1. **State the problem:** We have the parabola $y = x^2 - 6x + 8$ and a horizontal line $PQ$ where $P$ lies on the $y$-axis and both $P$ and $Q$ lie on the parabola. We need to find the coordinates of $Q$. 2. **Find coordinates of $P$:** Since $P$ lies on the $y$-axis, its $x$-coordinate is 0. Substitute $x=0$ into the equation: $$y = 0^2 - 6(0) + 8 = 8$$ So, $P = (0, 8)$. 3. **Find coordinates of $Q$:** Since $PQ$ is horizontal, $Q$ has the same $y$-coordinate as $P$, i.e., $y=8$. Substitute $y=8$ into the parabola equation: $$8 = x^2 - 6x + 8$$ Simplify: $$0 = x^2 - 6x + 8 - 8$$ $$0 = x^2 - 6x$$ Factor: $$x(x - 6) = 0$$ So, $x=0$ or $x=6$. 4. Since $P$ is at $x=0$, $Q$ is the other point at $x=6$. Calculate $Q$'s $y$ to verify: $$y = 6^2 - 6(6) + 8 = 36 - 36 + 8 = 8$$ So, $Q = (6, 8)$. **Final answer:** The coordinates of $Q$ are $(6, 8)$.