Question

the following function on the axes provided.
$f(x) = \

$$\begin{cases} -x - 6 & \\text{for} & x < 0 \\\\ 3x - 18 & \\text{for} & x > 5 \\end{cases}$$

$
click and drag to make a line. click the line to delete it.
click on an endpoint of a line to change it.

Explanation:

Step1: Analyze the first piece ($x < 0$)

The first piece of the function is $f(x) = -x - 6$ for $x < 0$. This is a linear function in the form $y = mx + b$, where the slope $m = -1$ and the y-intercept $b = -6$. But since the domain is $x < 0$, we need to find a point on this line. When $x = 0$, $f(0) = -0 - 6 = -6$, but since $x < 0$, the endpoint at $x = 0$ is not included (open circle). Let's take another point, say $x = -2$: $f(-2) = -(-2) - 6 = 2 - 6 = -4$. So we have a line with slope -1, passing through points like $(-2, -4)$ and approaching $(0, -6)$ (open circle).

Step2: Analyze the second piece ($x > 5$)

The second piece is $f(x) = 3x - 18$ for $x > 5$. This is also a linear function with slope $m = 3$ and y-intercept $b = -18$. When $x = 5$, $f(5) = 3(5) - 18 = 15 - 18 = -3$, but since $x > 5$, the endpoint at $x = 5$ is not included (open circle). Let's take $x = 6$: $f(6) = 3(6) - 18 = 18 - 18 = 0$. So we have a line with slope 3, passing through points like $(6, 0)$ and approaching $(5, -3)$ (open circle).

Step3: Plot the lines

For the first piece ($x < 0$), draw a line with slope -1, starting from the left (since $x < 0$) and approaching the point $(0, -6)$ (open circle). For the second piece ($x > 5$), draw a line with slope 3, starting from the right of $x = 5$ (open circle at $(5, -3)$) and going through $(6, 0)$ and beyond.

Answer:

To graph the piecewise function:

• For $x < 0$, draw the line $y = -x - 6$ with an open circle at $(0, -6)$ and extending to the left (e.g., through $(-2, -4)$).
• For $x > 5$, draw the line $y = 3x - 18$ with an open circle at $(5, -3)$ and extending to the right (e.g., through $(6, 0)$).