Finding the Rule
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.