Question: Express the function graphed on the axes below as a piecewise function.
Answer
f(x) = {
__ for __
__ for __
}
Graph: a horizontal ray extends leftward along y = 3 with an open circle at (0, 3), and a downward-sloping ray starts at an open circle at (0, 2) and goes rightward through about (3, -1), so the breakpoint is at x = 0; position_hint = center
1. **State the problem:** We need to express the given graph as a piecewise function $f(x)$.
2. **Analyze the graph:**
- For $x < 0$, the graph is a horizontal ray at $y = 3$ with an open circle at $(0,3)$, so the function does not include $x=0$ here.
- For $x > 0$, the graph is a downward-sloping ray starting at an open circle at $(0,2)$ and passing through approximately $(3,-1)$.
3. **Write the piecewise function:**
- For $x < 0$, $f(x) = 3$.
- For $x > 0$, find the equation of the line passing through $(0,2)$ and $(3,-1)$.
4. **Find the slope $m$ of the line for $x > 0$:**
$$m = \frac{-1 - 2}{3 - 0} = \frac{-3}{3} = -1$$
5. **Find the equation of the line:**
Using point-slope form with point $(0,2)$:
$$y - 2 = -1(x - 0)$$
$$y = -x + 2$$
6. **Write the final piecewise function:**
$$f(x) = \begin{cases} 3 & \text{if } x < 0 \\ -x + 2 & \text{if } x > 0 \end{cases}$$
7. **Note:** The function is undefined at $x=0$ because of open circles at $(0,3)$ and $(0,2)$.
**Final answer:**
$$f(x) = \begin{cases} 3 & x < 0 \\ -x + 2 & x > 0 \end{cases}$$