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Let's work through the function \( f(x) = \frac{x^2 - 4}{x^2 - 3x + 2} \) to identify intercepts, asymptotes, and the general shape for sketching.

### Step 1: Factor both the numerator and the denominator.

1. **Numerator**: 
   \[
   x^2 - 4 = (x - 2)(x + 2)
   \]

2. **Denominator**:
   \[
   x^2 - 3x + 2 = (x - 1)(x - 2)
   \]

So the function becomes:
\[
f(x) = \frac{(x - 2)(x + 2)}{(x - 1)(x - 2)}
\]

### Step 2: Identify the Vertical Asymptotes

Vertical asymptotes occur where the denominator equals zero, except for any factors that cancel with the numerator.

- The denominator \( (x - 1)(x - 2) = 0 \) gives two possible points: \( x = 1 \) and \( x = 2 \).
- Notice that \( (x - 2) \) cancels with the numerator, so **there is no vertical asymptote at \( x = 2 \)**. However, \( x = 2 \) will be a **hole** in the graph.
- Therefore, the only vertical asymptote is at \( x = 1 \).

### Step 3: Identify the Horizontal Asymptote

To find the horizontal asymptote, compare the degrees of the numerator and the denominator:
- Both the numerator and denominator are degree 2 polynomials.
- When the degrees are equal, the horizontal asymptote is found by dividing the leading coefficients. In this case, both leading coefficients are 1.

Thus, the horizontal asymptote is:
\[
y = \frac{1}{1} = 1
\]

### Step 4: Identify the Intercepts

- **x-intercept(s)**: Set the numerator equal to zero.
  \[
  (x - 2)(x + 2) = 0
  \]
  So, \( x = 2 \) and \( x = -2 \). However, \( x = 2 \) is a hole, not an intercept. Thus, the only x-intercept is at \( x = -2 \).

- **y-intercept**: Set \( x = 0 \) in the original function.
  \[
  f(0) = \frac{0^2 - 4}{0^2 - 3(0) + 2} = \frac{-4}{2} = -2
  \]
  The y-intercept is at \( y = -2 \).

### Step 5: Sketch the Graph

- **Vertical Asymptote**: \( x = 1 \)
- **Horizontal Asymptote**: \( y = 1 \)
- **Hole**: At \( x = 2 \)
- **Intercepts**: x-intercept at \( x = -2 \), y-intercept at \( y = -2 \)

Would you like me to generate a sketch of this graph or provide further details? 

Here are some related questions for further understanding:

1. How do you determine where a function has a hole versus a vertical asymptote?
2. Why does the degree of the polynomial influence the horizontal asymptote?
3. What happens to the graph near a hole, like at \( x = 2 \) in this case?
4. How would the graph of this function behave at its intercepts?
5. Can you explain how to check the behavior of the graph near the asymptotes?

**Tip:** Canceling terms between the numerator and denominator simplifies the function, but remember that this creates a hole, not an asymptote, at the canceled point.

Identify any intercepts and asymptotes of the graph of the function. Then sketch the graph of the function f(x) = (x^2 - 4) / (x^2 - 3x + 2).

Explore how to identify intercepts and asymptotes of the rational function f(x) = (x^2 - 4)/(x^2 - 3x + 2). Learn step-by-step factoring techniques, how to find holes, vertical and horizontal asymptotes, and sketch the graph. Suitable for students in Grades 10-12.

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