1. **State the problem:** We need to graph the piecewise function $$f(x) = \begin{cases} 1 - 2x & \text{if } x < 1 \\ x - 3 & \text{if } x \geq 1 \end{cases}$$ 2. **Understand the pieces:** - For $x < 1$, the function is $f(x) = 1 - 2x$, a line with slope $-2$ and y-intercept $1$. - For $x \geq 1$, the function is $f(x) = x - 3$, a line with slope $1$ and y-intercept $-3$. 3. **Evaluate at the boundary $x=1$:** - Left side limit: $f(1^-) = 1 - 2(1) = 1 - 2 = -1$ (open circle since $x<1$). - Right side value: $f(1) = 1 - 3 = -2$ (closed circle since $x \geq 1$). 4. **Plot the first piece ($x<1$):** - Line $y = 1 - 2x$ for $x < 1$. - At $x=0$, $f(0) = 1 - 0 = 1$. - At $x=1$, open circle at $(1, -1)$. 5. **Plot the second piece ($x \geq 1$):** - Line $y = x - 3$ for $x \geq 1$. - At $x=1$, closed circle at $(1, -2)$. - At $x=2$, $f(2) = 2 - 3 = -1$. 6. **Interpretation:** The graph has a jump discontinuity at $x=1$ with the first line ending at $(1, -1)$ open circle and the second line starting at $(1, -2)$ closed circle. **Final answer:** The correct graph is the one with a decreasing line $y=1-2x$ for $x<1$ ending at an open circle at $(1,-1)$ and an increasing line $y=x-3$ for $x \geq 1$ starting at a closed circle at $(1,-2)$.