Find the vertex of the parabola. \[ h(x)=-x^{2}+4 x-4 \]

Step 1 ：The given function is \(h(x)=-x^{2}+4 x-4\).

Step 2 ：The vertex of a parabola given in the form \(f(x) = ax^2 + bx + c\) is at the point \((-b/2a, f(-b/2a))\).

Step 3 ：In this case, \(a = -1\), \(b = 4\), and \(c = -4\).

Step 4 ：So, the x-coordinate of the vertex is \(-b/2a = -4/(-2) = 2\).

Step 5 ：We can substitute \(x = 2\) into the equation to find the y-coordinate of the vertex.

Step 6 ：Substituting \(x = 2\) into the equation gives \(y = 0\).

Step 7 ：\(\boxed{\text{Final Answer: The vertex of the parabola is at the point }(2, 0)}\)