https://file.aiquickdraw.com/mathbot/file/1726796173231ir64ypox.jpg

Let's go step by step to solve this.

### Part a: Complete the table

The given functions are:
- \( f(x) = 2x + 1 \)
- \( g(x) = -f(x) = -(2x + 1) = -2x - 1 \)
- \( h(x) = f(-x) = 2(-x) + 1 = -2x + 1 \)

Now, calculate the values for \( x = -4, -2, 0, 2, 4 \):

1. **For \( x = -4 \):**
   - \( f(-4) = 2(-4) + 1 = -8 + 1 = -7 \)
   - \( g(-4) = -2(-4) - 1 = 8 - 1 = 7 \)
   - \( h(-4) = -2(-(-4)) + 1 = -8 + 1 = -7 \)

2. **For \( x = -2 \):**
   - \( f(-2) = 2(-2) + 1 = -4 + 1 = -3 \)
   - \( g(-2) = -2(-2) - 1 = 4 - 1 = 3 \)
   - \( h(-2) = -2(-(-2)) + 1 = -4 + 1 = -3 \)

3. **For \( x = 0 \):**
   - \( f(0) = 2(0) + 1 = 1 \)
   - \( g(0) = -2(0) - 1 = -1 \)
   - \( h(0) = -2(0) + 1 = 1 \)

4. **For \( x = 2 \):**
   - \( f(2) = 2(2) + 1 = 4 + 1 = 5 \)
   - \( g(2) = -2(2) - 1 = -4 - 1 = -5 \)
   - \( h(2) = -2(2) + 1 = -4 + 1 = -3 \)

5. **For \( x = 4 \):**
   - \( f(4) = 2(4) + 1 = 8 + 1 = 9 \)
   - \( g(4) = -2(4) - 1 = -8 - 1 = -9 \)
   - \( h(4) = -2(4) + 1 = -8 + 1 = -7 \)

Here’s the completed table:

\[
\begin{array}{|c|c|c|c|}
\hline
x & f(x) = 2x + 1 & g(x) = -f(x) & h(x) = f(-x) \\
\hline
-4 & -7 & 7 & -7 \\
-2 & -3 & 3 & -3 \\
0 & 1 & -1 & 1 \\
2 & 5 & -5 & -3 \\
4 & 9 & -9 & -7 \\
\hline
\end{array}
\]

### Part b: Sketch the graphs
The graphs of \( f(x) \), \( g(x) \), and \( h(x) \) are straight lines:
- \( f(x) = 2x + 1 \) is a straight line with slope 2 and y-intercept at 1.
- \( g(x) = -2x - 1 \) is a straight line with slope -2 and y-intercept at -1.
- \( h(x) = -2x + 1 \) is a straight line with slope -2 and y-intercept at 1.

These lines would be sketched together on the same set of axes.

### Part c: Explain transformations and list invariant points
- \( g(x) \) is a reflection of \( f(x) \) in the x-axis. This transformation flips the graph of \( f(x) \) over the x-axis, changing the sign of all the \( y \)-coordinates.
- \( h(x) \) is a reflection of \( f(x) \) in the y-axis. This transformation flips the graph of \( f(x) \) horizontally, changing the sign of the \( x \)-coordinates but keeping the slopes the same.

**Invariant Points**: Invariant points are those that remain unchanged after a transformation. In this case, no points remain unchanged in the reflection over the y-axis or x-axis for the given functions.

### Part d: How is each function related to \( f(x) = 2x + 1 \)?
- \( g(x) \) is the reflection of \( f(x) \) over the x-axis.
- \( h(x) \) is the reflection of \( f(x) \) over the y-axis.

Would you like me to go into more details, or do you have any further questions?

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Here are 5 related questions to explore further:
1. What is the effect of multiplying a function by a negative value on its graph?
2. How does reflecting a function over the y-axis affect its domain and range?
3. Can you find the intersection points of \( f(x) \) and \( g(x) \)?
4. How would the graph change if \( f(x) \) was \( f(x) = 2x - 1 \) instead of \( 2x + 1 \)?
5. What happens to the invariant points when the function is shifted vertically?

**Tip**: When sketching transformations, it's helpful to plot a few key points (like the y-intercept and some other easy-to-calculate points) before drawing the full graph.

Copy and complete the table of values for the given functions f(x) = 2x + 1, g(x) = -f(x), h(x) = f(-x). Sketch the graphs of f(x), g(x), and h(x) on the same set of coordinate axes. Explain how the points on the graphs of g(x) and h(x) relate to the transformation of the function f(x) = 2x + 1. List any invariant points. How is each function related to the graph of f(x) = 2x + 1?

This lesson covers how to complete a table of values for functions f(x) = 2x + 1, g(x) = -f(x), and h(x) = f(-x), followed by sketching their graphs. We explore the effects of reflections over the x-axis and y-axis and identify any invariant points.

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