1. **State the problem:** We are given a piecewise function: $$f(x) = \begin{cases} -x - 4, & x < -3 \\ x, & -3 \leq x < 3 \\ -x + 1, & x \geq 3 \end{cases}$$ We want to understand the behavior of this function and confirm the correct graph representation. 2. **Analyze each piece:** - For $x < -3$, the function is $f(x) = -x - 4$. This is a line with slope $-1$ and y-intercept $-4$. - For $-3 \leq x < 3$, the function is $f(x) = x$. This is a line with slope $1$ passing through the origin. - For $x \geq 3$, the function is $f(x) = -x + 1$. This is a line with slope $-1$ and y-intercept $1$. 3. **Check continuity at boundaries:** - At $x = -3$: - Left limit: $f(-3^-) = -(-3) - 4 = 3 - 4 = -1$ - Right limit: $f(-3^+) = -3$ (since $f(x) = x$ for $x \geq -3$) - Since $-1 \neq -3$, there is a jump discontinuity at $x = -3$. - At $x = 3$: - Left limit: $f(3^-) = 3$ - Right limit: $f(3^+) = -3 + 1 = -2$ - Since $3 \neq -2$, there is a jump discontinuity at $x = 3$. 4. **Plot points at boundaries:** - At $x = -3$, the function value is $f(-3) = -3$ (from the middle piece, since $-3 \leq x < 3$). - At $x = 3$, the function value is $f(3) = -3 + 1 = -2$ (from the last piece, since $x \geq 3$). 5. **Summary:** - The graph has three linear segments with slopes $-1$, $1$, and $-1$ respectively. - There are jump discontinuities at $x = -3$ and $x = 3$. - The first graph described matches these properties exactly. **Final answer:** The correct graph is the first graph from the left, showing the piecewise linear function with the described slopes and jump discontinuities.