corners.
Problem 2. Marisol's \((N+4)^2 - N^2\) contains an \(N^2\) that the final answer does not. Explain where it goes and why that is not a mistake.
Solution
Step 1 — Expand: \((N+4)^2 - N^2 = N^2 + 8N + 16 - N^2\), and the two \(N^2\) terms cancel, leaving \(8N + 16\).
Step 2 — Where the \(N^2\) goes, and why it's not a mistake: The first \(N^2\) is the interior floor counted as part of the whole \((N+4)\)-by-\((N+4)\) square. Subtracting the second \(N^2\) removes exactly that same interior — the border, by definition, excludes it. Since the border expression is only supposed to describe the ring of tiles and not the interior, the interior term has to disappear for the expression to be counting the right thing.
Answer: The \(N^2\) cancels because it is added once (as part of the whole floor) and removed once (as the subtracted center hole); losing it is the whole point of the subtraction, not an error.
Problem 3. Section 1.1 argued that a different-looking answer is not automatically a disagreement. Using \(8N + 16\) and \(4 \cdot 2(N+2)\), explain what has to be checked before two expressions can be called the same.
Practice — this section's own work.
Solution
Step 1 — Simplify both: \(8N + 16\) is already simplified. For the other, \(4 \cdot 2(N+2) = 8(N+2) = 8N + 16\).
Step 2 — What has to be checked: Both expressions must be fully expanded (not just glanced at) and compared term by term, or checked against several matching values of \(N\) as in §1.1's \(N=5\) check. Two expressions can look completely different on the page and still be equal, or look similar and not be — appearance proves nothing; only expansion (or repeated numerical agreement) does.
Answer: Both simplify to \(8N + 16\), so they are the same rule; sameness is established by expanding and comparing, never by how similar the two expressions look.
Problem 4. Name the four representations of a growing pattern, and give one question each is best at answering.
Solution
Step 1 — List the four representations and what each is best at, using the section's own summary:
- Picture / diagram — best at showing why a rule works, since you can point at the new pieces.
- Table — best at showing what happens at the very next step.
- Graph — best at showing the overall shape of the growth (speeding up, slowing down) at a glance.
- Rule (equation) — best at jumping straight to a far-away step without listing everything before it.
Answer: Picture/diagram answers "why is the rule that?"; Table answers "what happens next?"; Graph answers "is it speeding up or slowing down?"; Rule answers "what is the value at a far-away step?"
Problem 5. For Growing Dots, write the number of dots at 0, 1, 2, 3, 4, and 5 minutes, and state the constant difference.
Solution
Step 1 — Apply the recursive rule: Starting from \(D(0) = 1\) and adding 4 each minute (\(D(t) = D(t-1) + 4\)):
$$ 1,\ 5,\ 9,\ 13,\ 17,\ 21 $$Step 2 — State the constant difference: Each entry is exactly 4 more than the one before it, so the constant difference is 4.
Answer: \(D(0)\) through \(D(5)\) are \(1, 5, 9, 13, 17, 21\); the constant difference is 4.
Problem 6. Explain where the 4 and the \(+1\) in \(D(t) = 4t + 1\) come from in the picture.
Solution
Step 1 — Where the 4 comes from: The figure is an X with four diagonal arms, and every arm grows by exactly one dot each minute. Four arms gaining one dot apiece is 4 new dots every minute — that total is the 4 in \(D(t) = 4t + 1\).
Step 2 — Where the \(+1\) comes from: The single center dot is present already at 0 minutes, before any arm has grown at all. It is never one of the dots being added each minute, so it appears once, added on top of the growing part.
Answer: The 4 is the number of new dots gained per minute (one per arm, times four arms); the \(+1\) is the fixed starting center dot.
Problem 7. Use \(D(t) = 4t + 1\) to find the number of dots at 45 minutes.
Solution
Step 1 — Substitute \(t = 45\) into the explicit rule:
$$ D(45) = 4(45) + 1 $$Step 2 — Compute: \(4 \times 45 = 180\), so \(D(45) = 180 + 1 = 181\).
Answer: 181 dots.
Problem 8. A Growing Dots figure has 217 dots. How many minutes have passed?
Solution
Step 1 — Set up the equation: We want the \(t\) for which \(D(t) = 217\):
$$ 4t + 1 = 217 $$Step 2 — Solve for \(t\):
$$ 4t = 217 - 1 = 216 $$ $$ t = \frac{216}{4} = 54 $$Answer: 54 minutes.
Problem 9. Write the recursive rule for Growing, Growing Dots, and state its constant ratio.
Solution
Step 1 — Write the recursive rule: The whole figure doubles every minute, starting from 1 dot:
$$ N(0) = 1, \qquad N(t) = 2 \cdot N(t-1) $$Step 2 — State the constant ratio: Every term is exactly twice the one before it (that is what "doubles" means), so the constant ratio is 2.
Answer: \(N(0) = 1\), \(N(t) = 2 \cdot N(t-1)\); constant ratio 2.
Problem 10. Use \(N(t) = 2^{t}\) to find the number of dots at 10 minutes.
Solution
Step 1 — Substitute \(t = 10\) into the explicit rule:
$$ N(10) = 2^{10} $$Step 2 — Compute: \(2^{10} = 1024\).
Answer: 1024 dots.
Problem 11. At what minute does Growing, Growing Dots first pass Growing Dots? Show the two counts at that minute and at the one before it.
Solution
Step 1 — Build both counts side by side, using \(D(t) = 4t+1\) and \(N(t) = 2^t\):
| \(t\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| \(D(t)\) | 1 | 5 | 9 | 13 | 17 | 21 |
| \(N(t)\) | 1 | 2 | 4 | 8 | 16 | 32 |
Step 2 — Find the first minute where \(N(t) > D(t)\): At \(t = 4\), \(N(4) = 16\) is still less than \(D(4) = 17\). At \(t = 5\), \(N(5) = 32\) is greater than \(D(5) = 21\). So the crossing happens between minute 4 and minute 5.
Answer: Growing, Growing Dots first passes Growing Dots at minute 5: at minute 4 it trails, 16 to 17, and at minute 5 it leads, 32 to 21.
Problem 12. State the two-test procedure for classifying a table, in your own words.
Solution
Step 1 — First test, differences: Subtract each output from the next one. If every difference is the same, the pattern grows by adding and you are done.
Step 2 — Second test, ratios: If the differences were not all equal, divide each output by the one before it instead. If every ratio is the same, the pattern grows by multiplying.
Step 3 — Neither: If neither the differences nor the ratios are constant, the pattern is classified as "neither" — it may still be perfectly orderly, just not one of those two families.
Answer: Subtract consecutive terms first; constant differences mean it grows by adding. If that fails, divide consecutive terms; constant ratios mean it grows by multiplying. If neither is constant, the verdict is neither.
Problem 13. Explain why two matching differences are not enough to conclude that a pattern grows by adding.
Solution
Step 1 — What "constant difference" actually claims: It claims that every consecutive pair in the sequence has the same difference, not just the pair you happened to check.
Step 2 — Why two matches aren't proof: A sequence like \(1, 3, 5, 10\) has two matching differences of 2 in a row and then breaks. Checking only the first two differences would have wrongly called it the adding kind. Every available consecutive difference has to agree before the verdict is safe.
Answer: Because "constant" describes the whole sequence, and two agreeing differences only rule out a mismatch in that one pair — a later pair could still disagree.
Problem 14. Classify and write an explicit rule: \(7, 11, 15, 19, 23\).
Solution
Step 1 — Test for a constant difference: \(11 - 7 = 4\), \(15 - 11 = 4\), \(19 - 15 = 4\), \(23 - 19 = 4\). All equal, so the pattern grows by adding.
Step 2 — Write the explicit rule: Starting at \(7\) (the 1st term) and adding 4 a total of \(n-1\) times to reach the \(n\)th term:
$$ t(n) = 7 + 4(n-1) = 4n + 3 $$Check: \(t(1) = 7\), \(t(5) = 23\). Both match.
Answer: Grows by adding; \(t(n) = 4n + 3\).
Problem 15. Classify and write an explicit rule: \(4, 12, 36, 108\).
Solution
Step 1 — Test for a constant ratio: \(12/4 = 3\), \(36/12 = 3\), \(108/36 = 3\). All equal, so the pattern grows by multiplying.
Step 2 — Write the explicit rule: Starting at \(4\) and multiplying by 3 a total of \(n-1\) times:
$$ t(n) = 4 \cdot 3^{\,n-1} $$Check: \(t(1) = 4\), \(t(4) = 4 \cdot 27 = 108\). Both match.
Answer: Grows by multiplying; \(t(n) = 4 \cdot 3^{\,n-1}\).
Problem 16. Classify and write an explicit rule: \(100, 75, 50, 25\).
Solution
Step 1 — Test for a constant difference: \(75 - 100 = -25\), \(50 - 75 = -25\), \(25 - 50 = -25\). All equal (a constant difference can be negative), so the pattern grows by adding.
Step 2 — Write the explicit rule: Starting at \(100\) and adding \(-25\) a total of \(n-1\) times:
$$ t(n) = 100 - 25(n-1) = 125 - 25n $$Check: \(t(1) = 100\), \(t(4) = 125 - 100 = 25\). Both match.
Answer: Grows by adding; \(t(n) = 125 - 25n\).
Problem 17. Classify and write an explicit rule: \(1, 4, 9, 16, 25\).
Solution
Step 1 — Test one, differences: \(4-1=3\), \(9-4=5\), \(16-9=7\), \(25-16=9\). Not constant, so it does not grow by adding.
Step 2 — Test two, ratios: \(4/1=4\), \(9/4=2.25\), \(16/9\approx1.78\), \(25/16=1.5625\). Not constant either, so it does not grow by multiplying — the pattern is neither.
Step 3 — An explicit rule anyway: These are the perfect squares, so \(t(n) = n^2\) fits every term exactly even though the sequence is neither kind.
Answer: Neither; \(t(n) = n^2\).
Problem 18. A sequence starts at 6 and has constant ratio \(\tfrac{1}{3}\). Write the first four terms and say what happens to the values as the sequence continues.
Solution
Step 1 — Multiply by \(\tfrac13\) three times, starting at 6:
$$ 6,\quad 6\cdot\tfrac13 = 2,\quad 2\cdot\tfrac13 = \tfrac23,\quad \tfrac23\cdot\tfrac13 = \tfrac29 $$Step 2 — Describe what happens as the sequence continues: Each term is only a third of the one before it, so the values keep shrinking. They get closer and closer to 0 without ever actually reaching it.
Answer: \(6,\ 2,\ \tfrac23,\ \tfrac29\); as the sequence continues, the terms shrink toward 0.
Problem 19. For the sequence \(10, 13, 16, \ldots\) with explicit rule \(y = 3x + 7\), say which rule you would use to find the 5th term and which for the 100th term, and give both answers.
Solution
Step 1 — Check the rule's indexing: \(y = 3x+7\) at \(x=1\) gives \(10\), matching the first listed term, so the terms are \(t(1), t(2), t(3), \dots\)
Step 2 — 5th term, using the recursive rule (nearby step): Continue the table by adding 3 each time: \(10, 13, 16, 19, 22\). The 5th term is 22 — checked against the rule, \(3(5)+7=22\). Recursive is the natural choice here because the 5th term is right next to terms already listed.
Step 3 — 100th term, using the explicit rule (far-away step): Adding 3 ninety-five times by hand would be slow; substitute directly instead:
$$ y = 3(100) + 7 = 307 $$Answer: Recursive rule for the 5th term, \(22\); explicit rule for the 100th term, \(307\).
Problem 20. Looking at the Growing Dots figure at \(t\) minutes, one student counts the four arms one at a time and writes \(D(t) = t + t + t + t + 1\). Another counts them together and writes \(D(t) = 4t + 1\). Show that the two rules are the same expression, and say what the \(+1\) counts in each.
Solution
Step 1 — expand the first rule: Counting the arms one at a time just adds \(t\) four times, and four copies of the same quantity is what multiplication is for.
$$D(t) = t + t + t + t + 1 = 4t + 1$$Step 2 — compare: That is the second student's rule exactly, so the two are the same expression. Neither student is more right; one wrote the counting down and the other wrote the shortcut.
Step 3 — say what the \(+1\) counts: In both rules it is the same single dot — the centre dot the pattern started with, which never belongs to any arm and never changes. The first rule reaches it after four separate arm counts; the second reaches it after one multiplication. It is the same dot on the page either way.
Answer: \(t + t + t + t + 1 = 4t + 1\), so the rules are equivalent, and in both the \(+1\) counts the centre dot.
Problem 21. Here is a third way to see the same figure. The Growing Dots X is made of two full diagonals crossing at the centre, and each diagonal is \(2t + 1\) dots long.
a) Explain why \(2(2t + 1)\) is not the right count.
b) Fix it, and show that your corrected rule simplifies to \(4t + 1\).
Solution
Step 1 — check the claim about one diagonal: A single diagonal runs from the end of one arm, through the centre, to the end of the opposite arm. That is \(t\) dots on one side, \(t\) on the other, and the centre dot between them:
$$t + t + 1 = 2t + 1$$So each diagonal really is \(2t + 1\) dots long.
Part (a) — why \(2(2t+1)\) is wrong: The two diagonals cross at the centre dot, so that dot lies on both of them. Adding the diagonals counts it once for each — twice in total — but there is only one centre dot on the page. The count is too big by exactly 1.
This is the same error §1.1 found in the four-full-length-sides method, where the corner tiles were counted by two sides each. Overlapping pieces double-count whatever they share.
Part (b) — fix it and simplify: Subtract the one dot that was counted twice.
$$2(2t + 1) - 1 = 4t + 2 - 1 = 4t + 1$$Check at \(t = 2\): two diagonals of \(2(2) + 1 = 5\) dots each is 10, minus the doubled centre gives 9 — and the table says 9 dots at 2 minutes. Agrees.
Answer: (a) the centre dot lies on both diagonals, so \(2(2t+1)\) counts it twice; (b) \(2(2t + 1) - 1 = 4t + 1\), the same rule a third way.
Problem 22. A graph of distance from home is flat for ten minutes. Explain what the person is doing, and explain why "flat" does not mean "nothing is happening."
Solution
Step 1 — What the flat piece means: The graph is tracking distance from home, and for ten minutes that quantity does not change — the person is not moving.
Step 2 — Why flat does not mean "nothing is happening": The graph only records one quantity, distance from home. The person could be doing plenty at that fixed spot — working, talking, resting — without any of it moving them closer to or farther from home. Flat means the graphed quantity is unchanged, not that the person is idle.
Answer: The person has stopped moving (is staying in one place); "flat" describes distance from home holding steady, which says nothing about whether the person is otherwise active.
Problem 23. Wen rides to a friend's house at a steady speed, taking 20 minutes. Wen stays 20 minutes, then rides for 20 more minutes to a shop that sits between the two houses. Sketch the graph of distance from home against time, and explain why the third piece falls rather than rises.
Solution
Step 1 — Break the trip into its three pieces, distance from home vs. time:
| Minutes | What is happening | Distance from home | Shape |
|---|---|---|---|
| 0–20 | riding out to the friend's house at a steady pace | rises from 0 up to the friend's-house distance | a straight segment rising |
| 20–40 | staying at the friend's house | stays at the friend's-house distance | a horizontal segment |
| 40–60 | riding to the shop | falls from the friend's-house distance down to the shop's distance | a straight segment falling |
Step 2 — Why the third piece falls: The shop sits between the two houses, so it is closer to home than the friend's house is. Riding from the friend's house to the shop moves Wen closer to home even though Wen is moving away from the friend — so distance from home decreases, and the graph falls.
Answer: Rising for 20 minutes, flat for 20 minutes, then falling for 20 minutes; the fall happens because the shop is nearer to home than the friend's house is, so heading to the shop reduces the distance-from-home reading.
Problem 24. A running total of items shipped to a store climbs steeply and steadily for several days, then goes flat while the factory is shut, then climbs again at half the old steepness when only half the workforce returns. Sketch the graph, and explain why the flat piece is flat rather than falling.
Solution
Step 1 — Break the running total into its three pieces:
| Period | What is happening | Shape |
|---|---|---|
| several days of normal shipping | steady output, several days in a row | a steep straight segment rising |
| the shutdown | no items shipped | a horizontal segment |
| half workforce returns | shipping resumes, at half the old rate | a straight segment rising, at half the steepness of the first |
Step 2 — Why the flat piece is flat, not falling: This is a running total — a cumulative count that only ever adds items, never removes them. With zero items shipped during the shutdown, nothing is added, so the total simply holds at whatever it reached before the shutdown. It has no mechanism to decrease, so the graph cannot fall; it can only stay level.
Answer: Steep rise, then flat, then a gentler rise at half the original steepness; the flat piece is flat because a running total cannot decrease, and zero shipped means zero added, not a loss.
Problem 25. Sketch a graph of the water depth in a bucket left outside during a rainstorm that stops, followed by a hole opening in the bottom. Label what each piece of your graph means.
Solution
Step 1 — Break the story into its pieces, water depth vs. time:
| Period | What is happening | Shape | What it means |
|---|---|---|---|
| during the rainstorm | rain adds water to the bucket | rising | depth increasing as rain falls |
| after the rain stops, before the hole opens | no rain in, no way out yet | flat | depth holds steady — nothing is entering or leaving |
| after the hole opens | water drains out through the hole | falling | depth decreasing as water leaks out |
| once the bucket is empty | no water left to drain | flat at zero | depth stays at 0 |
Answer: Rising while it rains, flat once the rain stops and before the hole opens, falling once the hole opens and drains the bucket, then flat at zero once it's empty — each piece labeled by what is adding, removing, or doing nothing to the water.
Problem 26. A graph of total pages read against time has a steep straight piece, then a gentle straight piece, then a flat piece. Write a three-part story that matches it.
Solution
Step 1 — Match each piece of the graph to a stage of reading: A steep straight rise means pages are being read quickly; a gentle straight rise means reading has slowed down but not stopped; a flat piece means no more pages are being read for a while.
Step 2 — Write the three-part story: "At first I read quickly, tearing through pages (steep rise). After a while I slowed down, reading more carefully or getting tired (gentle rise). Then I stopped reading altogether — maybe I finished a chapter and set the book down, or ran out of time (flat)."
Answer: Fast reading, then slower reading, then no reading at all — matching the steep-rise, gentle-rise, flat shape of the graph.
Problem 27. On a graph of cost against hours parked, explain what the point \((3, 9)\) means and what it does not mean.
Solution
Step 1 — Read the point correctly: On a graph of cost against hours parked, the horizontal coordinate is hours and the vertical coordinate is cost, so \((3, 9)\) means: 3 hours of parking costs $9.
Step 2 — What it does not mean: It is a single input-output pair, not a statement about the rate. It does not mean parking costs a flat $3 for every hour, and it does not by itself prove that 6 hours would cost $18 — that would require knowing the graph is a straight line through the origin, which one point cannot establish.
Answer: \((3,9)\) means 3 hours of parking costs $9; it does not mean the price is $3 per hour for every duration.
Problem 28. A graph passes through \((0, 2)\) and \((60, 8)\). Sketch three graphs through those same two points that describe three completely different situations, and say what this shows about how much two data points can settle.
Solution
Step 1 — Three different shapes through the same two points \((0,2)\) and \((60,8)\):
- A straight line rising steadily the whole way from 2 to 8 — a quantity changing at a constant rate.
- A curve that rises quickly at first and then levels off as it approaches 8 — a quantity that grows fast early and slows down later.
- A curve that stays near 2 for a long stretch and then rises sharply near the end to reach 8 — a quantity that barely changes at first and then changes fast at the end.
Step 2 — What this shows: All three graphs agree at the two given points but tell three completely different stories about what happens in between. Two data points fix only two locations on a graph — they say nothing about the shape connecting them, so a huge range of very different situations can share the exact same two points.
Answer: A straight line, an early-fast/late-slow curve, and a late-fast/early-slow curve all pass through \((0,2)\) and \((60,8)\); this shows two points alone cannot settle the shape of a graph, only two of its locations.
Problem 29. Sketch a graph of \(B\) satisfying all five conditions at once: \(B(0) = 0\), \(B(1) < B(7)\), \(B(9) = 11\), \(B(10) = B(20)\), and \(B(20) > B(40)\). Describe the shape in words, and say why many different sketches can satisfy all five.
Solution
Step 1 — Place the five known facts on the timeline: \(B(0)=0\) — starts at 0. \(B(1) < B(7)\) — the value is higher at time 7 than at time 1, so it climbs somewhere across that stretch. \(B(9) = 11\) — by time 9 it has reached 11. \(B(10) = B(20)\) — the value is unchanged from time 10 to time 20, a flat stretch. \(B(20) > B(40)\) — the value at time 40 is lower than at time 20, so it falls somewhere across that stretch.
Step 2 — Describe the overall shape: Starting at 0, \(B\) climbs overall (though not necessarily at every instant) up to 11 by time 9, holds flat from time 10 to time 20, then drops to some lower value by time 40. In story terms: fills, holds, then drains.
Step 3 — Why many sketches work: The five conditions pin down only a few specific points and a few comparisons between them — not the entire path of the graph. Any curve that rises somewhere between 1 and 7, reaches exactly 11 at time 9, stays level from 10 to 20, and ends up lower by time 40 satisfies every condition, and there are infinitely many such curves.
Answer: \(B\) rises from 0 to 11 by time 9, stays flat from 10 to 20, then falls by time 40 — a fill-hold-drain shape; many sketches work because the conditions fix only a handful of points and comparisons, leaving the rest of the path free.
Problem 30. Starting from the rule \(y = 5x - 1\), produce a table for \(x = 0\) through \(4\), then describe the graph.
Solution
Step 1 — Build the table by substituting each \(x\):
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(y\) | \(-1\) | 4 | 9 | 14 | 19 |
Step 2 — Describe the graph: The differences between consecutive \(y\)-values are all 5, so the points climb 5 units for every 1 unit across — a straight line. It crosses the vertical axis at \((0, -1)\), the value the rule gives at \(x=0\).
Answer: Table: \(-1, 4, 9, 14, 19\). Graph: a straight line rising 5 units for every 1 unit, crossing the vertical axis at \((0,-1)\).
Problem 31. Explain why checking an answer in a second representation catches mistakes that rechecking the first representation does not.
Preview — pointing at §1.3.
Solution
Step 1 — What rechecking the same representation risks: If a mistake was made while carrying out a process — say, mis-expanding an algebraic expression, or miscounting a picture — redoing that exact same process is likely to repeat the exact same slip, because nothing about the method has changed.
Step 2 — Why a second representation is different: Building a second representation (say, a table, when the first mistake happened in a rule) requires an entirely different sequence of steps to arrive at an answer. An error made while doing algebra essentially never reproduces itself while counting a picture or reading a graph, because the two processes don't share the same failure points.
Answer: A mistake tends to repeat itself when you redo the same steps in the same representation, but it almost never survives being checked by a completely different process — so agreement between two independently-built representations is much stronger evidence than repeating the same one twice.
Problem 32. Use the bagel sign at the end of this section to find the cheapest way to buy 15 bagels, and give the total.
Solution
Step 1 — List possible ways to make 15 bagels from the sign's priced quantities (1, 6, 9, 12) and price each:
- \(12 + 1 + 1 + 1\): \(\$10.00 + 3(\$1.25) = \$13.75\)
- \(9 + 6\): \(\$8.00 + \$6.00 = \$14.00\)
- \(9 + 1 \times 6\): \(\$8.00 + \$7.50 = \$15.50\)
- \(6 + 6 + 1 + 1 + 1\): \(\$6.00 + \$6.00 + \$3.75 = \$15.75\)
- \(1 \times 15\): \(15 \times \$1.25 = \$18.75\)
Step 2 — Compare the totals: The smallest of these is $13.75, from one 12-pack plus three singles.
Answer: Buy one 12-pack and three singles, for a total of $13.75.
Problem 33. Explain why the bagel sign gives four different totals for 13 bagels without anyone making an arithmetic mistake. What would the shop have to do to make the sign give exactly one answer?
Solution
Step 1 — Why four different totals are all "correct": The sign only prices four specific bundle sizes — 1, 6, 9, and 12. Thirteen bagels can be broken into a sum of those sizes in more than one way (13 singles; two 6-packs plus a single; a 9-pack plus four singles; a 12-pack plus a single), and each breakdown gives a different price because the bundles discount the per-bagel cost differently. No arithmetic is wrong in any of the four totals — each one correctly prices a valid combination that adds up to 13 bagels; the sign simply never said which combination to use.
Step 2 — What would fix it: The input "13 bagels" needs to map to exactly one output. The shop could do this by posting an explicit price for every possible quantity (not just 1, 6, 9, and 12), or by stating a single rule for filling any order (for example, "always use the combination that costs the least" or "always use the largest packs first"). Either way, the fix is the same idea: force each input to determine exactly one output.
Answer: The four totals arise because 13 bagels can be split into the sign's priced bundles in more than one way, each giving a different (but individually correct) price; to fix it, the shop must specify one definite rule — such as a full price list or a stated tie-breaking method — so every quantity has exactly one price, which is exactly the property §1.3 will call being a function.
Key Terms
constant difference — the fixed number added to get from each output of a pattern to the next.
constant ratio — the fixed number multiplied to get from each output of a pattern to the next.
recursive rule — a rule giving a pattern's starting value plus a way to build each value from the one before it.
explicit rule — a rule computing the value at any step directly from the step number.