Line equation | Drawn over the domain interval given by: |
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$y=12-2x$y=12−2x | $-2\le x<2$−2≤x<2 |
$y=\frac{1}{2}x+7$y=12x+7 | $2\le x\le10$2≤x≤10 |
$y=52-4x$y=52−4x | $10 |
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.
What is the function definition of the graph?
$y$y | $=$= | $\frac{3}{2}x+3$32x+3 | if $x\le0$x≤0 | |
$3-x$3−x | if $0 |
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$2x-6$2x−6 | if $x>3$x>3 |
$y$y | $=$= | $\frac{3}{2}x+3$32x+3 | if $x\le0$x≤0 | |
$3-x$3−x | if $0 |
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$2x-6$2x−6 | if $3 |
$y$y | $=$= | $\frac{3}{2}x+3$32x+3 | if $-5\le x$−5≤x$\le$≤$0$0 | |
$3-x$3−x | if $0 |
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$2x-6$2x−6 | if $x>3$x>3 |
$y$y | $=$= | $\frac{3}{2}x+3$32x+3 | if $-5\le x$−5≤x$\le$≤$0$0 | |
$3-x$3−x | if $0 |
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$2x-6$2x−6 | if $3 |
$y$y | $=$= | $\frac{3}{2}x+3$32x+3 | if $x\le0$x≤0 | |
$3-x$3−x | if $0 |
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$2x-6$2x−6 | if $x>3$x>3 |
$y$y | $=$= | $\frac{3}{2}x+3$32x+3 | if $x\le0$x≤0 | |
$3-x$3−x | if $0 |
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$2x-6$2x−6 | if $3 |
$y$y | $=$= | $\frac{3}{2}x+3$32x+3 | if $-5\le x$−5≤x$\le$≤$0$0 | |
$3-x$3−x | if $0 |
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$2x-6$2x−6 | if $x>3$x>3 |
$y$y | $=$= | $\frac{3}{2}x+3$32x+3 | if $-5\le x$−5≤x$\le$≤$0$0 | |
$3-x$3−x | if $0 |
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$2x-6$2x−6 | if $3 |
Consider the function graphed below.
What is the value of the function at $x=-1$x=−1?
Is the graph continuous or discontinuous at $x=-1$x=−1?
Continuous
Discontinuous
Continuous
Discontinuous
What is the value of the function at $x=0$x=0?
Is the graph continuous or discontinuous at $x=0$x=0?
Continuous
Discontinuous
Continuous
Discontinuous
What is the value of the function at $x=2$x=2?
Is the graph continuous or discontinuous at $x=2$x=2?
Continuous
Discontinuous
Continuous
Discontinuous
The function $f\left(x\right)$f(x) is defined below:
$f\left(x\right)$f(x) | $=$= | $x+2$x+2 | when $x\ge3$x≥3 | |
$3$3 | when $x<3$x<3 |
Graph the function.
Which of the following statements is true?
The function is defined and continuous at $x=3$x=3.
The function is undefined and not continuous at $x=3$x=3.
The function is defined but not continuous at $x=3$x=3.
The function is undefined but continuous at $x=3$x=3.
The function is defined and continuous at $x=3$x=3.
The function is undefined and not continuous at $x=3$x=3.
The function is defined but not continuous at $x=3$x=3.
The function is undefined but continuous at $x=3$x=3.
Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs
Apply graphical methods in solving problems