To determine where the function $f(x)$ is discontinuous, let's first rewrite the function clearly:

$f(x) = \frac{x^2 - 4x + 4}{x^2 - 2x}$

Step 1: Factor the numerator and the denominator.

Numerator: $x^2 - 4x + 4$

This is a perfect square trinomial, so we can factor it as: $x^2 - 4x + 4 = (x - 2)^2$

Denominator: $x^2 - 2x$

We can factor this as: $x^2 - 2x = x(x - 2)$

So the function can now be written as: $f(x) = \frac{(x - 2)^2}{x(x - 2)}$

Step 2: Simplify the function.

We can cancel out the common factor $(x - 2)$ from both the numerator and denominator, but we must keep in mind that $x - 2 = 0$ when $x = 2$. This creates a potential issue at $x = 2$.

After cancellation, the simplified function is: $f(x) = \frac{x - 2}{x}, \quad \text{for} \ x \neq 2$

Step 3: Identify points of discontinuity.

1. At $x = 0$:

The denominator $x$ becomes zero, so the function is undefined at $x = 0$.

2. At $x = 2$:

Although we canceled the factor $x - 2$, the original function had a zero in the denominator when $x = 2$, so the function is also undefined at $x = 2$.

Conclusion:

The function is discontinuous at $x = 0$ and $x = 2$.

Thus, the values of $x$ at which the function is discontinuous are:

$\boxed{0, 2}$

Would you like more details or further clarifications?

Here are 5 related questions to deepen your understanding:

1. How do you determine if a function is continuous at a point?
2. What is the significance of factoring when simplifying rational functions?
3. How do you find the domain of a rational function?
4. What is the difference between removable and non-removable discontinuities?
5. How would you graph this function and show its discontinuities?

Tip: Always check for common factors that can cancel out, but remember to account for the points where the denominator equals zero.