Question

$f(x)=\

$$\begin{cases} x + 7 & \\text{for } x \\leq -2 \\\\ -3x - 4 & \\text{for } -2 < x \\leq 1 \\\\ x - 9 & \\text{for } x > 1 \\end{cases}$$

$
line closed circle open circle
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: Plot first piece: $x+7, x\leq-2$

Find endpoints:
When $x=-2$, $f(-2)=-2+7=5$ (closed circle, since $x\leq-2$)
To show the line extends left, pick $x=-5$: $f(-5)=-5+7=2$
Draw line from $(-5,2)$ to $(-2,5)$, closed circle at $(-2,5)$.

Step2: Plot second piece: $-3x-4, -2

Find endpoints:
When $x=-2$, $f(-2)=-3(-2)-4=6-4=2$ (open circle, since $x>-2$)
When $x=1$, $f(1)=-3(1)-4=-3-4=-7$ (closed circle, since $x\leq1$)
Draw line from $(-2,2)$ (open circle) to $(1,-7)$ (closed circle).

Step3: Plot third piece: $x-9, x>1$

Find endpoints:
When $x=1$, $f(1)=1-9=-8$ (open circle, since $x>1$)
To show the line extends right, pick $x=5$: $f(5)=5-9=-4$
Draw line from $(1,-8)$ (open circle) to $(5,-4)$.

Answer:

The graph consists of three segments:

1. A line from $(-\infty, 2)$ (extending left) to $(-2, 5)$ with a closed circle at $(-2, 5)$ (for $f(x)=x+7, x\leq-2$)
2. A line from $(-2, 2)$ (open circle) to $(1, -7)$ (closed circle) (for $f(x)=-3x-4, -2
3. A line from $(1, -8)$ (open circle) to $(+\infty, -4)$ (extending right) (for $f(x)=x-9, x>1$)