Let's address the given problem step by step.

The function provided is:

$f(x) = \frac{x^2 - 2x + 6}{2x - 1}$

We are tasked with: a. Finding the domain of $f(x)$.
b. Evaluating $f(0)$ and $f(-4)$.

Part a: Finding the Domain of $f(x)$

The domain of a rational function is all real numbers except where the denominator equals zero, because division by zero is undefined.

We set the denominator equal to zero to find the values of $x$ to exclude:

$2x - 1 = 0$ $x = \frac{1}{2}$

Thus, the function is undefined at $x = \frac{1}{2}$.

Domain: All real numbers except $x = \frac{1}{2}$, or in interval notation: $\text{Domain: } (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

Part b: Evaluating $f(0)$ and $f(-4)$

1. Evaluating $f(0)$:

Substitute $x = 0$ into the function:

$f(0) = \frac{0^2 - 2(0) + 6}{2(0) - 1} = \frac{6}{-1} = -6$

So, $f(0) = -6$.

2. Evaluating $f(-4)$:

Substitute $x = -4$ into the function:

$f(-4) = \frac{(-4)^2 - 2(-4) + 6}{2(-4) - 1} = \frac{16 + 8 + 6}{-8 - 1} = \frac{30}{-9} = -\frac{10}{3}$

So, $f(-4) = -\frac{10}{3}$.

Summary:

• Domain: $(-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$
• $f(0) = -6$
• $f(-4) = -\frac{10}{3}$

Let me know if you need further details or have any questions! Here are five related questions to expand on this topic:

1. What is the range of the function $f(x)$?
2. Can you simplify the function $f(x)$ further by factoring the numerator?
3. How would you determine vertical and horizontal asymptotes for $f(x)$?
4. What happens to $f(x)$ as $x$ approaches infinity?
5. How would you graph $f(x)$?

Tip: Always check for undefined points in rational functions by looking at where the denominator equals zero!