Piecewise Graphs

Consider the graph of lines described by the following equations and domains.

Line equation	Drawn over the domain interval given by:
$y=12-2x$y=12−2x	$-2\le x<2$−2≤x<2
$y=\frac{1}{2}x+7$y=12​x+7	$2\le x\le10$2≤x≤10
$y=52-4x$y=52−4x	$1010<x≤13

We consider this as a piecewise graph and this specific example is defined across the restricted domain $-2\le x\le13$−2≤x≤13. The complete graph is shown here. As you read the following description, keep referring to it so that you get a visual understanding of how it is constructed.

While not a requirement of piecewise graphs, this example also shows another interesting feature. It happens to be continuous across the entire domain. We can think of continuity as being able to trace across the three line segments without lifting a pencil off the paper. See this lesson for more information.

Remember, however, that there is no requirement for a piecewise graph to be continuous.

The left most line segment begins when $x=-2$x=−2 so that $y=16$y=16. From $\left(-2,16\right)$(−2,16), it falls linearly with a gradient of $-2$−2, crossing the $y$y-axis at $y=12$y=12. It continues on toward the point where $x=2$x=2.

The domain of this portion of the piecewise graph does not include the end point $x=2$x=2. But, if it did include it, we would have determined the corresponding $y$y value as $y=12-2\left(2\right)=8$y=12−2(2)=8.

However the domain of the middle line segment includes $x=2$x=2. So using the middle equation, the second line segment starts with the y value given by $y=\frac{1}{2}\left(2\right)+7=8$y=12​(2)+7=8.

This means that the first segment connects to the middle segment.

The middle segment rises from $\left(2,8\right)$(2,8) with a gradient of $\frac{1}{2}$12​ continuing to the point where $x=10$x=10. This time, according to the table above, $x=10$x=10 is included as part of the domain of the middle segment. Therefore, at $x=10$x=10, we can see that $y=\frac{1}{2}\left(10\right)+7=12$y=12​(10)+7=12.

The final segment does not include the point $\left(10,12\right)$(10,12), but we note that the corresponding equation for the segment is $y=52-4x$y=52−4x, and if we were to put $x=10$x=10 into this, we would see that $y=52-4\left(10\right)=12$y=52−4(10)=12.

In other words, the middle segment connects to the last segment.

This last segment has the very steep negative gradient of $-4$−4, and we note that at the endpoint $x=13$x=13, we have $y=52-4\left(13\right)=0$y=52−4(13)=0, and this means the piecewise graph ends on the $x$x-axis.

Worked Examples

question 1

question 2

question 3