Question

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$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.

Explanation:

Step1: Plot first segment endpoint

For $f(x)=-x+1$, $x\leq-3$:
When $x=-3$, $f(-3)=-(-3)+1=4$. This is a closed circle at $(-3, 4)$ since $x\leq-3$ includes the endpoint.

Step2: Extend first segment left

The line $f(x)=-x+1$ has a slope of $-1$. For $x<-3$, as $x$ decreases, $f(x)$ increases. For example, $x=-4$ gives $f(-4)=-(-4)+1=5$, so the line extends left from $(-3,4)$ with slope $-1$.

Step3: Plot second segment endpoint

For $f(x)=-3x+2$, $x>0$:
When $x=0$, $f(0)=-3(0)+2=2$. This is an open circle at $(0, 2)$ since $x>0$ excludes the endpoint.

Step4: Extend second segment right

The line $f(x)=-3x+2$ has a slope of $-3$. For $x>0$, as $x$ increases, $f(x)$ decreases. For example, $x=1$ gives $f(1)=-3(1)+2=-1$, so the line extends right from $(0,2)$ with slope $-3$.

Answer:

1. Draw a line with a closed circle at $(-3, 4)$ extending left with slope $-1$ (for $x\leq-3$).
2. Draw a line with an open circle at $(0, 2)$ extending right with slope $-3$ (for $x>0$).