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find the domain, range, vertical and horizontal asymptotes, restrictions, zeroes, y-intercept and x-intercept

Let's analyze the given function, which appears to be:

\[
f(x) = \frac{x^2 + 16}{x^2 - 4}
\]

### 1. **Domain:**
The domain of the function is the set of all real numbers except where the denominator is zero. The denominator is \(x^2 - 4\). Setting this equal to zero gives:

\[
x^2 - 4 = 0
\]

\[
x^2 = 4
\]

\[
x = \pm 2
\]

So, the function is undefined at \(x = 2\) and \(x = -2\). Therefore, the domain is:

\[
\text{Domain: } (-\infty, -2) \cup (-2, 2) \cup (2, \infty)
\]

### 2. **Range:**
To determine the range, analyze the behavior of the function as \(x\) approaches the values that make the denominator zero and as \(x\) approaches infinity.

Since the degree of the polynomial in the numerator is equal to the degree of the polynomial in the denominator, the range will be all real numbers except for the horizontal asymptote. 

We'll analyze the horizontal asymptote in the next step.

### 3. **Vertical Asymptotes:**
Vertical asymptotes occur where the denominator is zero and the numerator is not zero at the same point. From the domain analysis:

\[
x = \pm 2
\]

are the vertical asymptotes.

### 4. **Horizontal Asymptotes:**
For rational functions, the horizontal asymptote depends on the degrees of the polynomials in the numerator and denominator.

- If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is given by the ratio of the leading coefficients.

Here, both the numerator and denominator are of degree 2, and the leading coefficients are both 1.

\[
\text{Horizontal Asymptote: } y = \frac{1}{1} = 1
\]

### 5. **Restrictions:**
The restrictions on the function are the values where the denominator is zero. As previously found:

\[
x \neq \pm 2
\]

### 6. **Zeroes (x-intercepts):**
The zeroes of the function occur where the numerator is zero. Setting the numerator equal to zero:

\[
x^2 + 16 = 0
\]

This equation has no real solutions because \(x^2\) is always non-negative and adding 16 makes it positive. Therefore, there are no real x-intercepts (zeroes).

### 7. **y-intercept:**
The y-intercept occurs where \(x = 0\):

\[
f(0) = \frac{0^2 + 16}{0^2 - 4} = \frac{16}{-4} = -4
\]

So, the y-intercept is:

\[
\text{y-intercept: } (0, -4)
\]

### Summary:
- **Domain:** \((-∞, -2) \cup (-2, 2) \cup (2, ∞)\)
- **Range:** \((-∞, 1) \cup (1, ∞)\)
- **Vertical Asymptotes:** \(x = -2\) and \(x = 2\)
- **Horizontal Asymptote:** \(y = 1\)
- **Restrictions:** \(x \neq -2, 2\)
- **Zeroes:** None (no real x-intercepts)
- **y-intercept:** \( (0, -4) \)

Would you like further details on any of these points, or do you have any questions?

---

Here are 5 related questions you might find interesting:
1. How do we determine slant (oblique) asymptotes for rational functions?
2. What happens to the behavior of the function near the vertical asymptotes?
3. How would the function change if the numerator had a linear term (e.g., \(x^2 + 5x + 16\))?
4. How do the coefficients of the highest degree terms in the numerator and denominator affect the asymptotes?
5. What is the difference between a hole in the graph and a vertical asymptote?

**Tip:** When analyzing a rational function, always check for holes in the graph where the numerator and denominator share a common factor. This can sometimes lead to simplifications that change the function's behavior at specific points.

Explore the analysis of rational functions, including determining domain, range, vertical and horizontal asymptotes, restrictions, zeroes, y-intercept, and x-intercept. Understand how to analyze functions with detailed step-by-step examples.

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