Question

the function f(x) is to be graphed on a coordinate plane.
f(x)=\begin{cases}-x, & x < 0\\1, & xgeq0end{cases}
at what point should an open - circle be drawn?
(-1, 0)
(0, 0)
(1, 0)

Explanation:

Step1: Analyze piece - wise function

The function $f(x)=

$$\begin{cases}-x, &x < 0\\1, &x\geq0\end{cases}$$

$ has a break at $x = 0$.

Step2: Determine open - circle position

For $x<0$, the function is $y=-x$. When $x$ approaches $0$ from the left - hand side, $y = 0$. But the function is not defined as $-x$ at $x = 0$. The value of the function for $x\geq0$ is $y = 1$. So, an open circle should be drawn at the point where the left - hand side function "stops" being applicable at $x = 0$ and $y=0$.

Answer:

B. $(0,0)$