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To find the equation of the tangent line to the curve \( y = 8\sqrt{x} \) at the point \( (16, 32) \), we will follow these steps:

### 1. **Differentiate the curve:**
We need to compute the derivative of the function \( y = 8\sqrt{x} \) to get the slope of the tangent line at any point.

Rewriting \( y = 8\sqrt{x} \) as \( y = 8x^{1/2} \), we differentiate with respect to \( x \):

\[
\frac{dy}{dx} = 8 \cdot \frac{1}{2}x^{-1/2} = \frac{4}{\sqrt{x}}
\]

### 2. **Find the slope at \( x = 16 \):**
To get the slope of the tangent line at \( x = 16 \), substitute \( x = 16 \) into the derivative:

\[
\text{slope} = \frac{4}{\sqrt{16}} = \frac{4}{4} = 1
\]

So, the slope of the tangent line at \( (16, 32) \) is \( 1 \).

### 3. **Use point-slope form to find the equation of the tangent line:**
The point-slope form of the equation of a line is:

\[
y - y_1 = m(x - x_1)
\]

where \( m \) is the slope, and \( (x_1, y_1) \) is the given point. Here, the slope \( m = 1 \) and the point is \( (16, 32) \).

Substitute these values into the equation:

\[
y - 32 = 1(x - 16)
\]

Simplifying:

\[
y = x + 16
\]

### 4. **Conclusion:**
The equation of the tangent line to the curve \( y = 8\sqrt{x} \) at the point \( (16, 32) \) is:

\[
y = x + 16
\]

Would you like me to generate the sketch of the curve and the tangent line as requested? If yes, I can proceed with that.

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Do you have any questions about the solution? Here are some related questions that could help deepen understanding:

1. How would the tangent line change if we considered a different point on the curve?
2. What is the general formula for the slope of the tangent line to the function \( y = 8\sqrt{x} \) at any point?
3. Can you derive the equation of a tangent line if the curve was \( y = 8x^{3/2} \)?
4. What happens to the tangent line's equation if the point provided was \( (4, 16) \) instead of \( (16, 32) \)?
5. How would you approach finding the tangent line to a curve where \( y \) is a more complex function of \( x \), like \( y = x^2 \ln(x) \)?

**Tip:** Always confirm the accuracy of both the point and the derivative when solving for the equation of a tangent line. Missteps in either can lead to incorrect equations.

Find an equation for the tangent line to the curve y = 8√x at the point (16, 32), then sketch the curve and the tangent line together.

Find the equation of the tangent line to the curve y = 8√x at the point (16, 32) using differentiation. This problem involves finding the slope of the curve at the given point and using the point-slope form of the equation of a line. Suitable for high school calculus students.

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