Find an equation of the tangent plane to the given surface at the specified point.
z = 4(x − 1)2 + 3(y + 3)2 + 5,   (2, −1, 21)

To find the equation of the tangent plane to the surface \( z = 4(x - 1)^2 + 3(y + 3)^2 + 5 \) at the point \( (2, -1, 21) \), follow these steps:

### Step 1: Verify the given point
We first verify that the point \( (2, -1, 21) \) lies on the surface by substituting \( x = 2 \) and \( y = -1 \) into the given equation and checking if we get \( z = 21 \).

\[
z = 4(2 - 1)^2 + 3(-1 + 3)^2 + 5
\]
\[
z = 4(1)^2 + 3(2)^2 + 5 = 4 + 12 + 5 = 21
\]

So, the point \( (2, -1, 21) \) lies on the surface.

### Step 2: Compute the partial derivatives
To find the tangent plane, we need the partial derivatives of \( z \) with respect to \( x \) and \( y \).

\[
\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \left( 4(x - 1)^2 + 3(y + 3)^2 + 5 \right) = 8(x - 1)
\]
\[
\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} \left( 4(x - 1)^2 + 3(y + 3)^2 + 5 \right) = 6(y + 3)
\]

### Step 3: Evaluate the partial derivatives at \( (x, y) = (2, -1) \)
Now, substitute \( x = 2 \) and \( y = -1 \) into the partial derivatives to get the slopes at the point of tangency:

\[
\frac{\partial z}{\partial x} (2, -1) = 8(2 - 1) = 8
\]
\[
\frac{\partial z}{\partial y} (2, -1) = 6(-1 + 3) = 12
\]

### Step 4: Write the equation of the tangent plane
The equation of the tangent plane at \( (x_0, y_0, z_0) \) is given by:

\[
z - z_0 = \frac{\partial z}{\partial x} (x - x_0) + \frac{\partial z}{\partial y} (y - y_0)
\]

Substitute \( (x_0, y_0, z_0) = (2, -1, 21) \), \( \frac{\partial z}{\partial x} = 8 \), and \( \frac{\partial z}{\partial y} = 12 \) into the equation:

\[
z - 21 = 8(x - 2) + 12(y + 1)
\]

Simplifying this:

\[
z - 21 = 8(x - 2) + 12(y + 1)
\]
\[
z - 21 = 8x - 16 + 12y + 12
\]
\[
z = 8x + 12y - 4 + 21
\]
\[
z = 8x + 12y + 17
\]

Thus, the equation of the tangent plane is:

\[
z = 8x + 12y + 17
\]

### Do you want more details or have any questions?

Here are some related questions to deepen your understanding:
1. How do you derive the general form of the tangent plane for a surface \( z = f(x, y) \)?
2. Can you find the equation of the normal line to the surface at the same point?
3. What is the geometric interpretation of the gradient vector in this context?
4. How do second-order partial derivatives relate to the curvature of the surface?
5. Can you find the tangent planes to this surface at other points?

**Tip**: When working with surfaces, always verify the point lies on the surface before proceeding with the tangent plane equation.

Find an equation of the tangent plane to the given surface at the specified point. z = 4(x − 1)^2 + 3(y + 3)^2 + 5, (2, −1, 21)

Find the equation of the tangent plane to the surface z = 4(x − 1)^2 + 3(y + 3)^2 + 5 at the point (2, −1, 21). This solution explains the process step-by-step, including partial derivatives and the tangent plane equation. Suitable for students studying multivariable calculus.

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