Question: Graph the image of rhombus $EFGH$ after the following sequence of transformations:
Reflection across the line $y = -1$
Translation 14 units left and 1 unit up
Rhombus $EFGH$ is in the top-right of the coordinate plane; the vertices appear to be $E(2,12)$, $F(8,10)$, $G(10,4)$, and $H(4,6)$.
1. **State the problem:** We need to find the image of rhombus $EFGH$ after two transformations: a reflection across the line $y = -1$ and then a translation 14 units left and 1 unit up.
2. **Original vertices:**
- $E(2,12)$
- $F(8,10)$
- $G(10,4)$
- $H(4,6)$
3. **Reflection across the line $y = -1$:**
The reflection of a point $(x,y)$ across the line $y = k$ is given by $(x, 2k - y)$.
Here, $k = -1$, so the reflected point is $(x, 2(-1) - y) = (x, -2 - y)$.
Calculate reflected vertices:
- $E': (2, -2 - 12) = (2, -14)$
- $F': (8, -2 - 10) = (8, -12)$
- $G': (10, -2 - 4) = (10, -6)$
- $H': (4, -2 - 6) = (4, -8)$
4. **Translation 14 units left and 1 unit up:**
Translation moves a point $(x,y)$ to $(x - 14, y + 1)$.
Calculate translated vertices:
- $E'': (2 - 14, -14 + 1) = (-12, -13)$
- $F'': (8 - 14, -12 + 1) = (-6, -11)$
- $G'': (10 - 14, -6 + 1) = (-4, -5)$
- $H'': (4 - 14, -8 + 1) = (-10, -7)$
5. **Final answer:** The image of rhombus $EFGH$ after the transformations has vertices:
$$E''(-12, -13), F''(-6, -11), G''(-4, -5), H''(-10, -7)$$