The given equation is:

$y = \frac{4x^2 + 4x + 1}{x^2 - 2x}$

Step 1: Domain of the function

The domain of the function is determined by the values of $x$ for which the denominator is non-zero. The denominator is:

$x^2 - 2x = x(x - 2)$

This is equal to zero when:

$x(x - 2) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 2$

So, the function is undefined at $x = 0$ and $x = 2$. Therefore, the domain is all real numbers except these points:

$\text{Domain: } x \in (-\infty, 0) \cup (0, 2) \cup (2, \infty)$

Step 2: Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (and the numerator is non-zero at the same points). From Step 1, we know that the denominator is zero at $x = 0$ and $x = 2$. Therefore, the vertical asymptotes are at:

$x = 0 \quad \text{and} \quad x = 2$

Step 3: Horizontal Asymptotes

To find the horizontal asymptote, we look at the degrees of the numerator and the denominator. Both the numerator $4x^2 + 4x + 1$ and the denominator $x^2 - 2x$ are degree 2 polynomials. When the degrees are the same, the horizontal asymptote is given by the ratio of the leading coefficients:

$\frac{\text{Leading coefficient of numerator}}{\text{Leading coefficient of denominator}} = \frac{4}{1} = 4$

Thus, the horizontal asymptote is:

$y = 4$

Final Summary:

• Domain: $(-\infty, 0) \cup (0, 2) \cup (2, \infty)$
• Vertical asymptotes: $x = 0$ and $x = 2$
• Horizontal asymptote: $y = 4$

Let me know if you'd like more details or have any questions.

Here are some related questions you might find helpful:

1. How do we find slant (oblique) asymptotes if the degree of the numerator is one higher than the denominator?
2. How do vertical asymptotes relate to the factorization of the denominator?
3. What happens to the function as $x \to 0$ and $x \to 2$?
4. Can the function ever cross its horizontal asymptote?
5. How does the numerator affect the behavior near the vertical asymptotes?

Tip: Always check the factorization of both the numerator and denominator to simplify rational functions where possible.