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The problem asks to find the derivative \( f'(2) \), given that the equation of the tangent line at \( (2, 6) \) to the graph of the function \( f \) is \( y = -3x + 14 \).

### Solution:
The derivative \( f'(x) \) at a particular point \( x \) represents the slope of the tangent line at that point. In this case, the equation of the tangent line is given as \( y = -3x + 14 \).

This is a linear equation, and the slope of this line is the coefficient of \( x \), which is \( -3 \). Since the tangent line is the line that touches the curve at the point \( (2, 6) \), the slope of the tangent line at \( x = 2 \) is \( f'(2) \).

Therefore, the value of \( f'(2) \) is:
\[
f'(2) = -3
\]

Would you like more details on the steps, or do you have any further questions?

### Related Questions:
1. What is the geometric interpretation of a tangent line to a curve?
2. How do you compute the derivative of a non-linear function at a point?
3. How can the derivative be interpreted as the rate of change of a function?
4. If given the function itself, how would you find the equation of the tangent line?
5. What happens when the tangent line is horizontal? What does that imply about the derivative?

**Tip:** The slope of the tangent line to a function at a point gives the instantaneous rate of change of the function at that point.

An equation of the tangent line to the graph of a function f at (2, 6) is y = -3x + 14. What is f'(2)?

Learn how to find the derivative f'(2) from the given tangent line equation y = -3x + 14. This problem involves concepts from calculus, including derivatives and tangent lines. Suitable for students in Grades 11-12.

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