Graph of Piecewise Function f(x) = -2, x-1, 0

Answer

The graph of the piecewise function $f(x)$ consists of a horizontal ray starting at $(-1, -2)$ extending left, a linear segment connecting $(-1, -2)$ to $(2, 1)$, and a horizontal ray starting from an open circle at $(2, 0)$ extending right. The final image provided in your prompt correctly illustrates these three distinct behaviors based on the given domain restrictions.

Explanation

The image shows a Cartesian coordinate system where a piecewise function is plotted in blue. I observe three distinct parts: a solid horizontal line for $x \leq -1$, a slanted line with a positive slope for $-1 < x \leq 2$, and another horizontal line along the x-axis for $x > 2$.

1. Analyze the first piece: Constant Function For the interval $x \leq -1$, the function is defined as $f(x) = -2$. $f(x) = -2 \text{ for } x \in (-\infty, -1]$ This represents a horizontal line at height $-2$ that includes a solid dot at $x = -1$ because of the "less than or equal to" symbol ($\leq$). ⚠️ This step is required on exams: Always use a solid circle for inclusive boundaries ($\leq$ or $\geq$).

2. Analyze the second piece: Linear Function For the interval $-1 < x \leq 2$, the function is $f(x) = x - 1$. To graph this, evaluate the endpoints: $f(-1) \to -1 - 1 = -2$ $f(2) = 2 - 1 = 1$ This formula calculates the $y$-values at the boundaries of the middle segment. Since $f(-1) = -2$, this piece starts exactly where the first piece ended (continuity at $x = -1$). Because $-1 < x$ is a strict inequality, we technically place an open circle, but it is "filled in" by the previous piece. At $x = 2$, we use a solid dot because of the $\leq$ sign.

3. Analyze the third piece: Zero Function For the interval $x > 2$, the function is $f(x) = 0$. $f(x) = 0 \text{ for } x \in (2, \infty)$ This describes a horizontal ray lying on the x-axis starting from $x=2$. ⚠️ This step is required on exams: Since the condition is $x > 2$ (strict inequality), you must place an open circle at $(2, 0)$ to show that the function does not actually equal $0$ at $x = 2$.

4. Summary of Endpoint Behavior The following table summarizes the key points for the graph:

Interval	Function	Left Endpoint	Right Endpoint
$(-\infty, -1]$	$-2$	$(-\infty, -2)$	$(-1, -2)$ [Solid]
$(-1, 2]$	$x-1$	$(-1, -2)$ [Open]	$(2, 1)$ [Solid]
$(2, \infty)$	$0$	$(2, 0)$ [Open]	$(\infty, 0)$

Final Answer

The graph is correctly represented by a horizontal ray at $y=-2$ for $x \le -1$, a line segment from $(-1, -2)$ to $(2, 1)$, and a horizontal ray at $y=0$ for $x > 2$ starting with an open circle. $\boxed{f(x) \text{ has a jump discontinuity at } x = 2}$

Common Mistakes

• Incorrect Circle Types: Students often forget the distinction between solid circles (inclusive: $\leq, \geq$) and open circles (exclusive: $<, >$). Always check the inequality sign at the boundaries.
• Vertical Line Errors: Occasionally, students try to "connect" the jump at $x=2$ with a vertical line. This is incorrect as it would violate the Vertical Line Test, meaning the relation would no longer be a function.