Question: Write an inequality for the function graphed above
Example: y < a|x - h| + k
y = a|x - h| + k
< and > dashed
≤ and ≥ solid
Graph shape and math: An inverted V-shaped absolute value graph with solid boundary, vertex at approximately (3, 1), passing through approximately (-2, 0) and (8, 0), and the region shaded above the graph; position_hint: centeranswer
1. **State the problem:**
We need to write an inequality for an absolute value function graphed as an inverted V with vertex at approximately $$(3, 1)$$, passing through points $$( -2, 0 )$$ and $$( 8, 0 )$$, with the region shaded above the graph and a solid boundary.
2. **Identify the function form:**
The general form of an absolute value function is:
$$y = a|x - h| + k$$
where $$(h, k)$$ is the vertex.
3. **Use the vertex:**
Given vertex $$(3, 1)$$, the function is:
$$y = a|x - 3| + 1$$
4. **Find the slope $a$ using a point on the graph:**
Use point $$( -2, 0 )$$:
$$0 = a| -2 - 3| + 1$$
$$0 = a| -5| + 1$$
$$0 = 5a + 1$$
$$5a = -1$$
$$a = -\frac{1}{5}$$
5. **Write the function:**
$$y = -\frac{1}{5}|x - 3| + 1$$
6. **Determine inequality direction and boundary type:**
- The graph is solid boundary, so inequality is $$\geq$$ or $$\leq$$.
- The region shaded is above the graph, so:
$$y \geq -\frac{1}{5}|x - 3| + 1$$
7. **Final inequality:**
$$\boxed{y \geq -\frac{1}{5}|x - 3| + 1}$$
This inequality represents the shaded region above the inverted V-shaped absolute value graph with vertex at $$(3, 1)$$.