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We are asked to find the equation of the tangent line to the graph of the function \( y = \sqrt{2x - 3} \) at the point where the abscissa \( x_0 = 2 \).

### Steps to solve:

1. **Find the function's value at \( x = 2 \):**

   First, evaluate the function at \( x_0 = 2 \):

   \[
   y(2) = \sqrt{2(2) - 3} = \sqrt{4 - 3} = \sqrt{1} = 1
   \]

   So, the point of tangency is \( (2, 1) \).

2. **Find the derivative of the function:**

   The derivative of the function \( y = \sqrt{2x - 3} \) is found using the chain rule. Let \( u = 2x - 3 \), so \( y = \sqrt{u} \), and the derivative of \( \sqrt{u} \) is:

   \[
   \frac{dy}{dx} = \frac{1}{2\sqrt{2x - 3}} \cdot 2 = \frac{1}{\sqrt{2x - 3}}
   \]

3. **Evaluate the derivative at \( x = 2 \):**

   Now, we substitute \( x = 2 \) into the derivative to get the slope of the tangent line:

   \[
   \frac{dy}{dx} \Big|_{x=2} = \frac{1}{\sqrt{2(2) - 3}} = \frac{1}{\sqrt{1}} = 1
   \]

   Thus, the slope of the tangent line is \( 1 \).

4. **Write the equation of the tangent line:**

   The general form of the equation of a tangent line is given by:

   \[
   y - y_0 = m(x - x_0)
   \]

   where \( m \) is the slope and \( (x_0, y_0) \) is the point of tangency. Here, \( m = 1 \), \( x_0 = 2 \), and \( y_0 = 1 \). Substituting these values into the equation:

   \[
   y - 1 = 1(x - 2)
   \]

   Simplifying:

   \[
   y = x - 1
   \]

Thus, the equation of the tangent line is \( y = x - 1 \).

Would you like more details or have any questions?

Here are 5 related questions:
1. How can you find the derivative of more complex functions using the chain rule?
2. What is the geometric interpretation of a tangent line to a curve?
3. How do you differentiate functions involving square roots or radicals in general?
4. How would the process change if the function involved higher-order roots, such as cube roots?
5. What happens if you are asked to find the equation of a normal line instead of a tangent line?

**Tip:** The slope of the tangent line at any point on a curve gives you the rate of change of the function at that point.

Записати рівняння дотичної до графіка функції y = √(2x - 3) в точці з абсцисою x₀ = 2.

Learn how to find the equation of the tangent line to the graph of the function y = √(2x - 3) at the point where x₀ = 2. This solution uses derivatives and the chain rule. Suitable for students in Grades 10-12.

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