1. The problem involves analyzing the graph of the piecewise function $$f(x) = \begin{cases} -x + 1 & x < 0 \\ x - 2 & 0 \leq x \end{cases}$$ and identifying the error in the graph. 2. For piecewise functions, the type of dot (open or closed) at the boundary points indicates whether the function value is included (closed dot) or excluded (open dot) at that point. 3. At $$x=0$$, since the function definition for $$x \geq 0$$ is $$f(x) = x - 2$$, the value at $$x=0$$ is $$f(0) = 0 - 2 = -2$$. This means the graph should have a closed dot at $$(0,-2)$$. 4. For $$x < 0$$, the function is $$f(x) = -x + 1$$. Approaching $$x=0$$ from the left, $$f(0^-) = -0 + 1 = 1$$, so the graph should have an open dot at $$(0,1)$$ because $$x=0$$ is not included in this piece. 5. The description says the solid line starts with a closed circle at $$(0,-2)$$ and the dotted line starts with an open circle at $$(0,1)$$, which matches the correct interpretation. 6. The error options are: - Option #1: Closed dot at $$(0,1)$$ (incorrect, should be open dot) - Option #2: Open dot at $$(0,-2)$$ (incorrect, should be closed dot) - Option #3: Point at $$(0,-2)$$ should be an arrow to the left (incorrect, no arrow needed here) - Option #4: Point at $$(-2,3)$$ should be an arrow to the left (incorrect, this point is an endpoint) 7. Since the graph shows a closed circle at $$(0,-2)$$ and an open circle at $$(0,1)$$, the error is that the graph should have a closed dot at $$(0,1)$$ instead of an open dot. Final answer: Option #1 describes the error in the graph.