1. **State the problem:** Given the function $f(x)$ defined by the piecewise points from the graph: - Segment 1 from $(−9,4)$ to $(−5,−1)$ - Segment 2 from $(−5,−1)$ to $(0,−9)$ - Segment 3 from $(0,−9)$ to $(3,1)$ We need to find values for: - Table 1: $f(x) + 2$ for $x = -5, 0, 3, -9$ - Table 2: $f(x) - 2$ for $x = -5, 0, 3, -9$ Then sketch the graphs of $y = f(x) + 2$ and $y = f(x) - 2$. 2. **Find original $f(x)$ values at points:** From the graph points: - $f(-9) = 4$ - $f(-5) = -1$ - $f(0) = -9$ - $f(3) = 1$ 3. **Calculate values for Table 1 ($f(x) + 2$):** - $f(-9) + 2 = 4 + 2 = 6$ - $f(-5) + 2 = -1 + 2 = 1$ - $f(0) + 2 = -9 + 2 = -7$ - $f(3) + 2 = 1 + 2 = 3$ 4. **Calculate values for Table 2 ($f(x) - 2$):** - $f(-9) - 2 = 4 - 2 = 2$ - $f(-5) - 2 = -1 - 2 = -3$ - $f(0) - 2 = -9 - 2 = -11$ - $f(3) - 2 = 1 - 2 = -1$ 5. **Describe the graph shifts:** - The graph of $y = f(x) + 2$ is the original graph of $f(x)$ shifted **up by 2 units**. - The graph of $y = f(x) - 2$ is the original graph of $f(x)$ shifted **down by 2 units**. **Final answer:** Table 1: $\begin{array}{c|c} x & f(x) + 2 \\ \hline -9 & 6 \\ -5 & 1 \\ 0 & -7 \\ 3 & 3 \end{array}$ Table 2: $\begin{array}{c|c} x & f(x) - 2 \\ \hline -9 & 2 \\ -5 & -3 \\ 0 & -11 \\ 3 & -1 \end{array}$