You are here:

Graphing linear functions

This section focuses on the key features and methods for working with linear graphs. It demonstrates how to sketch graphs from rules, derive rules from graphs, and calculate key features such as the gradient and \(x\)- and \(y\)-axis intercepts.

Use this page to revise the following concepts of graphing linear functions:

Features of linear graphs
Plotting linear graphs

Features of linear graphs

Linear graphs are straight-line graphs that visually represent a constant rate of change in the relationship between two variables, showing how one changes in response to the other. They are expressed in the general form \(y = a + bx\), where a is the \(y\)-intercept and \(b\) is the gradient.

Key features of linear graphs are:

gradient or slope : Represents the steepness and direction of the line
\(x\)- and \(y\)-axis intercepts: The points where the line crosses the \(x\)-axis and \(y\)-axis, respectively.

Determining the gradient

The gradient of a linear function can be calculated from its rule, derived from a table of values, determined using two points, or found using rise over run on a graph.

Below are four methods that can be used to determine the gradient.

From a rule

To establish the gradient between two points, use the formula \(b=\dfrac{y_2-y_1}{x_2 - x_1}\) where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the points.
Worked Example
Calculate the gradient of the line passing through the points \((-2, 5)\) and \((6, -13)\).
Use the forumla \(b=\dfrac{y_2-y_1}{x_2 - x_1}\).
\[\begin{align}b &= \frac{-13 - 5}{6 - (-2)} \\ b &= \frac{-18}{8} \\ b &= \frac{-9}{4}\end{align}\]
The gradient is \(\dfrac{-9}{4}\).
To establish the gradient from a table of values, ensure the \(x\)-values increase by a consistent step.
The constant difference in consecutive \(y\)-values (for example, \(y_{n+1}-y_n\)) provides the gradient of the linear function.
Worked Example
Example 1
The table below shows the values of and for a linear function. Determine the gradient of the function.
The \(x\)-values are increasing by a consistent step of \(1\) and there is a constant difference of \(4\) between consecutive \(y\)-values, therefore the gradient of the function is \(4\).
Example 2
The table below shows the values of \(x\) and \(y\) for a linear function. Determine the gradient of the function.
The \(x\) values are increasing by a consistent step of \(1\) and there is a constant difference of \(-2\) between consecutive \(y\)-values, therefore the gradient of the function is \(-2\).
To find the gradient from a graph, pick two points on the line, calculate the vertical change (rise), and divide it by the horizontal change (run) between them.
\[b = \dfrac{\text{rise}}{\text{run}}\]
Worked Example
The graph of a straight line is shown below. Determine the gradient of the line.
Select two points on the line, such as \((0,1)\) and \((1,4)\).
Note any two points on the line can be used, ideally, select ones that are easy to work with.
\(b = \dfrac{\text{rise}}{\text{run}} = \dfrac{3}{1}\)
The gradient of the line is \(3\).

Determining \(x\)- and \(y\) -axis intercepts

The \(x\)-intercept is the point where a graph crosses the \(x\)-axis \((y=0)\), and the \(y\)-intercept is where it crosses the \(y\)-axis \((x=0)\).

For an equation in the form \(y = a + bx\) the:

\(x\)-axis intercept is calculated by substituting \(y = 0\) into the rule and solving for \(x\)

\[0 = a + bx \quad \Rightarrow \quad x = -\frac{a}{b}\]

The \(x\)-intercept is \(\left({-}\frac{a}{b},0\right)\).

Sometimes it can be helpful to rearrange the rule into the standard form \(y = a + bx\) first, however this is not required.

\(y\)-axis intercept is calculated by substituting \(x = 0\) into the rule and solving for \(y\)

\[y = a + b(0) \quad \Rightarrow \quad y = a\]

The \(y\)-inercept is \((0,a)\).

Worked Example

Determine \(x\)- and \(y\)-axis intercepts for the linear function with rule \(2.5y = 0.5 - 1.5x\).

\(x\)-axis intercept

Substitute \(y = 0\) into the rule

\[2.5(0) = 0.5 - 1.5x\]

Solve for \(x\)

\[\begin{align}0 &= 0.5 - 1.5x \\ 1.5x &= 0.5 \\ x &= \frac{1}{3}\end{align}\]

The \(x\)-intercept is \((\frac{1}{3},0)\).

\(y\)-axis intercept

Subsitute \(x = 0\) into the rule

\[2.5y = 0.5 -1.5(0)\]

Solve for \(y\)

\[y = 0.2\]

The \(y\)-intercept is \((0,0.2)\).

Check your understanding View

Plotting linear graphs

Going from rule to graph

To sketch a linear graph from a rule, determine two points that satisfy the rule and connect them with a straight line.

Two commonly used methods are:

calculate both axes intercepts and draw a straight line between them
use the \(y\)-intercept and gradient to locate a second point.

Worked Example
Sketch the graph of the linear function given by the rule \(5y = 2 - 10x\).
Step 1: Determine the x-axis intercept
Substitute \(y = 0\) into the rule
\[5(0) = 2 - 10x\]
Solve for \(x\):
\[\begin{align}0 &= 2 - 10x \\ 10x &= 2 \\ x &= \frac{1}{5} \text{ or } x = 0.2\end{align}\]
The \(x\)-intercept is \((0.2,0)\)
Step 2: Determine the \(y\)-axis intercept
Substitute \(x = 0\) into the rule
\[5y = 2 - 10(0)\]
Solve for \(y\)
\[\begin{align}5y &= 2 \\ y &=\frac{2}{5}\text{ or }x = 0.4\end{align}\]
The \(y\)-intercept is \((0,0.4)\)
Step 3: Sketch the graph
Plot the points \((0.2,0)\) and \((0,0.4)\) on the coordinate plane and draw a straight line through them.
Worked Example
Sketch the graph of the linear function given by the rule \(3x+y=5\).
Step 1: Rearrange the rule
Rearrange to the form \(y = a + bx\), where a is the \(y\)-intercept and \(b\) is the gradient:
\[y = 5 - 3x\]
Step 2: Identify the \(y\)-intercept and gradient
The \(y\)- intercept is at \((0, 5)\).
The gradient is \(-3\).
Step 3: Sketch the graph
Plot the \(y\)-intercept at \((0,5)\).
Use the gradient of \(-3\) to locate a second point by moving \(3\) units down and \(1\) unit to the right from \((0,5)\).
Remember the gradient \(=\dfrac{\text{rise}}{\text{run}} = \dfrac{-3}{1}\)

Going from graph to rule

To determine the rule for a given linear graph, locate two points on the graph and use these to calculate the gradient and establish the \(y\)-intercept.

Two commonly used methods are:

read the \(y\)-intercept directly from the graph and locate a second convenient point to calculate the gradient
select two convenient points on the line to calculate the gradient and \(y\)-intercept

Worked Example
Determine the rule of the linear function shown in the graph below.
Step 1: Read the y-intercept from the graph
The \(y\)-intercept is at \((0,2)\).
Step 2: Select a second convenient point
Select a second convenient point on the graph where the coordinates can be read precisely.
Second point selected is \((2,3)\).
Step 3: Calculate the gradient
Use the y-intercept and second point to calculate the gradient.
The gradient can be found by either reading it directly from the graph using \(b = \dfrac{\text{rise}}{\text{run}}\) or by calculating it with the formula \(b = \dfrac{y_2-y_1}{x_2-x_1}\).
Using \(b = \dfrac{\text{rise}}{\text{run}}\) the rise is 1 unit up and the run is 2 units to the right, therefore the gradient is \(\dfrac{1}{2}\).
Using the formula \(b = \dfrac{y_2-y_1}{x_2-x_1}\)
\[\begin{align}\left(x_1,y_1\right) &= (0,2) \\ \left(x_2,y_2\right) &= (2,3)\end{align}\]
\[b =\frac{3-2}{2-0} = \frac{1}{2}\]
Therefore, the gradient is \(\frac{1}{2}\).
Step 4: Substitute values into the general rule
The general form for a linear equation is \(y = a + bx\) where \(a\) is the \(y\)-intercept and \(b\) is the gradient.
For this example:
- the \(y\)-intercept is at \((0,2)\), therefore \(a = 2\)
- the gradient is \(\frac{1}{2}\), therefore \(b = \frac{1}{2}\)
The rule of the linear function shown in the graph is \(y = 2 +\frac{1}{2}x\).
Worked Example
Determine the rule of the linear function shown in the graph below.
Step 1: Select two points
Select two points on the graph where the coordinates can be readily identified.
Points selected are \((60,2.5)\) and \((110,2)\).
Step 2: Calculate the gradient
\[b = \dfrac{y_2-y_1}{x_2-x_1}\]
\[\begin{align}(x_1,y_1) &= (60,2.5) \\ (x_2,y_2) &= (110,2)\end{align}\]
\[\begin{align}B &= \frac{2 - 2.5}{110 - 60} \\ b &= \frac{-0.5}{50} = -0.01\end{align}\]
Therefore, the gradient is \(-0.01\)z.
Step 3: Determine the \(y\)-intercept
Substitute the gradient, \(b = -0.01\), into the general rule \(y = a + bx\):
\[y = a - 0.01x\]
Select one of the two points from Step 1 and substitute it into the equation to solve for \(a\):
Substituting the point \((110,2)\) the equation \(y = a - 0.01x\):
\[\begin{align}2 &= a - 0.01(110) \\ 2 &= a - 1.1 \\ a &= 3.1\end{align}\]
Therefore the \(y\)-intercept is at \((0,3.1)\).
Step 4: Substitute values into the general rule
The general form for a linear equation is \(y = a + bx\) where a is the \(y\)-intercept and b is the gradient.
For this example:
- the \(y\)-intercept is at \((0,3.1)\), therefore \(a = 3.1\)
- the gradient is -0.01, therefore \(b = -0.01\)
The rule of the linear function shown in the graph is \(y = 3.1 - 0.01x\).

Check your understanding View

Consolidate your knowledge: Practice questions View

Back Next

\(-5x + \frac{3}{4}y =2\)
\(\frac{3}{4}y = 2 + 5x\)	added \(5x\) to both sides
\(y = 2 \times \frac{4}{3} + 5x \times \frac {4}{3}\)	multiplied both sides by \(\frac{4}{3}\)
\(y = \frac{8}{3} + \frac{20x}{3}\)	simplified

Student Academic Success

Student Academic Success

Graphing linear functions

Features of linear graphs

Determining the gradient

Worked Example

Worked Example

Worked Example

Worked Example

Determining \(x\)- and \(y\) -axis intercepts

Worked Example

Check your understanding View

Plotting linear graphs

Going from rule to graph

Worked Example

Worked Example

Going from graph to rule

Worked Example

Worked Example

Check your understanding View

Consolidate your knowledge: Practice questions View