Question

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

$$\begin{cases} -x + 1 & \\text{for} & x \\leq -3 \\\\ -3x + 2 & \\text{for} & x > 0 \\end{cases}$$

$
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.
graph axes with x from -10 to 10 and y from -2 to 10

Explanation:

Step1: Find endpoint for $x \leq -3$

Substitute $x=-3$: $f(-3) = -(-3)+1 = 4$. So closed circle at $(-3, 4)$.

Step2: Plot line for $x \leq -3$

Choose $x=-10$: $f(-10) = -(-10)+1 = 11$. Draw line from $(-3, 4)$ to $(-10, 11)$, extending left.

Step3: Find endpoint for $x > 0$

Substitute $x=0$: $f(0) = -3(0)+2 = 2$. So open circle at $(0, 2)$.

Step4: Plot line for $x > 0$

Choose $x=2$: $f(2) = -3(2)+2 = -4$. Draw line from $(0, 2)$ (open circle) to $(2, -4)$, extending right.

Answer:

(Graph description: Two line segments:

1. A line segment for $f(x) = -x + 1$ with a closed circle at $(-3, 4)$ extending leftwards, passing through $(-10, 11)$.
2. A line segment for $f(x) = -3x + 2$ with an open circle at $(0, 2)$ extending rightwards, passing through $(2, -4)$.)