[Geometry]
Problem: The parabola $y=x^{2}-2 x+4$ is translated $p$ units to the right and $q$ units down. The $x$-intercepts of the resulting parabola are 3 and 5 . What are the values of $p$ and $q$ ?

Solution: Completing the square on the original parabola, we obtain

$$
y=x^{2}-2 x+4=x^{2}-2 x+1-1+4=(x-1)^{2}+3
$$

Therefore, the vertex of the original parabola is $(1,3)$.

Since the new parabola is a translation of the original parabola and has $x$-intercepts 3 and 5 , then its equation is $y=1(x-3)(x-5)=x^{2}-8 x+15$.

Completing the square here, we obtain

$$
y=x^{2}-8 x+15=x^{2}-8 x+16-16+15=(x-4)^{2}-1
$$

Therefore, the vertex of the new parabola is $(4,-1)$.

Thus, the point $(1,3)$ is translated $p$ units to the right and $q$ units down to reach $(4,-1)$, so $p=3$ and $q=4$.

Final answer: 3,4