6.04 Linear graphs

Sketching a graph using a table of values

A table of values, created using an equation, forms a set of points that can be plotted on a number plane. A line, drawn through the points, becomes the graph of the equation.

Exploration

We'll begin by creating a table of values for the following equation:

$y=3x-5$y=3x−5

The first row of the table will contain values for the independent variable (in this case, $x$x). The choice of $x$x-value is often determined by the context, but in many cases they will be given. To find the corresponding $y$y-value, we substitute each $x$x-value into the equation $y=3x-5$y=3x−5.

$x$x	$1$1	$2$2	$3$3	$4$4
$y$y

Substituting $x=1$x=1:

$y$y	$=$=	$3\times1-5$3×1−5
	$=$=	$3-5$3−5
	$=$=	$-2$−2

Substituting the remaining values of $x$x, allows us to complete the table:

$x$x	$1$1	$2$2	$3$3	$4$4
$y$y	$-2$−2	$1$1	$4$4	$7$7

We can see here that the $y$y-values are creating a growing (increasing) pattern that is going up by $3$3 each time the $x$x-value increases by $1$1. This means that the corresponding line should be increasing as $x$x-increases (increasing from left to right).

 

Plotting points from a table of values

The $x$x and $y$y value in each column of the table can be grouped together to form the coordinates of a single point, $\left(x,y\right)$(x,y).

 

 

If we sketch a straight line through the points, we get the graph of $y=3x-5$y=3x−5.

Notice that when sketching a straight line through a set of points, the line should not start and end at the points, but continue beyond them, across the entire coordinate plane. As predicted the line is increasing from left to right. 

 

 

Sketching a graph using its intercepts

The word intercept in mathematics refers to a point where a line or curve crosses or intersects with the axes.

• We can have $x$x-intercepts: where the line or curve crosses the $x$x-axis.
• We can have $y$y intercepts: where the line or curve crosses the $y$y-axis.

Consider what happens as a point moves up or down along the $y$y-axis. It will eventually reach the origin $\left(0,0\right)$(0,0) where $y=0$y=0. Now, if the point moves along the $x$x-axis in either direction, the $y$y-value is still $0$0.

Similarly, consider what happens as a point moves along the $x$x-axis. It will eventually reach the origin where $x=0$x=0. Now, if the point moves along the $y$y-axis in either direction, the $x$x-value is still $0$0.

This interactive demonstrates the idea behind the coordinates of $x$x and $y$y-intercepts.  

 

 

How do I find an $x$x-intercept?

$x$x-intercepts occur when the $y$y-value is $0$0. So let $y=0$y=0 and then solve for $x$x. 

 

How do I find a $y$y-intercept?

$y$y-intercepts occur when the $x$x-value is $0$0. So let $x=0$x=0 and then solve for $y$y.

Alternatively we can read the $y$y-intercept value from the equation when it is in the form $y=ax+b$y=ax+b. The value of $b$b is the value of the $y$y-intercept.

 

Worked example

example 1

Sketch the graph of the line with equation $3x-y=6$3x−y=6.

Think: When the equation of a line is not in the form $y=ax+b$y=ax+b, it is easiest to sketch the line by findings its intercepts.

Do: 

$x=0:$x=0:

$0-y$0−y	$=$=	$6$6
$-y$−y	$=$=	$6$6
$y$y	$=$=	$-6$−6

$y=0:$y=0:

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

 

So the intercepts are $x=2$x=2 and $y=-6$y=−6. By plotting these points and drawing a straight line through them we get the following graph: 

 

 

Sketching a graph using its slope and one point

We can also graph a line by identifying the slope and the $y$y-intercept from the equation when it is in the form $y=ax+b$y=ax+b.

We know that the $y$y-intercept occurs at $\left(0,b\right)$(0,b), and the slope is equal to $a$a. Using this information we can plot the point at the $y$y-intercept (or any other point by substituting in a value for $x$x and solving for $y$y) and then move right by $1$1, and up (or down if $a$a is negative) by $a$a. 
As as an example, if we have the equation $y=2x+3$y=2x+3, then we know the $y$y-intercept is at $\left(0,3\right)$(0,3) and as the slope is $2$2, another point will be at $\left(1,3+2\right)=\left(1,5\right)$(1,3+2)=(1,5). So then we just draw a straight line through these two points.

 

Practice questions

Question 1

Question 2

Question 3

Question 4

Lines parallel to x and y-axis

 

The line with equation given by $x=c$x=c, with $c$c as any real number, is drawn parallel to the $y$y axis passing through the point $\left(c,0\right)$(c,0) on the $x$x axis.  Three examples, specifically $x=-1$x=−1, $x=1$x=1, and $x=2\sqrt{2}$x=2√2 are shown in the diagram below:

A nice way to think about the line given by the equation $x=c$x=c is to realize that every point on it has the $x$x part of the coordinate address equal to $c$c, and the $y$y part can be any number. Thus points like $\left(c,-3\right),\left(c,0\right),\left(c,5\right),\left(c,23\right)$(c,−3),(c,0),(c,5),(c,23) are all on the line, one directly above the other.  Hence, the line is perpendicular to the $x$x-axis.

In a similar way, the line with equation $y=b$y=b, where $b$b is a constant, is parallel to the $x$x-axis and passes through the point $\left(0,b\right)$(0,b) on the $y$y-axis. The lines $y=-3,y=2$y=−3,y=2 and $y=5$y=5 are shown in the following diagram:

 

Practice questions

QUESTION 5

QUESTION 6

Equivalent expressions

We know that the expression $2(3x-4)$2(3x−4) is equivalent to the expression $6x-8$6x−8.  We can show this algebraically by expanding the brackets of the first expression to get the second expression. 

But another way we can show this is by graphing both linear functions and showing that we get the same line. 

 

Worked example

example 2

Show that the expression $3x+5$3x+5 is equivalent to the expression $3(x+2)-1$3(x+2)−1 by graphing the lines $y=3x+5$y=3x+5 and $y=3(x+2)-1$y=3(x+2)−1.

Think: We can use a table of values to find points that both lines pass through. 

Do: 

For $y=3x+5$y=3x+5: 

$x$x	$0$0	$1$1
$y$y	$5$5	$8$8

 

For $y=3(x+2)-1$y=3(x+2)−1: 

$x$x	$0$0	$1$1
$y$y	$5$5	$8$8

Therefore both lines pass through the points $(0,5)$(0,5) and$(1,8)$(1,8), to get the following line:

Since both expressions produce the same line the expressions are equivalent.