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To find the equation of the tangent plane to the surface \( z = (x + 2)^2 - 2(y - 1)^2 - 5 \) at the point \( (2, 3, 3) \), we need to follow these steps:

1. **Calculate the Partial Derivatives** of \( z \) with respect to \( x \) and \( y \).
2. **Evaluate the Partial Derivatives** at the point \( (2, 3) \).
3. **Use the Tangent Plane Formula**.

The formula for the tangent plane to a surface \( z = f(x, y) \) at a point \( (x_0, y_0, z_0) \) is:
\[
z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)
\]

### Step 1: Partial Derivatives
Given:
\[
z = (x + 2)^2 - 2(y - 1)^2 - 5
\]

1. **Partial derivative with respect to \( x \):**
   \[
   f_x = \frac{\partial}{\partial x} \left( (x + 2)^2 - 2(y - 1)^2 - 5 \right) = 2(x + 2)
   \]

2. **Partial derivative with respect to \( y \):**
   \[
   f_y = \frac{\partial}{\partial y} \left( (x + 2)^2 - 2(y - 1)^2 - 5 \right) = -4(y - 1)
   \]

### Step 2: Evaluate at the Point \( (2, 3) \)
Now, substitute \( x = 2 \) and \( y = 3 \) into the partial derivatives:

- \( f_x(2, 3) = 2(2 + 2) = 2 \times 4 = 8 \)
- \( f_y(2, 3) = -4(3 - 1) = -4 \times 2 = -8 \)

### Step 3: Tangent Plane Equation
The point \( (x_0, y_0, z_0) \) is \( (2, 3, 3) \), so we substitute into the tangent plane formula:

\[
z - 3 = 8(x - 2) - 8(y - 3)
\]

Expanding and simplifying:

\[
z - 3 = 8x - 16 - 8y + 24
\]
\[
z = 8x - 8y + 11
\]

### Final Answer
The equation of the tangent plane is:
\[
z = 8x - 8y + 11
\]

Would you like further details on any step, or have additional questions?

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### Related Questions
1. How do you find the equation of a tangent plane given a function of three variables?
2. What is the geometric interpretation of the gradient in the context of tangent planes?
3. How would the equation of the tangent plane change if the surface equation had an additional term in \( z \)?
4. Can the method of partial derivatives be used to find tangent planes for implicit surfaces?
5. What role do the partial derivatives play in determining the direction of the tangent plane?

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### Tip
For functions with more variables, the gradient vector provides a quick way to find the normal vector for the tangent plane at a given point.

Find the equation of the tangent plane to the surface z = (x + 2)^2 - 2(y - 1)^2 - 5 at the point (2,3,3).

This page provides a step-by-step solution for finding the equation of the tangent plane to the surface z = (x + 2)^2 - 2(y - 1)^2 - 5 at the point (2,3,3). Key concepts include partial derivatives and the tangent plane formula, suitable for college-level calculus students.

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