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.
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).
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).
Each point can then be plotted on a $xy$xy-plane.
To plot a point, $\left(a,b\right)$(a,b), on a number plane, we first identify where $x=a$x=a lies along the $x$x-axis, and where $y=b$y=b lies along the $y$y axis.
For example, to plot the point $\left(3,4\right)$(3,4), we identify $x=3$x=3 on the $x$x-axis and construct a vertical line through this point. Then we identify $y=4$y=4 on the $y$y-axis and construct a horizontal line through this point. The point where the two lines meet has the coordinates $\left(3,4\right)$(3,4).
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.
To sketch a straight line graph we actually only need to identify two points!
The word intercept in mathematics refers to a point where a line or curve crosses or intersects with the axes.
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.
The $x$x-intercept occurs at the point where $y=0$y=0.
The $y$y-intercept occurs at the point where $x=0$x=0.
$x$x-intercepts occur when the $y$y-value is $0$0. So let $y=0$y=0 and then solve for $x$x.
$y$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.
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:
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.
Consider the equation $y=2x-4$y=2x−4.
Complete the table of values.
$x$x | $0$0 | $1$1 | $2$2 | $3$3 |
---|---|---|---|---|
$y$y | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Using the table of values, plot the points that correspond to when $x=0$x=0 and $y=0$y=0:
Using the points plotted above, sketch the line that passes through the two points:
Consider the linear equation $y=2x-2$y=2x−2.
What are the coordinates of the $y$y-intercept?
Give your answer in the form $\left(a,b\right)$(a,b).
What are the coordinates of the $x$x-intercept?
Give your answer in the form $\left(a,b\right)$(a,b).
Now, sketch the line $y=2x-2$y=2x−2:
Sketch the line $y=-x-5$y=−x−5 using the $y$y-intercept and any other point on the line.
Sketch the line that has a slope of $-3$−3 and an $x$x-intercept of $-5$−5.
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:
Plot the line $x=-8$x=−8 on the number plane.
Find the intersection of the lines $x=6$x=6 and the line $y=-3$y=−3 .
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.
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.