1.06 Graphs of linear functions

Horizontal lines

For horizontal lines, the $y$y-value is always the same at every point on the line. In other words, there is no rise - the graph is completely "flat". For example, consider the graph below with the given coordinates:

$A=\left(-4,4\right)$A=(−4,4)

$B=\left(2,4\right)$B=(2,4)

$C=\left(4,4\right)$C=(4,4)

All the $y$y-coordinates are the same. Every point on the line has a $y$y value equal to $4$4, regardless of the $x$x-value.

The equation of this line is $y=4$y=4.

Since the gradient is calculated by $\frac{\text{rise }}{\text{run }}$rise run ​ and there is no rise (as $\text{rise }=0$rise =0), the gradient of a horizontal line is always zero.

Vertical lines

For vertical lines, the $x$x value is always the same at every point on the line.

Look at the coordinates for A, B and C on the vertical line.  

$A=\left(5,-4\right)$A=(5,−4)

$B=\left(5,-2\right)$B=(5,−2)

$C=\left(5,4\right)$C=(5,4)

All the $x$x-coordinates are the same, $x=5$x=5, regardless of the $y$y value.

The equation of this line is $x=5$x=5.

Vertical lines have no "run" (that is, $\text{run }=0$run =0). If this run of zero were to be substituted into the $\frac{\text{rise }}{\text{run }}$rise run ​ equation, there would be a $0$0 as the denominator of the fraction. However, fractions with a denominator of $0$0 are undefined.

So, the gradient of vertical lines is always undefined.

Question 1

Points on a line

Think of the line as an infinite collection of points, all of which lie in a straight line and obey a given rule. The critical thing to remember is that the coordinates of every single point that is on the line, when substituted into the equation for the line, will make that equation true. In other words, points on the line will satisfy the equation. Points not on the line will not satisfy the equation. 

Practice question

Question 2

Plotting the graph of a linear relationship

To graph any liner relationship, only two points on the line are required.  Any two points could be chosen from a table of values, or substituting any two values of $x$x into the linear equation and solving for their corresponding $y$y-values can also provide the two points.  Often, using the two axis intercepts is one of the most convenient ways to sketch the line from two points.

Sketching linear graphs from a table of values

To sketch from a table of values, choose any two points from the table.  From this table there are 4 coordinates, $\left(1,3\right)$(1,3), $\left(2,5\right)$(2,5), $\left(3,7\right)$(3,7), $\left(4,9\right)$(4,9).  

Drag $2$2 of the points on this interactive to the correct positions and graph this linear relationship.  

Sketching linear graphs from any two points

The equation of a linear relationship, like $y=3x+5$y=3x+5, can be sketched using any two points.  

Start by picking any two $x$x-values you like, often the $x$x-value of $0$0 is a good choice because the calculation for $y$y can be quite simple.  For this example, $y=3x+5$y=3x+5 becomes $y=0+5$y=0+5, so $y=5$y=5.  This gives the point $\left(0,5\right)$(0,5)

Similarly look for other easy values to calculate such as  $x=1$x=1, $10$10, $2$2.  Here, pick $x=1$x=1.  Then for $y=3x+5$y=3x+5, this implies that $y=3\times1+5$y=3×1+5, and so $y=8$y=8. This provides the second point $\left(1,8\right)$(1,8). Now, plot the two points and create a line.  

Sketching linear graphs from intercepts

The general form of a line is great for identifying both the $x$x and $y$y intercepts easily. 

For example, consider the line $3y+2x-6=0$3y+2x−6=0

The $x$x-intercept happens when the $y$y value is $0$0. 

The $y$y-intercept happens when the $x$x value is $0$0.

From here, plot both the $x$x intercept and $y$y intercept, and draw the line through both points.

Sketching linear graphs using the gradient and a point

Start by plotting the single point provided.  

Remember that the gradient is a measure of the change of rise with respect to run. Step out one measure of the gradient from the original point given to sketch. The point can be any point $\left(x,y\right)$(x,y), or it could be an intercept.  Either way, plot the point, step out the gradient and draw the line.

A line with gradient $4$4. Move $1$1 unit across and $4$4 units up.

A line with gradient $-3$−3. Move $1$1 unit across and $3$3 units down.

A line with gradient $\frac{1}{2}$12​. Move $2$2 units across and $1$1 unit up.

For example, plot the line with gradient $-2$−2 and $y$y intercept of $4$4.  

Practice questions

Question 3

Question 4

Question 5

Question 6