Question: Graph the image of $\triangle XYZ$ after the following glide reflection:
Translation 19 units down
Reflection across the y-axis
Graph shape: triangle $\triangle XYZ$ in the top-left of the coordinate plane, with vertices approximately $X(-5, 12)$, $Y(-11, 9)$, and $Z(-8, 6)$; after translating 19 units down and reflecting across the y-axis, the image would be a congruent triangle in the bottom-left/right depending on reflection, centered far below the x-axis.
1. **State the problem:** We need to find the image of triangle $\triangle XYZ$ with vertices $X(-5, 12)$, $Y(-11, 9)$, and $Z(-8, 6)$ after a glide reflection consisting of a translation 19 units down followed by a reflection across the y-axis.
2. **Recall the transformations:**
- Translation down by 19 units means subtracting 19 from the $y$-coordinate: $ (x, y) \to (x, y - 19) $.
- Reflection across the y-axis means changing the sign of the $x$-coordinate: $ (x, y) \to (-x, y) $.
3. **Apply the translation to each vertex:**
- $X(-5, 12) \to X'(-5, 12 - 19) = (-5, -7)$
- $Y(-11, 9) \to Y'(-11, 9 - 19) = (-11, -10)$
- $Z(-8, 6) \to Z'(-8, 6 - 19) = (-8, -13)$
4. **Apply the reflection across the y-axis to the translated points:**
- $X'(-5, -7) \to X''(5, -7)$
- $Y'(-11, -10) \to Y''(11, -10)$
- $Z'(-8, -13) \to Z''(8, -13)$
5. **Final image vertices:**
- $X''(5, -7)$
- $Y''(11, -10)$
- $Z''(8, -13)$
6. **Explanation:** The glide reflection moves the triangle down 19 units, then flips it over the y-axis, resulting in a congruent triangle located in the bottom-right quadrant of the coordinate plane.
**Answer:** The image of $\triangle XYZ$ after the glide reflection has vertices $X''(5, -7)$, $Y''(11, -10)$, and $Z''(8, -13)$.