Let's analyze the given function:

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

Step 1: Simplify the Function

First, factor the numerator: $f(x) = \frac{(x + 1)(x + 1)}{x + 1} = \frac{(x + 1)^2}{x + 1}$

Now, simplify the function by canceling out the common factor $x + 1$: $f(x) = x + 1, \quad \text{for} \ x \neq -1$

However, since the original function had $x + 1$ in the denominator, there is a hole at $x = -1$.

Step 2: Identify the Key Features of the Graph

1. Domain: The function is undefined at $x = -1$, so the domain is: $\text{Domain: } (-\infty, -1) \cup (-1, \infty)$

2. Vertical Asymptote: There is no vertical asymptote because the denominator cancels out. Instead, there is a hole at $x = -1$.

3. Horizontal Asymptote: Since the simplified form is a linear function $f(x) = x + 1$, there is no horizontal asymptote.

4. Intercepts:

   • x-intercept: Set $f(x) = 0$: $x + 1 = 0 \quad \Rightarrow \quad x = -1$ However, since $x = -1$ is where the hole is located, the function does not cross the x-axis.
   • y-intercept: Set $x = 0$: $f(0) = 0 + 1 = 1$ So, the y-intercept is $(0, 1)$.

5. Behavior Near the Hole: As $x$ approaches -1 from either side, the function approaches $f(-1) = -1 + 1 = 0$, but it doesn't reach 0 due to the hole.

Conclusion:

• Domain: $(-\infty, -1) \cup (-1, \infty)$
• Hole: at $x = -1$
• No vertical or horizontal asymptotes
• Y-intercept: $(0, 1)$

Would you like more details or have any other questions?

Here are some related questions to explore further:

1. How would the graph look with the hole at $x = -1$?
2. What are the implications of having a hole rather than a vertical asymptote?
3. How does the domain of the function affect its graph?
4. How can you identify holes in a function graphically?
5. What other types of discontinuities can occur in rational functions?

Tip: Always factor and simplify rational functions to identify key features like holes and asymptotes!