Find the vertex of the parabola.
\[
h(x)=-x^{2}+4 x-4
\]
\(\boxed{\text{Final Answer: The vertex of the parabola is at the point }(2, 0)}\)
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 :\(\boxed{\text{Final Answer: The vertex of the parabola is at the point }(2, 0)}\)