To determine the domain and range of the equation \( xy^2 = 16 \), let's break it down:

### Domain
The domain of a function consists of all possible values of \( x \) that make the equation valid. In this case, the equation \( xy^2 = 16 \) does not impose any restrictions on \( x \). For any real number \( x \), we can find a corresponding value of \( y \) that satisfies the equation. Therefore, the domain of the equation is all real numbers.

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

### Range
The range consists of all possible values of \( y \) that the function can output. To find the range, solve the equation for \( y^2 \):

\[
y^2 = \frac{16}{x}
\]

For \( y^2 \) to be non-negative (since the square of any real number is non-negative), \( x \) must be positive or negative (non-zero). The equation shows that for any non-zero \( x \), \( y^2 \) is a positive number, implying that \( y \) can take any positive or negative real value. Therefore, \( y \) can be any real number except zero.

\[
\text{Range: } (-\infty, 0) \cup (0, \infty)
\]

### Summary
- **Domain:** \( (-\infty, \infty) \)
- **Range:** \( (-\infty, 0) \cup (0, \infty) \)

Would you like more details or have any questions?

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Here are five related questions:
1. What are the domain and range for the equation \( x^2 + y^2 = 16 \)?
2. How would the domain and range change if the equation were \( xy = 16 \)?
3. What is the domain and range of \( y = \sqrt{\frac{16}{x}} \)?
4. How do you determine the domain and range for a function defined implicitly?
5. What happens to the range if the equation \( xy^2 = 16 \) is modified to \( xy^2 = 0 \)?

**Tip:** When solving for the range, consider the possible outputs based on the equation's structure and restrictions.

Explore the domain and range of the equation xy^2 = 16 through detailed analysis. Learn about the unrestricted domain and non-zero range of the equation, highlighting key concepts in equations and functions. Suitable for advanced high school students.

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