Let's analyze each part of the problem to find any issues:

(a) Domain of the Function

The answer provided for the domain: $(-\infty, -4) \cup (-4, -1) \cup (-1, \infty)$ is correct, as it excludes the values $x = -4$ and $x = -1$ where the denominator $x^2 + 5x + 4$ is zero.

(b) Intercepts

1. x-intercepts: The x-intercepts are the values of $x$ where $f(x) = 0$, which occurs when the numerator $2x^3 - x^2 - 32x + 16 = 0$.
   • Smaller x-value: You entered $(-4, 0)$, but it was marked incorrect.
   • Larger x-value: The answer $(4, 0)$ is correct.
2. y-intercept: The y-intercept occurs when $x = 0$, yielding $f(0) = \frac{16}{4} = 4$. So, the answer $(0, 4)$ is also correct.

(c) Vertical and Slant Asymptotes

1. Vertical Asymptotes: Vertical asymptotes occur where the denominator is zero (and the numerator is not zero at those points), so we have $x = -4$ and $x = -1$, which you entered correctly.
2. Slant Asymptote: A slant asymptote exists if the degree of the numerator is exactly one more than the degree of the denominator. Since the numerator has degree 3 and the denominator has degree 2, this qualifies for a slant asymptote.

To find the slant asymptote, we perform polynomial division of $2x^3 - x^2 - 32x + 16$ by $x^2 + 5x + 4$, which should yield the correct slant asymptote equation.

Would you like me to go through the polynomial division process to find the correct slant asymptote, or do you have specific questions on any part?

Here are some related questions you might find helpful:

1. How do you determine the domain of a rational function?
2. What is the process for finding intercepts of a function?
3. How do you identify vertical asymptotes in a rational function?
4. When does a function have a slant asymptote instead of a horizontal asymptote?
5. Can we have both a horizontal and a slant asymptote?

Tip: When finding a slant asymptote, always check if the degree of the numerator is exactly one more than that of the denominator. This confirms a slant asymptote can exist.