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To analyze the behavior of the function \( y = \frac{x - 2}{(x - 1)(x - 2)} \) as \( x \) approaches 2 and 1, we'll complete the tables with the function values, denoted as \( y \), corresponding to each \( x \).

### 1. Behavior as \( x \to 2^- \) (approaches from the left)

For values of \( x \) approaching 2 from the left, the function \( y = \frac{x - 2}{(x - 1)(x - 2)} \) can be evaluated as follows:

1. \( x = 1.9 \):
   \[ y = \frac{1.9 - 2}{(1.9 - 1)(1.9 - 2)} = \frac{-0.1}{(0.9)(-0.1)} = \frac{-0.1}{-0.09} = 1.11111 \]

2. \( x = 1.99 \):
   \[ y = \frac{1.99 - 2}{(1.99 - 1)(1.99 - 2)} = \frac{-0.01}{(0.99)(-0.01)} = \frac{-0.01}{-0.0099} = 1.01010 \]

3. \( x = 1.999 \):
   \[ y = \frac{1.999 - 2}{(1.999 - 1)(1.999 - 2)} = \frac{-0.001}{(0.999)(-0.001)} = \frac{-0.001}{-0.000999} = 1.00100 \]

These values match the given ones. Now, let's calculate for \( x \) approaching 2 from the right.

### 2. Behavior as \( x \to 2^+ \) (approaches from the right)

1. \( x = 2.1 \):
   \[ y = \frac{2.1 - 2}{(2.1 - 1)(2.1 - 2)} = \frac{0.1}{(1.1)(0.1)} = \frac{0.1}{0.11} \approx 0.90909 \]

2. \( x = 2.01 \):
   \[ y = \frac{2.01 - 2}{(2.01 - 1)(2.01 - 2)} = \frac{0.01}{(1.01)(0.01)} = \frac{0.01}{0.0101} \approx 0.99010 \]

3. \( x = 2.001 \):
   \[ y = \frac{2.001 - 2}{(2.001 - 1)(2.001 - 2)} = \frac{0.001}{(1.001)(0.001)} = \frac{0.001}{0.001001} \approx 0.99900 \]

### 3. Behavior as \( x \to 1^- \) (approaches from the left)

1. \( x = 0.9 \):
   \[ y = \frac{0.9 - 2}{(0.9 - 1)(0.9 - 2)} = \frac{-1.1}{(-0.1)(-1.1)} = \frac{-1.1}{0.11} = -10 \]

2. \( x = 0.99 \):
   \[ y = \frac{0.99 - 2}{(0.99 - 1)(0.99 - 2)} = \frac{-1.01}{(-0.01)(-1.01)} = \frac{-1.01}{0.0101} = -100 \]

3. \( x = 0.999 \):
   \[ y = \frac{0.999 - 2}{(0.999 - 1)(0.999 - 2)} = \frac{-1.001}{(-0.001)(-1.001)} = \frac{-1.001}{0.001001} \approx -1000 \]

### 4. Behavior as \( x \to 1^+ \) (approaches from the right)

1. \( x = 1.1 \):
   \[ y = \frac{1.1 - 2}{(1.1 - 1)(1.1 - 2)} = \frac{-0.9}{(0.1)(-0.9)} = \frac{-0.9}{-0.09} = 10 \]

2. \( x = 1.01 \):
   \[ y = \frac{1.01 - 2}{(1.01 - 1)(1.01 - 2)} = \frac{-0.99}{(0.01)(-0.99)} = \frac{-0.99}{-0.0099} = 100 \]

3. \( x = 1.001 \):
   \[ y = \frac{1.001 - 2}{(1.001 - 1)(1.001 - 2)} = \frac{-0.999}{(0.001)(-0.999)} = \frac{-0.999}{-0.000999} = 1000 \]

The behavior of the function at \( x = 2 \) indicates that there is a removable discontinuity (a hole) because the function approaches a finite value from both sides but is not defined at \( x = 2 \). At \( x = 1 \), the function shows a vertical asymptote as the values approach \( \pm\infty \) from both sides, indicating a non-removable discontinuity.

Let me know if you have any questions or need further details!

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Here are eight related questions to expand on this topic:

1. How do removable and non-removable discontinuities differ in terms of graph behavior?
2. What is the significance of calculating function values near a point of interest?
3. How does the limit definition help in understanding function behavior at specific points?
4. What are the implications of a vertical asymptote in the real-world interpretation of functions?
5. How can you identify a hole in the graph of a function analytically?
6. What does it mean for a function to be undefined at a point but behave normally elsewhere?
7. How do you use limits to describe the behavior of functions around discontinuities?
8. What strategies can be used to determine whether a discontinuity is removable?

**Tip:** When analyzing functions near points of interest, always consider both left-hand and right-hand limits to fully understand the behavior and potential discontinuities.

Explore the behavior of the function y = (x - 2)/((x - 1)(x - 2)) as x approaches 1 and 2. Understand the concepts of limits, discontinuities, and function behavior with detailed examples and analysis. Ideal for advanced high school students studying calculus.

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