2.04 Linear functions

Identifying key features

The graph of a linear relationship will create a line. All linear functions can be written in either of these two common forms:

Gradient intercept form	General form
$y=mx+c$y=mx+c	$ax+by+c=0$ax+by+c=0

 If we sketch a linear relationship on a plane, the straight line formed will be one of the following types:

• An increasing graph means that as $x$x values increase, the $y$y values increase.
• A decreasing graph means that as $x$x values increase, the $y$y values decrease.
• A horizontal graph means that as $x$x values change the $y$y values remain the same
• In a vertical graph the $x$x value is constant. 

Regardless of all different shapes all linear functions have some common characteristics.

 

Intercepts

They all have at least one intercept.  Linear functions might have

• an $x$x intercept only (in the case of a vertical line)
• a $y$y intercept only (in the case of horizontal lines) 
• or some have $2$2 intercepts, both an $x$x and a $y$y (in the case of increasing or decreasing functions)

 

Gradient

The gradient (slope) of a line is a measure of how steep the line is. For a linear function the gradient is constant. That is as the $x$x-value increases by a constant amount, the $y$y-value also increases by a constant amount. We can calculate the gradient from any two points $\left(x_1,y_1\right)$(x1​,y1​), $\left(x_2,y_2\right)$(x2​,y2​) on a line:

The  gradient is often represented by the letter $m$m and has the following properties:

• If $m<0$m<0, the gradient is negative and the line is decreasing
• If $m>0$m>0, the gradient is positive and the line is increasing
• If $m=0$m=0 the gradient is $0$0 and the line is horizontal
• For vertical lines $m$m is undefined
• It can be read off directly, as the coefficient of $x$x, from the gradient intercept form ($y=mx+c$y=mx+c)
• By rearranging general form ($ax+by+c=0$ax+by+c=0 ) we can see the gradient will be $m=-\frac{a}{b}$m=−ab​
• The gradient tells us as the $x$x-value increases by $1$1 the $y$y-value changes by $m$m
• Also, the larger the value of $\left|m\right|$|m| the steeper the line

 

Horizontal lines

On horizontal lines, the $y$y value is always the same for every point on the line. In other words, there is no rise- it's completely flat. 

$A=\left(-4,4\right)$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 gradient is calculated by $\frac{\text{rise }}{\text{run }}$rise run ​ and there is no rise (ie. $\text{rise }=0$rise =0), the gradient of a horizontal line is always zero.

 

 

 

 

Vertical lines

On vertical lines, the $x$x value is always the same for every point on the line.

Let's look at the coordinates for A,B and C on this 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" (ie. $\text{run }=0$run =0). l If we substituted this into the $\frac{\text{rise }}{\text{run }}$rise run ​ equation, we'd have 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.

 

 

 

Practice questions

Question 1

Question 2

Question 3

Question 4

Points on a line

We can 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 lines equation, 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 5

 

Plotting the graph of a linear relationship

To graph any liner relationship you only need two points that are on the line.  You can use any two points from a table of values, or substitute in any two values of $x$x into the equation and solve for corresponding $y$y-value to create your own two points.  Often, using the intercepts is one of the most convenient ways to sketch the line.

Example - sketch from table of values

To sketch from a table of values, we need just any two points from the table.  From this table we have 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 the $2$2 of the points on this interactive to the correct positions and graph this linear relationship.  

 

Created with Geogebra

 

 

Example - sketch from any two points

If we are given the equation of a linear relationship, like $y=3x+5$y=3x+5, then to sketch it we need two points.  We can pick any two points we like.  

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

Similarly look for other easy values to calculate such as $1$1, $10$10, $2$2.  I'll pick $x=1$x=1.  Then for $y=3x+5$y=3x+5, we have $y=3\times1+5$y=3×1+5, $y=8$y=8.This gives us the point $\left(1,8\right)$(1,8)

Now we plot the two points and create a line.  

 

Example - sketch from the intercepts

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

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

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

$3y+2x-6$3y+2x−6	$=$=	$0$0
$0+2x-6$0+2x−6	$=$=	$0$0
$2x$2x	$=$=	$6$6
$x$x	$=$=	$3$3

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

$3y+2x-6$3y+2x−6	$=$=	$0$0
$3y+0-6$3y+0−6	$=$=	$0$0
$3y$3y	$=$=	$6$6
$y$y	$=$=	$2$2

From here it is pretty easy to sketch, we find the $x$x intercept $3$3, and the $y$y intercept $2$2, and draw the line through both.

Example - sketch from the gradient and a point

Start by plotting the single point that you are given.  

Remembering that gradient is a measure of change in the rise per change in run, we can step out one measure of the gradient from the original point given. 

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 your line!

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

Practice questions

Question 6

Question 7

Question 8

Question 9