To solve the problem, we need to find:

1. The domain of the function $f(x) = \frac{4x^2 - 4x - 1}{2x - 3}$.
2. The equation of the slant asymptote.

Step 1: Domain in interval notation

The domain of a rational function is determined by the values that make the denominator equal to zero, as division by zero is undefined. The denominator of the function is $2x - 3$. Set it equal to zero:

$2x - 3 = 0 \implies x = \frac{3}{2}$

Thus, $x = \frac{3}{2}$ is not in the domain. The domain excludes this value and is expressed in interval notation as:

$(-\infty, \frac{3}{2}) \cup (\frac{3}{2}, \infty)$

Step 2: Slant asymptote

A slant asymptote exists when the degree of the numerator is exactly one higher than the degree of the denominator. In this case, the numerator is quadratic ($4x^2 - 4x - 1$) and the denominator is linear ($2x - 3$).

To find the equation of the slant asymptote, we perform polynomial long division of $4x^2 - 4x - 1$ by $2x - 3$.

Division process:

1. Divide the leading term of the numerator ($4x^2$) by the leading term of the denominator ($2x$) to get $2x$.
2. Multiply $2x$ by $2x - 3$ to get $4x^2 - 6x$.
3. Subtract $4x^2 - 6x$ from the original numerator $4x^2 - 4x - 1$, resulting in $2x - 1$.
4. Divide $2x$ by $2x$ to get $1$.
5. Multiply $1$ by $2x - 3$ to get $2x - 3$.
6. Subtract $2x - 3$ from $2x - 1$, resulting in $2$.

So the quotient is $2x + 1$ and the remainder is 2. The equation of the slant asymptote is given by the quotient, which is:

$y = 2x + 1$

Final answers:

• The domain in interval notation is: $(-\infty, \frac{3}{2}) \cup (\frac{3}{2}, \infty)$
• The equation of the slant asymptote is: $y = 2x + 1$

Would you like further clarification on any of these steps?

Here are five related questions you might explore:

1. How do you find vertical asymptotes for rational functions?
2. What is the difference between a horizontal and slant asymptote?
3. How can we determine whether a rational function has a hole?
4. What happens