Finding the Rule
Finding the rule means working out the equation that links \(x\) and \(y\) in a table, a set of points or a graph, such as \(y = 3x + 2\). In NSW Year 8 maths this is done from the change in \(y\) and the starting value, without gradient formulas (NESA outcome MA4‑LIN‑C‑01).
You will find the number that multiplies \(x\) from how much \(y\) changes each step, and the number added from the value when \(x = 0\), including decreasing tables, negative values and tables where \(x\) goes up in twos. You will fill in missing entries, check that a rule works for every column, decide between two suggested rules, read a rule from plotted points, and write rules for real costs such as bike hire. It is part of Year 8 Graphing Linear Equations.
Every question with a fully worked solution.
- Finding the Rule - Video - Visual Pattern - Writing an Equation and Creating a Table Watch
Theory
A linear rule \(y = ax + k\) has two numbers: \(a\), the change in \(y\) each time \(x\) goes up \(1\), and \(k\), the \(y\)-value when \(x = 0\). For \(x = 0, 1, 2, 3\) and \(y = 6, 4, 2, 0\): \(a = -2\) and \(k = 6\), so \(y = -2x + 6\). This Year 8 maths page shows how.
Every rule on this page is \(y = ax + k\) (also written \(y = mx + b\)).
Find \(a\). It is the change in \(y\) each time \(x\) goes up \(1\). A table whose \(y\)-values fall gives a negative \(a\).
Find \(k\). It is the \(y\)-value when \(x = 0\). If there is no \(x = 0\) column, substitute a column you know and solve: for \(y = 5x + k\) with \(x = 2\), \(y = 6\): \(6 = 10 + k\), so \(k = -4\).
\(x\)-values that jump. If \(x\) goes up \(2\) at a time, the change in \(y\) covers \(2\) steps, so halve it. Divide by whatever the jump is. If \(y\) goes up \(1\) every time \(x\) goes up \(4\), then \(a = \dfrac{1}{4}\) and the rule starts \(y = \dfrac{x}{4}\).
Part of the rule given. In \(y = 8 + \square\,x\) the box is the change in \(y\) for each \(1\) in \(x\). In \(w = 9 - \square\,v\) the minus sign is already written, so the box is the size of the drop.
Filling a gap. Going right along a table, add the step; going left, take it away.
From a graph. Read the plotted points into a table of values, then carry on as above.
Check every column. A rule that fits one column can still be wrong. Substitute a column you did not use.
In \(y = ax + k\):
| Number | What it is | How to find it |
|---|---|---|
| \(a\) | change in \(y\) when \(x\) goes up \(1\) | change in y ÷ jump in x |
| \(k\) | \(y\)-value when \(x = 0\) | read it, or substitute a column and solve |
Finding y = ax + k from a table
- Find \(a\): the change in \(y\) divided by the jump in \(x\).
- Find \(k\): read \(y\) at \(x = 0\), or substitute a column into \(y = ax + k\) and solve.
- Write the rule.
- Check it in every column.
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \(y\) | \(5\) | \(6\) | \(7\) | \(8\) |
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \(y\) | \(5\) | \(7\) | \(9\) | \(11\) |
| \(5\) | \(=\) | \(0 + b\) |
| \(b\) | \(=\) | \(5\) |
| \(7\) | \(=\) | \(m + 5\) |
| \(m\) | \(=\) | \(2\) |
(i) \(b = 5\), so \(y = x + 5\) (using \(x = 0\), \(y = 5\))
(ii) \(m = 2\), so \(y = 2x + 5\) (using \(x = 1\), \(y = 7\))
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \(y\) | \(6\) | \(4\) | \(2\) | \(0\) |
| \(6\) | \(=\) | \(0 + b\) |
| \(b\) | \(=\) | \(6\) |
| \(4\) | \(=\) | \(m + 6\) |
| \(m\) | \(=\) | \(-2\) |
\(m = -2\) and \(b = 6\), so \(y = -2x + 6\). Check \(x = 3\): \(-6 + 6 = 0\).
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|
| \(y\) | \(-4\) | \(1\) | \(6\) | \(11\) |
The box is the change in \(y\) each time \(x\) goes up \(1\): \(1 - (-4) = 5\), \(6 - 1 = 5\), \(11 - 6 = 5\).
| \(5 \times 3 - 4\) | \(=\) | \(11\) |
The box holds \(5\): \(y = 5x - 4\).
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|
| \(y\) | \(a\) | \(8\) | \(2\) | \(-4\) | \(b\) |
The step is \(2 - 8 = -6\). \(a\) is one step left of \(8\); \(b\) is one step right of \(-4\).
| \(a\) | \(=\) | \(8 + 6 = 14\) |
| \(b\) | \(=\) | \(-4 - 6 = -10\) |
| \(y\) | \(=\) | \(-6x + 14\) |
\(a = 14\), \(b = -10\) and \(y = -6x + 14\). Check \(x = 3\): \(-18 + 14 = -4\).
(ii) Find \(x\) when \(y = 12\).
| \(x\) | \(0\) | \(4\) | \(8\) | \(12\) |
|---|---|---|---|---|
| \(y\) | \(5\) | \(6\) | \(7\) | \(8\) |
\(x\) jumps by \(4\) while \(y\) goes up \(1\), so \(a = \dfrac{1}{4}\). The \(x = 0\) column gives \(k = 5\).
| \(y\) | \(=\) | \(\dfrac{x}{4} + 5\) |
| \(\dfrac{x}{4}\) | \(=\) | \(12 - 5 = 7\) |
| \(x\) | \(=\) | \(28\) |
(i) \(y = \dfrac{x}{4} + 5\) (check \(x = 12\): \(3 + 5 = 8\))
(ii) \(x = 28\)
(i) Find the rule.
(ii) Find \(y\) when \(x = 7\).
(iii) Find \(x\) when \(y = 12\).
The points are \((-2, 6)\), \((-1, 4)\), \((0, 2)\), \((1, 0)\) and \((2, -2)\). Each step right, \(y\) goes down \(2\); at \(x = 0\), \(y = 2\).
| \(y\) | \(=\) | \(-2x + 2\) |
| \(-2 \times 7 + 2\) | \(=\) | \(-12\) |
| \(-2x + 2\) | \(=\) | \(12\) |
| \(x\) | \(=\) | \(-5\) |
(i) \(y = -2x + 2\)
(ii) \(-12\)
(iii) \(x = -5\)
| \(x\) | \(2\) | \(4\) | \(6\) | \(8\) |
|---|---|---|---|---|
| \(y\) | \(1\) | \(7\) | \(13\) | \(19\) |
(i) \(x\) jumps by \(2\) and \(y\) goes up \(6\), so \(a = 6 \div 2 = 3\). Substitute \(x = 2\), \(y = 1\).
(ii) \(w\) drops by \(2\) for each \(1\) in \(v\); the minus sign is already written.
| \(1\) | \(=\) | \(3 \times 2 + k\) |
| \(k\) | \(=\) | \(-5\) |
| \(y\) | \(=\) | \(3x - 5\) |
(i) \(y = 3x - 5\) (check \(x = 8\): \(24 - 5 = 19\))
(ii) \(2\), so \(w = 9 - 2v\).
Common pitfalls
Frequently asked questions
How do I find the rule for a table of values?
Find a, the change in y when x goes up by 1, and k, the y-value when x = 0. The rule is y = ax + k. Check it in every column.
What if the table has no x = 0 column?
Find a first, then substitute any column into y = ax + k and solve for k.
What if the x-values go up by 2?
Divide the change in y by 2 to get a.
Can a be a fraction?
Yes. If y goes up 1 when x goes up 4, then a = 1/4.
How do I find the rule from a graph?
Read the coordinates of the points into a table, then find a and k as usual.