[Algebra]
Problem: The equations $y=a(x-2)(x+4)$ and $y=2(x-h)^{2}+k$ represent the same parabola. What are the values of $a, h$ and $k$ ?

Solution: We expand the right sides of the two equations, collecting like terms in each case:

$$
\begin{aligned}
& y=a(x-2)(x+4)=a\left(x^{2}+2 x-8\right)=a x^{2}+2 a x-8 a \\
& y=2(x-h)^{2}+k=2\left(x^{2}-2 h x+h^{2}\right)+k=2 x^{2}-4 h x+\left(2 h^{2}+k\right)
\end{aligned}
$$

Since these two equations represent the same parabola, then the corresponding coefficients must be equal. That is, $a=2$ and $2 a=-4 h$ and $-8 a=2 h^{2}+k$.

Since $a=2$ and $2 a=-4 h$, then $4=-4 h$ and so $h=-1$.

Since $-8 a=2 h^{2}+k$ and $a=2$ and $h=-1$, then $-16=2+k$ and so $k=-18$.

Thus, $a=2, h=-1$, and $k=-18$.
From the equation $y=a(x-2)(x+4)$, we can find the axis of symmetry by calculating the midpoint of the $x$-intercepts.

Since the $x$-intercepts are 2 and -4 , the axis of symmetry is at $x=\frac{1}{2}(2+(-4))=-1$.

Since the vertex of the parabola lies on the axis of symmetry, then the $x$-coordinate of the vertex is -1 .

To find the $y$-coordinate of the vertex, we substitute $x=-1$ back into the equation $y=a(x-2)(x+4)$ to obtain $y=a(-1-2)(-1+4)=-9 a$.

Thus, the vertex of the parabola is $(-1,-9 a)$.

Since the second equation for the same parabola is in vertex form, $y=2(x-h)^{2}+k$, we can see that the vertex is at $(h, k)$ and $a=2$.

Since $a=2$, the vertex has coordinates $(-1,-18)$, which means that $h=-1$ and $k=-18$. Thus, $a=2, h=-1$ and $k=-18$.

Final answer: $2,-1,-18$