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 easiest ways to sketch the line.
x | 1 | 2 | 3 | 4 |
y | 3 | 5 | 7 | 9 |
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.
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.
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 intercept happens when the $y$y value is $0$0. $3y+2x-6=0$3y+2x−6=0 $0+2x-6=0$0+2x−6=0 $2x=6$2x=6 $x=3$x=3 | The y intercept happens when the $x$x value is $0$0. $3y+2x-6=0$3y+2x−6=0 $3y+0-6=0$3y+0−6=0 $3y=6$3y=6 $y=2$y=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.
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.
For a gradient of $4$4 $1$1 unit across and $4$4 units up. | For a gradient of $-3$−3 $1$1 unit across and $3$3 units down. | For a gradient of $\frac{1}{2}$12 $1$1 unit across and $\frac{1}{2}$12 unit up. |
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$4.
Start with the point, ($y$y intercept of $4$4) | Step out the gradient, (-$2$2 means $2$2 units down) |
Draw the line |
To sketch linear graphs, it's easiest to substitute in values to find coordinates to put it in gradient-intercept form.
$y=mx+b$y=mx+b
where $m$m is the gradient and $b$b is the $y$y-intercept
Our graphs may not always be in this form so we may need to rearrange the equation to make $y$y the subject (that means $y$y is on one side of the equation and everything else is on the other side).
Sometimes, it doesn't matter. We can sketch a straight line on a graph just by knowing a couple of its features such as a point that lies on the line and it's gradient. At other times, we may need to generate an equation before we sketch it. So other than the gradient-intercept form, we can use:
Ok let's look at this in action with some examples.
Plot the graph of the line whose gradient is $-3$−3 and passes through the point $\left(-2,4\right)$(−2,4).
Consider the linear equation $y=3x+1$y=3x+1.
State the $y$y-value of the $y$y-intercept of this line.
Using the point $Y$Y as the $y$y-intercept, sketch a graph of the equation $y=3x+1$y=3x+1.
Graph the linear equation $-6x+3y+24=0$−6x+3y+24=0 by finding any two points on the line.
On horizontal lines, the $y$y value is always the same for every point on the line.
On vertical lines, the $x$x value is always the same for every point on the line.
Draw a graph of the line $y=-3$y=−3.