Let's break down and solve the problems from the image you uploaded.

Part (a)

The problem asks to find an equation for the family of linear functions with slope 2. The general form of a linear equation is: $y = mx + b$ where:

• $m$ is the slope (in this case, $m = 2$),
• $b$ is the y-intercept.

Therefore, the equation for the family of linear functions with slope 2 is: $y = 2x + b$ Here, $b$ is a parameter that can take different values, which would result in different members of this family.

Graphs in Part (a):

Each of the lines drawn represents the equation $y = 2x + b$, with different values of $b$. Each line is parallel because they all have the same slope (2), but they have different intercepts.

Part (b)

The second part asks to find an equation for the family of linear functions such that $f(2) = 1$.

This condition means that the line passes through the point $(2, 1)$. The equation of a line still takes the form $y = mx + b$, but we use the point $(2, 1)$ to find a relationship between $m$ and $b$: $1 = 2m + b$ or $b = 1 - 2m$

Thus, the equation for this family is: $y = mx + (1 - 2m)$ where $m$ is the slope and $b = 1 - 2m$, ensuring the line passes through $(2, 1)$.

Graphs in Part (b):

These lines pass through $(2, 1)$, but they have different slopes, leading to different angles for each line.

Part (c)

The final part asks: What equation belongs to both families?

To solve this, we need an equation that satisfies both conditions:

1. It has a slope of 2 (from part (a)),
2. It passes through the point $(2, 1)$ (from part (b)).

From part (a), the equation is $y = 2x + b$, and from part (b), we know it must pass through $(2, 1)$. Substituting $x = 2$ and $y = 1$ into $y = 2x + b$: $1 = 2(2) + b \quad \Rightarrow \quad 1 = 4 + b \quad \Rightarrow \quad b = -3$

Thus, the equation that belongs to both families is: $y = 2x - 3$

Summary of Answers:

1. Part (a): The family of linear functions is $y = 2x + b$.
2. Part (b): The family of linear functions is $y = mx + (1 - 2m)$.
3. Part (c): The equation that belongs to both families is $y = 2x - 3$.

Do you have any questions about these solutions? Would you like further details on any part?

Here are 5 related questions to further your understanding:

1. What happens if the slope is changed from 2 to another number in Part (a)?
2. Can you find another point that each family of lines in Part (b) would pass through?
3. How would the graph look if the slope was negative in either part?
4. Can you graph a system of two families of lines with different slopes and intersections?
5. How would the equations change if the condition $f(2) = 1$ in Part (b) were modified to $f(3) = 4$?

Tip: In linear equations, parallel lines always have the same slope but differ by their intercepts. This can help you visualize and understand their relationships.