1. **Problem Statement:** Reflect the triangle with vertices at $A(0,-4)$, $B(2,-4)$, and $C(2,-6)$ over the vertical line $p$ given by $x = -2$. 2. **Reflection Formula:** For a point $(x,y)$ reflected over the vertical line $x = a$, the reflected point $(x',y')$ is given by: $$x' = 2a - x, \quad y' = y$$ This means the $x$-coordinate is mirrored across the line $x=a$, while the $y$-coordinate remains the same. 3. **Apply the formula to each vertex:** - For $A(0,-4)$: $$x'_A = 2(-2) - 0 = -4, \quad y'_A = -4$$ So, $A' = (-4,-4)$. - For $B(2,-4)$: $$x'_B = 2(-2) - 2 = -6, \quad y'_B = -4$$ So, $B' = (-6,-4)$. - For $C(2,-6)$: $$x'_C = 2(-2) - 2 = -6, \quad y'_C = -6$$ So, $C' = (-6,-6)$. 4. **Result:** The reflected triangle has vertices at $(-4,-4)$, $(-6,-4)$, and $(-6,-6)$. This reflection flips the triangle horizontally across the line $x=-2$ while keeping the vertical positions unchanged.