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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.

Graph the function f(x) = (1/4)x + 3. Use the line tool to draw the line x = -4 and then use the dot tool to graph the point (-4, f(-4)).

This problem focuses on graphing the linear function f(x) = (1/4)x + 3 and plotting the vertical line x = -4. Learn how to calculate f(-4) and graph the point (-4, 2) using key algebraic concepts. Ideal for Grades 9-10.

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