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To help with the problem shown in your image, I'll guide you through the necessary steps for calculating the tangent and normal lines for the function \( y = x \sqrt{x} - 3x \) at the given point \( (x_0 = 9, y_0 = x_0 \sqrt{x_0} - 3) \).

### Step-by-Step Explanation:

1. **Function and Derivative:**
   The function given is:
   \[
   y = x \sqrt{x} - 3x = x^{3/2} - 3x
   \]
   To find the tangent and normal lines, we first need the derivative of the function \( y \), which will give us the slope of the tangent line.

   Using the power rule:
   \[
   \frac{dy}{dx} = \frac{d}{dx} \left( x^{3/2} \right) - \frac{d}{dx}(3x) = \frac{3}{2}x^{1/2} - 3
   \]

2. **Slope of Tangent Line:**
   At \( x_0 = 9 \), plug in the value of \( x_0 \) into the derivative:
   \[
   \text{slope of tangent} = \frac{3}{2} \cdot 9^{1/2} - 3 = \frac{3}{2} \cdot 3 - 3 = \frac{9}{2} - 3 = \frac{3}{2}
   \]

3. **Equation of Tangent Line:**
   The equation of the tangent line can be found using the point-slope form of a line:
   \[
   y - y_0 = m(x - x_0)
   \]
   Here, \( m = \frac{3}{2} \), \( x_0 = 9 \), and we calculate \( y_0 \) as follows:
   \[
   y_0 = 9 \sqrt{9} - 3(9) = 9 \cdot 3 - 27 = 27 - 27 = 0
   \]
   So the equation becomes:
   \[
   y - 0 = \frac{3}{2}(x - 9) \quad \text{or} \quad y = \frac{3}{2}(x - 9)
   \]

4. **Slope of the Normal Line:**
   The slope of the normal line is the negative reciprocal of the tangent slope:
   \[
   \text{slope of normal} = -\frac{2}{3}
   \]
   Using the same point-slope form for the normal line, with the same point \( (9, 0) \):
   \[
   y - 0 = -\frac{2}{3}(x - 9) \quad \text{or} \quad y = -\frac{2}{3}(x - 9)
   \]

### Final Equations:

- **Tangent line equation**: \( y = \frac{3}{2}(x - 9) \)
- **Normal line equation**: \( y = -\frac{2}{3}(x - 9) \)

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Would you like me to go into more detail on any step?

Here are 5 questions related to this content:
1. How do you find the derivative of a function like \( y = x \sqrt{x} \)?
2. Why do we use the point-slope form for tangent lines?
3. How do you calculate the normal line’s slope?
4. What are the steps to find the equation of a line if given a point and slope?
5. How can you verify that the point \( (9, 0) \) lies on both the tangent and normal lines?

**Tip**: Always check that the point of tangency is correct by substituting it back into the original function.

Calculate the tangent and normal for the function y = x√x - 3x at the point (x₀ = 9, y₀ = x₀√x₀ - 3). What is the algebraic expression for the vertical coordinate in tangent and normal equation in terms of x?

This problem involves finding the tangent and normal lines to the function y = x√x - 3x at the point x₀ = 9. The solution includes differentiation and applying the point-slope form to determine the equations for both the tangent and normal lines.

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