To solve this problem, we are looking for the equation of the line tangent to the graph of $f$ at $x = 2$, using the graph of $f'$, which represents the derivative of $f$.

Here’s the process step-by-step:

1. Identify the slope of the tangent line at $x = 2$

The slope of the tangent line to the graph of $f$ at any point is given by the value of $f'(x)$. We are asked to find the slope at $x = 2$.

From the graph of $f'$, at $x = 2$, the value of $f'(2) = 5$ (from the height of the graph at $x = 2$).

Thus, the slope $m$ of the tangent line at $x = 2$ is: $m = f'(2) = 5$

2. Use the point-slope form of the tangent line

We are given the point $(2, 6)$ on the graph of $f$. The equation of a line can be written in point-slope form: $y - y_1 = m(x - x_1)$ where $m$ is the slope and $(x_1, y_1)$ is the point on the line. Substituting $m = 5$ and the point $(2, 6)$, we get: $y - 6 = 5(x - 2)$ Simplifying: $y - 6 = 5x - 10$ $y = 5x - 4$

Thus, the equation of the tangent line is $y = 5x - 4$.

3. Conclusion

The correct answer is B $y = 5x - 4$.

Would you like a deeper explanation, or do you have any follow-up questions?

Here are 5 related questions to further explore the topic:

1. How is the derivative related to the slope of the tangent line?
2. Why do we use the point-slope form for finding the tangent line?
3. What is the significance of $f'$ being positive, negative, or zero at a point?
4. How do you find the equation of a tangent line if the derivative is given as a function, not a graph?
5. Can the tangent line intersect the graph of the function at multiple points?

Tip: Always double-check the values from the graph when calculating the slope and ensure you are reading the derivative correctly for precise results.