Linear or not?
Tonight's idea
A relationship is linear when equal steps in \(x\) always produce equal steps in \(y\). The change in \(y\) for each extra unit of \(x\) is the rate of change. On a graph the rate is the slope, and a line has the same slope everywhere.
In a table, look at the gaps. If \(x\) goes up by 1 each row and \(y\) by 2, the rate is 2 and the rule is \(y = 2x + b\), with \(b\) the value of \(y\) at \(x = 0\). If the \(y\) gaps change, the points do not lie on a line.
| \(x\) | A | B |
|---|---|---|
| 0 | 3 | 0 |
| 1 | 5 | 1 |
| 2 | 7 | 4 |
| 3 | 9 | 9 |
| 4 | 11 | 16 |
In a story, ask: does each extra unit cost the same? A taxi with a fixed fee plus a fixed amount per kilometre is linear. The area of a square is not: side 3 to side 4 adds 7, side 4 to side 5 adds 9.
One trap: the test needs equal steps in \(x\). If the \(x\) values are uneven, divide each \(y\) gap by its \(x\) gap and compare the rates instead.
Watch one: \(x = 1, 2, 4, 7\) and \(y = 5, 8, 14, 23\)
The \(y\) gaps are 3, 6, 9: not equal. But neither are the \(x\) gaps, 1, 2, 3. Divide:
\[\frac{3}{1} = 3, \qquad \frac{6}{2} = 3, \qquad \frac{9}{3} = 3.\]
Every rate is 3, so the points lie on a line, \(y = 3x + b\). One point gives \(b\): \(5 = 3 \times 1 + b\), so \(b = 2\), and \(y = 3x + 2\). Check the last point: \(3 \times 7 + 2 = 23\).
Your turn
- Which tables are linear? For each linear one, give the rate and the rule \(y = mx + b\).
(a) \(x = 0, 1, 2, 3\) and \(y = 7, 4, 1, -2\). (b) \(x = 0, 1, 2, 3\) and \(y = 1, 3, 9, 27\). (c) \(x = 0, 2, 4, 6\) and \(y = 5, 6, 7, 8\). - Which of these stories is linear? Write a rule for each linear one.
(a) A phone plan costs $12 a month plus 5 cents per text. (b) A dropped ball falls \(5t^2\) metres in \(t\) seconds. (c) A candle 20 cm tall burns down 1.5 cm every hour. When does it burn out? - A table has \(x = 1, 3, 8, 12\) and \(y = 4, 10, 25, 37\). Is it linear? If so, find the rule and predict \(y\) when \(x = 20\).
- A square has side \(s\). (a) Make a table of its perimeter and its area for \(s = 1, 2, 3, 4\). (b) Which is linear in \(s\)? (c) Show that going from side \(s\) to side \(s + 1\) adds \(2s + 1\) to the area, and explain why that means the area can never be linear.
Stretch: the gaps of the gaps
A table has \(x = 0, 1, 2, 3, 4\) and \(y = 2, 3, 6, 11, 18\). Write down the gaps, then the gaps of the gaps. Guess a rule for \(y\) in terms of \(x\) and check it against all five rows. What kind of rule gives this pattern?
Answers
- (a) Linear: gaps of \(-3\), so \(y = 7 - 3x\). (b) Not linear: the gaps are 2, 6, 18. (c) Linear: \(y\) rises 1 for every 2 in \(x\), so the rate is \(\tfrac12\) and \(y = 5 + \tfrac12 x\).
- (a) Linear: \(C = 12 + 0.05n\) dollars for \(n\) texts. (b) Not linear: in successive seconds it falls 5, 15, 25 metres. (c) Linear: \(h = 20 - 1.5t\). It burns out when \(h = 0\), at \(t = 20/1.5 = 13\tfrac13\) hours, 13 h 20 min.
- The rates are \(6/2\), \(15/5\), \(12/4\), all 3, so yes. \(y = 3x + b\) with \(4 = 3 + b\), so \(y = 3x + 1\). At \(x = 20\), \(y = 61\).
- (a) Perimeter 4, 8, 12, 16; area 1, 4, 9, 16. (b) Perimeter: equal gaps of 4, \(P = 4s\). The area gaps are 3, 5, 7. (c) \((s+1)^2 - s^2 = 2s + 1\). The gap depends on \(s\) and grows by 2 every step, so no single rate fits; linear would need the same gap every time.
Stretch. Gaps 1, 3, 5, 7; gaps of the gaps 2, 2, 2. The rule is \(y = x^2 + 2\): check \(0 + 2, 1 + 2, 4 + 2, 9 + 2, 16 + 2\). Constant second gaps are the signature of a squared term (a quadratic), just as constant first gaps are the signature of a line.