In the given exercise 2.D1.55, you're asked to find the equation of the tangent line to the graph of the function $f$ at the point $(2, f(2))$. The information provided includes:

• $f(2) = 5$
• $f'(2) = -4$

1st Method:

The equation of a line is given by $y = ax + b$, where $a$ is the slope (in this case $f'(2)$) and $b$ is the y-intercept.

1. The slope of the tangent is $f'(2) = -4$.
2. Substituting $(x_0, f(x_0)) = (2, 5)$ into the equation $y = ax + b$: $5 = -4(2) + b$ $5 = -8 + b$ Solving for $b$, we get: $b = 13$ Therefore, the equation of the tangent line is: $y = -4x + 13$

2nd Method:

Using the point-slope formula for the equation of a tangent line: $y = f'(x_0) \cdot (x - x_0) + f(x_0)$ Substituting the values $f'(2) = -4$, $x_0 = 2$, and $f(2) = 5$: $y = -4(x - 2) + 5$ Simplifying: $y = -4x + 8 + 5$ $y = -4x + 13$ Thus, the equation of the tangent line is the same: $y = -4x + 13$

Would you like more details on the methods used or have any questions?

Here are 5 related questions to think about:

1. How do you find the slope of a tangent line using a derivative?
2. What is the significance of $f'(2) = -4$ in this context?
3. How do you apply the point-slope form to different types of functions?
4. Why is it important to find the y-intercept in the first method?
5. Can the second method be generalized to any function at any point?

Tip: The derivative gives the slope of the tangent line to a function at a specific point, which is key to understanding the behavior of the function at that point.