To make the function $f(x)$ continuous for all real numbers, including $x = 2$, we need to determine the appropriate value of $f(2)$.

1. Simplify the Function

First, simplify the expression for $f(x)$ when $x \neq 2$:

$f(x) = \frac{x^2 - 6x + 8}{x - 2}$

Factor the quadratic expression in the numerator:

$x^2 - 6x + 8 = (x - 2)(x - 4)$

So the function simplifies to:

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

For $x \neq 2$, the factor $(x - 2)$ cancels out, and we are left with:

$f(x) = x - 4 \quad \text{for } x \neq 2$

2. Determine the Limit as $x$ Approaches 2

To ensure continuity at $x = 2$, we need to find the limit of $f(x)$ as $x$ approaches 2:

$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x - 4) = 2 - 4 = -2$

3. Assign the Value of $f(2)$

For $f(x)$ to be continuous at $x = 2$, the value $f(2)$ must be equal to the limit found above:

$f(2) = -2$

Conclusion

To make $f(x)$ continuous for all real numbers, we must assign:

$f(2) = -2$

This ensures that the function is continuous at $x = 2$ and everywhere else.

Would you like further details or have any questions? Here are some related questions:

1. What is the definition of a continuous function?
2. How do you determine if a piecewise function is continuous at a specific point?
3. Can a function have a removable discontinuity? How does that relate to this problem?
4. What happens if we choose a different value for $f(2)$?
5. How do you find limits involving piecewise functions?

Tip: When a function has a potential discontinuity due to division by zero, try simplifying the function and checking the limit at the problematic point.