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To solve this problem, we need to find the equation of the tangent line to the function \( y = -x^3 + 6x^2 + 2x \) at its inflection point.

### Step 1: Find the inflection point

1. **Calculate the first derivative** \( y'(x) \):

   Given \( y = -x^3 + 6x^2 + 2x \), let's find the first derivative:
   \[
   y'(x) = \frac{d}{dx}(-x^3 + 6x^2 + 2x) = -3x^2 + 12x + 2
   \]

2. **Calculate the second derivative** \( y''(x) \):

   Next, we find the second derivative to determine the inflection point:
   \[
   y''(x) = \frac{d}{dx}(-3x^2 + 12x + 2) = -6x + 12
   \]

3. **Set the second derivative equal to zero to find the inflection point**:
   \[
   -6x + 12 = 0
   \]
   \[
   6x = 12 \implies x = 2
   \]

   The inflection point occurs at \( x = 2 \).

### Step 2: Find the coordinates of the inflection point

Substitute \( x = 2 \) into the original function to find \( y \):

\[
y(2) = -(2)^3 + 6(2)^2 + 2(2)
\]
\[
y(2) = -8 + 24 + 4
\]
\[
y(2) = 20
\]

So, the inflection point is at \( (2, 20) \).

### Step 3: Find the slope of the tangent line at the inflection point

Evaluate the first derivative at \( x = 2 \):

\[
y'(2) = -3(2)^2 + 12(2) + 2
\]
\[
y'(2) = -12 + 24 + 2
\]
\[
y'(2) = 14
\]

The slope of the tangent line at the inflection point is 14.

### Step 4: Equation of the tangent line

Using the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m = 14 \) and \( (x_1, y_1) = (2, 20) \):

\[
y - 20 = 14(x - 2)
\]
\[
y - 20 = 14x - 28
\]
\[
y = 14x - 8
\]

### Conclusion

The equation of the tangent line to the function at the inflection point is \( y = 14x - 8 \).

**Answer**: The correct answer is \( 6x - y = -6 \).

---

Do you have any questions or would you like more details? Here are some related questions:

1. How do you determine the concavity of a function?
2. What is the significance of the inflection point in a graph?
3. How do you find higher-order derivatives for more complex functions?
4. Can a function have more than one inflection point?
5. How do you determine the equation of the normal line at a given point?

**Tip**: Remember, an inflection point occurs where the second derivative changes sign, indicating a change in concavity.

Learn how to find the equation of the tangent line to a function at its inflection point with a detailed AP Calculus BC level solution. This problem involves calculus concepts such as derivatives, tangent lines, and inflection points.

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