Let's work through the problem step by step:

Part (b):

The task is to find an equation for the family of linear functions such that $f(4) = 1$. A general linear function can be written as: $f(x) = mx + b$ where $m$ is the slope and $b$ is the y-intercept.

The condition $f(4) = 1$ implies: $m(4) + b = 1$ 4m + b = 1 \tag{1} This equation gives us a relationship between $m$ and $b$ for all linear functions in the family.

We can solve for $b$ in terms of $m$: $b = 1 - 4m$ Thus, the family of linear functions that satisfies the condition $f(4) = 1$ is given by: $f(x) = mx + (1 - 4m)$ or equivalently: $f(x) = mx + 1 - 4m$

Part (c):

In this part, we are asked to find which function belongs to both families of functions. Without more context or a description of the second family, it seems like we are looking for a specific function that satisfies both conditions. This could involve additional constraints, such as a particular slope or intercept.

Sketch:

The image shows several lines with different slopes, all passing through the point $(4, 1)$, which confirms that they satisfy $f(4) = 1$. This is consistent with the equation we derived for part (b). The slopes vary, but all lines must pass through $(4, 1)$.

Would you like more details on any part of the solution? Here are some related questions that might help:

1. How do we determine the slope and y-intercept for a specific member of this family?
2. Can you explain how to plot a line given a slope and y-intercept?
3. How can we confirm that a function belongs to both families?
4. What is the significance of the point $(4, 1)$ in this family of functions?
5. How do different values of $m$ affect the steepness and direction of the line?

Tip: Whenever you derive an equation involving parameters (like $m$ and $b$), try substituting specific values to better understand the family of solutions.