Question 3 Consider $$ f(x)=-3 x+6 $$. Drag points that can represent the possible inputs and outputs of each function into the correct bin. $$ f(x) $$ $$ f^{-1}(x) $$ $$ (-9,5) $$ $$ (-1,9) $$ $$ (0,2) $$ $$ (0,6) $$ $$ (-2,0) $$ $$ (3,-3) $$ $$ (3,1) $$ $$ (4,-6) $$ (6, 0) (8,-1)

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To determine which points can represent the possible inputs and outputs of the function $$ f(x) = -3x + 6 $$ and its inverse $$ f^{-1}(x) $$, we need to analyze both functions.

### Step 1: Analyze the function $$ f(x) $$

The function $$ f(x) = -3x + 6 $$ is a linear function. We can find the output for various inputs by substituting values of $$ x $$ into the function.

### Step 2: Calculate outputs for given inputs

1. For $$ x = -9 $$:
   $$
   f(-9) = -3(-9) + 6 = 27 + 6 = 33
   $$
2. For $$ x = -1 $$:
   $$
   f(-1) = -3(-1) + 6 = 3 + 6 = 9
   $$
3. For $$ x = 0 $$:
   $$
   f(0) = -3(0) + 6 = 0 + 6 = 6
   $$
4. For $$ x = -2 $$:
   $$
   f(-2) = -3(-2) + 6 = 6 + 6 = 12
   $$
5. For $$ x = 3 $$:
   $$
   f(3) = -3(3) + 6 = -9 + 6 = -3
   $$
6. For $$ x = 4 $$:
   $$
   f(4) = -3(4) + 6 = -12 + 6 = -6
   $$
7. For $$ x = 6 $$:
   $$
   f(6) = -3(6) + 6 = -18 + 6 = -12
   $$
8. For $$ x = 8 $$:
   $$
   f(8) = -3(8) + 6 = -24 + 6 = -18
   $$

### Step 3: Identify valid points for $$ f(x) $$

From the calculations, we find:
- $$ f(-1) = 9 $$ → point $$ (-1, 9) $$
- $$ f(0) = 6 $$ → point $$ (0, 6) $$
- $$ f(3) = -3 $$ → point $$ (3, -3) $$
- $$ f(4) = -6 $$ → point $$ (4, -6) $$

### Step 4: Analyze the inverse function $$ f^{-1}(x) $$

To find the inverse function, we set $$ y = f(x) $$ and solve for $$ x $$:
$$
y = -3x + 6 \implies 3x = 6 - y \implies x = \frac{6 - y}{3}
$$
Thus, the inverse function is:
$$
f^{-1}(x) = \frac{6 - x}{3}
$$

### Step 5: Calculate outputs for given inputs in $$ f^{-1}(x) $$

1. For $$ x = 5 $$:
   $$
   f^{-1}(5) = \frac{6 - 5}{3} = \frac{1}{3}
   $$
2. For $$ x = 9 $$:
   $$
   f^{-1}(9) = \frac{6 - 9}{3} = \frac{-3}{3} = -1
   $$
3. For $$ x = 6 $$:
   $$
   f^{-1}(6) = \frac{6 - 6}{3} = 0
   $$
4. For $$ x = 0 $$:
   $$
   f^{-1}(0) = \frac{6 - 0}{3} = 2
   $$
5. For $$ x = -2 $$:
   $$
   f^{-1}(-2) = \frac{6 - (-2)}{3} = \frac{8}{3}
   $$
6. For $$ x = -3 $$:
   $$
   f^{-1}(-3) = \frac{6 - (-3)}{3} = \frac{9}{3} = 3
   $$
7. For $$ x = -6 $$:
   $$
   f^{-1}(-6) = \frac{6 - (-6)}{3} = \frac{12}{3} = 4
   $$
8. For $$ x = -1 $$:
   $$
   f^{-1}(-1) = \frac{6 - (-1)}{3} = \frac{7}{3}
   $$

### Step 6: Identify valid points for $$ f^{-1}(x) $$

From the calculations, we find:
- $$ f^{-1}(6) = 0 $$ → point $$ (6, 0) $$
- $$ f^{-1}(9) = -1 $$ → point $$ (9, -1) $$
- $$ f^{-1}(0) = 2 $$ → point $$ (0, 2) $$

### Final Bins

- **For $$ f(x) $$**: 
  - $$ (-1, 9) $$
  - $$ (0, 6) $$
  - $$ (3, -3) $$
  - $$ (4, -6) $$

- **For $$ f^{-1}(x) $$**: 
  - $$ (6, 0) $$
  - $$ (0, 2) $$
  - $$ (3, 1) $$ (not valid for either function)
  - $$ (-2, 0) $$ (not valid for either function)
  - $$ (8, -1) $$ (not valid for either function)

### Summary

- **$$ f(x) $$**: $$ (-1, 9), (0, 6), (3, -3), (4, -6) $$
- **$$ f^{-1}(x) $$**: $$ (6, 0), (0, 2) $$
For $$ f(x) = -3x + 6 $$: - Points: (-1, 9), (0, 6), (3, -3), (4, -6) For $$ f^{-1}(x) = \frac{6 - x}{3} $$: - Points: (6, 0), (0, 2)

Question 3 Consider \( f(x)=-3 x+6 \). Drag points that can represent the possible inputs and outputs of each function into the correct bin. \( f(x) \) \( f^{-1}(x) \) \( (-9,5) \) \( (-1,9) \) \( (0,2) \) \( (0,6) \) \( (-2,0) \) \( (3,-3) \) \( (3,1) \) \( (4,-6) \) (6, 0) (8,-1)

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