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Graphing a pattern

Patterns and sequences · level 4, lesson 3 of 6

A question to start

Take the sequence 5, 8, 11, 14, 17, 20 and mark each term as a dot on squared paper: go across by the position and up by the term. Lay a ruler on the dots. They line up perfectly, all six on one straight line.

Now do the same with the square numbers 1, 4, 9, 16, 25. No ruler will touch them all: they bend upwards in a curve. What makes one pattern straight and the other bent? You can tell before you draw a single dot.

The idea

Plotting a sequence

A graph of a sequence is a picture of its position table. Draw two lines at right angles, called axes. The line across (the horizontal axis) is for the position \(n\). The line going up (the vertical axis) is for the term \(t_n\). Each column of the table becomes one dot, written as a pair of numbers in brackets \((\text{position}, \text{term})\): the 3rd term of 5, 8, 11, ... becomes the point \((3, 11)\), found by going 3 across and 11 up. These pairs are called coordinates; the across number always comes first.

\(n\) 1 2 3 4 5 6
\(t_n = 3n + 2\) 5 8 11 14 17 20
Point (1, 5) (2, 8) (3, 11) (4, 14) (5, 17) (6, 20)
0 1 2 3 4 5 6 7 0 5 10 15 20 25 position n term 1, 4, 9, 16, 25: a curve 5, 8, 11, ...: a line

The dots are not joined with a solid line, because a sequence only has terms at whole-number positions: there is no 2.5th term. The faint line in the picture is drawn only to show that the orange dots sit on it.

Why a constant difference makes a straight line

Look at the dashed steps between the orange dots. To get from one dot to the next you go 1 across (the position goes up by 1) and 3 up (the term goes up by \(d = 3\)). Every step is exactly the same shape: 1 across, 3 up.

When you climb a staircase whose steps are all the same width and the same height, the tips of the steps all lie on one straight slope. If any step were taller or shorter, the slope would bend there. So:

Constant difference ⟹ every step is the same ⟹ the points lie on a straight line.

(The arrow ⟹ means "so" or "this makes it true that".) It works the other way round too: if the dots lie on a straight line, equally spaced across, the steps must all be the same, so the difference is constant.

For the square numbers the differences are 3, 5, 7, 9, ...: each step is 1 across but taller than the last. The slope keeps getting steeper, so the dots bend upwards in a curve.

This is why rules like \(t_n = 3n + 2\) are called linear (from "line"): their graphs are straight.

What the rule tells you about the line

The two numbers in \(t_n = dn + c\) each show up in the picture.

  • \(d\), the difference, is the steepness. It is how far the line climbs for every 1 across. \(t_n = 5n + 1\) climbs 5 per step and looks steeper than \(t_n = 2n + 20\), which climbs 2. If \(d\) is negative, the line goes downhill from left to right.
  • \(c\), the zero term, is where the line meets the vertical axis. Run the line back to \(n = 0\): there, \(d \times 0\) is 0, so the height is just \(c\). For \(3n + 2\), the faint line crosses the vertical axis at 2, the zero term from lesson 2.

Two sequences with the same \(d\) but different starts, like \(3n + 2\) and \(3n + 7\), have lines with the same steepness, one above the other. They never meet: they are parallel, always 5 apart.

Reading the rule from a graph

Given dots on a straight line, you can read the rule off:

  1. Pick two dots next to each other. How far up from one to the next? That is \(d\).
  2. Run back to \(n = 0\) (step back one more time from the first dot). The height there is the zero term.
  3. Write \(t_n = dn + (\text{zero term})\).

Worked examples

1. Plot \(t_n = 2n + 1\). The points are \((1, 3)\), \((2, 5)\), \((3, 7)\), \((4, 9)\). Each step is 1 across and 2 up, so they lie on a straight line. Run back one step from \((1, 3)\): 2 down gives \((0, 1)\). The line meets the vertical axis at 1, the zero term.

2. A rule from a graph. Dots are at \((1, 7)\), \((2, 10)\), \((3, 13)\), \((4, 16)\).

  • Each step goes 3 up, so \(d = 3\).
  • One step back from \((1, 7)\) is \((0, 4)\): the zero term is 4.
  • The rule is \(t_n = 3n + 4\). Check: \(3 \times 4 + 4 = 16\) ✓.

3. Is the point on the line? Does \((20, 62)\) belong to \(t_n = 3n + 2\)? Put \(n = 20\) into the rule: \(3 \times 20 + 2 = 62\). Yes. Does \((12, 40)\)? \(3 \times 12 + 2 = 38\), not 40, so no: that point is 2 above the line.

4. Straight or curved, without drawing. Which of these give dots on a straight line?

  • 2, 5, 8, 11: differences 3, 3, 3. Straight.
  • 1, 3, 6, 10 (triangular numbers): differences 2, 3, 4. Curved (bending upwards).
  • 40, 34, 28, 22: differences −6, −6, −6. Straight, going downhill.

Try it

  1. Write the points for \(t_n = 4n - 1\) for \(n = 1\) to \(4\). Describe each step. Where does the line meet the vertical axis?
  2. Dots are at \((1, 10)\), \((2, 8)\), \((3, 6)\), \((4, 4)\). Does the line go uphill or downhill? Find the rule.
  3. Which line is steeper, \(t_n = 5n + 1\) or \(t_n = 2n + 20\)? Which has the larger first term?
  4. Does \((15, 47)\) lie on \(t_n = 3n + 2\)? Does \((15, 45)\)?
  5. Plot 1, 2, 4, 8, 16 (in your head or on paper). Do the dots lie on a straight line? How can you tell from the numbers?
  6. A straight-line pattern meets the vertical axis at 6 and climbs 2 for every step across. Write its rule and its first three terms.
  7. Explain why the graphs of \(t_n = 3n + 2\) and \(t_n = 3n + 7\) never meet.
  8. Two dots of a straight-line pattern are \((2, 11)\) and \((6, 23)\). Find the rule.
Answers
  1. \((1, 3)\), \((2, 7)\), \((3, 11)\), \((4, 15)\). Each step is 1 across and 4 up. One step back from \((1, 3)\) is \((0, -1)\), so the line meets the vertical axis at \(-1\).
  2. Downhill: each step goes 2 down, so \(d = -2\). One step back from \((1, 10)\) goes up to \((0, 12)\). Rule: \(t_n = 12 - 2n\). Check: \(12 - 8 = 4\) ✓.
  3. \(5n + 1\) is steeper (5 up per step, against 2). But \(2n + 20\) has the larger first term: 22 against 6. (The steeper line catches up later; lesson 5 finds exactly when.)
  4. \(3 \times 15 + 2 = 47\), so \((15, 47)\) is on the line; \((15, 45)\) is not.
  5. No. The differences are 1, 2, 4, 8: each step is taller than the last, so the dots curve upwards (more and more steeply).
  6. \(t_n = 2n + 6\). First three terms: 8, 10, 12.
  7. Both climb 3 for every 1 across, so they have the same steepness. At every position, the second is exactly 5 more than the first (\(3n + 7 - (3n + 2) = 5\)), so the gap never closes: the lines are parallel.
  8. From \(n = 2\) to \(n = 6\) is 4 steps across and \(23 - 11 = 12\) up, so \(d = 12 \div 4 = 3\). Zero term: \(11 - 2 \times 3 = 5\). Rule: \(t_n = 3n + 5\). Check: \(18 + 5 = 23\) ✓.

Common mistake

Swapping the axes. The point \((3, 11)\) means 3 across and 11 up, not the other way round. Position goes across, term goes up. A quick check: as you move right along the graph, \(n\) should go 1, 2, 3, 4.

Reading the steepness over several steps as \(d\). From \((2, 11)\) to \((6, 23)\) the line rises 12, but that is over 4 steps. The difference \(d\) is the rise for one step across: \(12 \div 4 = 3\).

Where it leads