Explain how the graph of the function $f(x)=\frac{1}{(x-2)^{2}}$ can be obtained from the graph of $y=\frac{1}{x^{2}}$. Then graph $f$ and give the (a) domain and (b) range. Determine the largest open intervals of the domain over which the function is (c) increasing or (d) decreasing.

Step 1 ：The function \(f(x)=\frac{1}{(x-2)^{2}}\) can be obtained from the function \(y=\frac{1}{x^{2}}\) by replacing \(x\) with \(x-2\). This is a horizontal shift of the graph of \(y=\frac{1}{x^{2}}\) to the right by 2 units.

Step 2 ：To graph \(f(x)=\frac{1}{(x-2)^{2}}\), we plot the points and draw the curve. The graph is a parabola opening upwards, with the vertex at \(x=2\).

Step 3 ：(a) The domain of \(f(x)=\frac{1}{(x-2)^{2}}\) is all real numbers except \(x=2\), because the denominator cannot be zero. So, the domain is \(\boxed{(-\infty,2)\cup(2,\infty)}\).

Step 4 ：(b) Note that \(f(x)=\frac{1}{(x-2)^{2}} >0\) for all \(x\neq2\). That is, the range of \(f\) must only include positive numbers. Conversely, if \(a\) is a positive number, then \[f\left(2+\frac{1}{\sqrt{a}}\right)=\frac{1}{(2+1/\sqrt{a}-2)^2} = a,\]so \(a\) is indeed in the range of \(f\). Thus, the range of \(f\) is the set of all positive real numbers; in interval notation, that's \(\boxed{(0,\infty)}\).

Step 5 ：(c) The function \(f(x)=\frac{1}{(x-2)^{2}}\) is always decreasing on the interval \((-\infty,2)\) and always increasing on the interval \((2,\infty)\). This is because the function is a downward-opening parabola, and the vertex is at \(x=2\). So, the function decreases as \(x\) approaches 2 from the left, and increases as \(x\) approaches 2 from the right.

Step 6 ：(d) As mentioned in part (c), the function \(f(x)=\frac{1}{(x-2)^{2}}\) is always decreasing on the interval \((-\infty,2)\) and always increasing on the interval \((2,\infty)\).