It looks like you are working on a graphing problem based on the function $f(x) = \frac{1}{4}x + 3$. The problem asks you to:

1. Graph the line of the given function.
2. Draw the vertical line $x = -4$.
3. Graph the point $(-4, f(-4))$.

Let's break this down:

1. Function Graph $f(x) = \frac{1}{4}x + 3$:

   • The slope of the line is $\frac{1}{4}$, meaning that for every increase of 1 unit in $x$, $y$ increases by $\frac{1}{4}$.
   • The y-intercept is 3, so the line crosses the y-axis at $(0, 3)$.

2. Evaluate $f(-4)$: $f(-4) = \frac{1}{4}(-4) + 3 = -1 + 3 = 2$ So, the point $(-4, f(-4))$ is $(-4, 2)$.

3. Plotting the Line $x = -4$:

   • This is a vertical line that passes through $x = -4$ on the graph. It will be a straight vertical line at that point.

4. Plot the Point $(-4, 2)$:

   • Once you've drawn the vertical line, plot the point $(-4, 2)$ on the graph, which lies on the line of the function.

Would you like a more detailed breakdown of any part of this process or further clarification?

Here are five follow-up questions related to this problem:

1. What does the slope $\frac{1}{4}$ tell us about the steepness of the line?
2. How would the graph change if the slope were negative?
3. What if the y-intercept were shifted up or down?
4. How do you interpret the point $(-4, 2)$ in relation to the graph?
5. Can you predict the value of $f(x)$ for other values of $x$, like $x = 0$ or $x = 4$?

Tip: When graphing linear functions, always start by plotting the y-intercept and then use the slope to find additional points.