The function given is:

$f(x) = \frac{1}{(x - 1)^2}$

To find the vertical asymptote, we need to determine where the denominator of the function becomes zero because a vertical asymptote occurs when the function tends to infinity due to division by zero.

The denominator is $(x - 1)^2$. Set the denominator equal to zero:

$(x - 1)^2 = 0$

Solve for $x$:

$x - 1 = 0 \quad \Rightarrow \quad x = 1$

Therefore, the vertical asymptote of the function is at:

$x = 1$

Would you like more details or have any questions?

Here are five related questions:

1. What is the behavior of the function near the vertical asymptote $x = 1$?
2. How would the graph of the function look near the asymptote?
3. Are there any horizontal asymptotes in this function?
4. How does the exponent of $(x - 1)^2$ affect the type of asymptote?
5. What happens to the function as $x$ approaches positive and negative infinity?

Tip: To find vertical asymptotes, always focus on when the denominator equals zero while ensuring the numerator is non-zero at those points.