what is the doman and range of:
y=square root of ((2x-3)/(x-2))

To find the **domain** and **range** of the function \( y = \sqrt{\frac{2x - 3}{x - 2}} \), we must first address the following constraints based on the properties of the square root and rational functions:

### 1. Domain:
The expression inside the square root, \( \frac{2x - 3}{x - 2} \), must be **non-negative** (i.e., \( \geq 0 \)) because the square root is only defined for non-negative values in the real number system. Additionally, the denominator \( x - 2 \) cannot be zero, as division by zero is undefined.

Thus, we need to solve for \( \frac{2x - 3}{x - 2} \geq 0 \), while ensuring that \( x \neq 2 \).

- First, find the critical points by solving:
  - \( 2x - 3 = 0 \) ⇒ \( x = \frac{3}{2} \)
  - \( x - 2 = 0 \) ⇒ \( x = 2 \)

Now, determine where the rational function is non-negative by analyzing intervals around these points:
- Consider the intervals \( (-\infty, \frac{3}{2}) \), \( (\frac{3}{2}, 2) \), and \( (2, \infty) \).

Evaluate the sign of \( \frac{2x - 3}{x - 2} \) in each interval:
- For \( x \in (-\infty, \frac{3}{2}) \), both \( 2x - 3 \) and \( x - 2 \) are negative, so their quotient is positive.
- For \( x \in (\frac{3}{2}, 2) \), \( 2x - 3 \) is positive and \( x - 2 \) is negative, so their quotient is negative.
- For \( x \in (2, \infty) \), both \( 2x - 3 \) and \( x - 2 \) are positive, so their quotient is positive.

Thus, \( \frac{2x - 3}{x - 2} \geq 0 \) holds in the intervals:
- \( (-\infty, \frac{3}{2}] \cup (2, \infty) \), excluding \( x = 2 \).

Therefore, the **domain** is:
\[ (-\infty, \frac{3}{2}] \cup (2, \infty) \]

### 2. Range:
Since the square root function outputs only non-negative values, we must analyze the range based on the domain.

- When \( x = \frac{3}{2} \), \( \frac{2x - 3}{x - 2} = 0 \), so \( y = \sqrt{0} = 0 \).
- As \( x \to -\infty \), \( \frac{2x - 3}{x - 2} \to 2 \), so \( y \to \sqrt{2} \).
- As \( x \to 2^+ \), \( \frac{2x - 3}{x - 2} \to +\infty \), so \( y \to +\infty \).

Thus, the **range** is:
\[ [0, \infty) \]

### Final Answer:
- **Domain**: \( (-\infty, \frac{3}{2}] \cup (2, \infty) \)
- **Range**: \( [0, \infty) \)

Would you like further details or have any questions?

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Here are 5 related questions to explore this concept further:
1. What is the domain of \( y = \sqrt{\frac{x^2 - 1}{x + 1}} \)?
2. How would the domain change if the square root was replaced with a cube root?
3. What happens to the domain if the function is instead \( y = \frac{2x - 3}{x - 2} \)?
4. How do you determine the range of a rational function without a square root?
5. How does the presence of a square root affect the behavior of functions?

**Tip:** Always check for points where the denominator becomes zero in rational functions—these are critical for identifying restrictions in the domain!

what is the domain and range of: y = square root of ((2x - 3) / (x - 2))

Explore how to determine the domain and range of the function y = sqrt((2x - 3) / (x - 2)). This step-by-step solution involves analyzing inequalities, identifying restrictions from the square root function, and performing sign analysis on a rational function. Ideal for students in Grades 10-12.

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