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To analyze the function \( f(x) = \frac{-x^2 + 2x - 1}{x - 1} \), let's go through each characteristic step-by-step.

### 1. Horizontal Asymptote
To find the horizontal asymptote of a rational function, compare the degrees of the polynomial in the numerator and the denominator:

- The numerator \( -x^2 + 2x - 1 \) is a degree 2 polynomial.
- The denominator \( x - 1 \) is a degree 1 polynomial.

Since the degree of the numerator (2) is greater than the degree of the denominator (1), there is **no horizontal asymptote**. Instead, there may be an oblique asymptote, which we can explore if needed.

### 2. Vertical Asymptote
A vertical asymptote occurs when the denominator equals zero, as long as the numerator does not also equal zero at that point.

- Setting \( x - 1 = 0 \), we find that \( x = 1 \).
- Substituting \( x = 1 \) in the numerator: \( -1^2 + 2 \cdot 1 - 1 = -1 + 2 - 1 = 0 \).

Since the numerator also equals zero when \( x = 1 \), there is **no vertical asymptote** at \( x = 1 \). Instead, there is a hole at \( x = 1 \).

### 3. x-Intercept
The x-intercepts occur where \( f(x) = 0 \), which means the numerator equals zero.

- Solving \( -x^2 + 2x - 1 = 0 \), we can factor this as:
  \[
  -x^2 + 2x - 1 = -(x - 1)^2 = 0
  \]
  So, \( x = 1 \) is a root.

However, we previously found that \( x = 1 \) leads to both the numerator and the denominator being zero, which indicates a **hole** rather than an x-intercept. So, there is **no x-intercept** for this function.

### 4. y-Intercept
The y-intercept is found by setting \( x = 0 \).

- Substituting \( x = 0 \) in \( f(x) \):
  \[
  f(0) = \frac{-0^2 + 2 \cdot 0 - 1}{0 - 1} = \frac{-1}{-1} = 1
  \]
  Therefore, the y-intercept is \( (0, 1) \).

### 5. Hole
As calculated earlier, a hole occurs at \( x = 1 \) because both the numerator and denominator equal zero at this point.

- To find the y-coordinate of the hole, simplify \( f(x) \) around \( x = 1 \):
  - Factor out \( (x - 1) \) from the numerator:
    \[
    f(x) = \frac{-(x - 1)(x - 1)}{x - 1} = -(x - 1)
    \]
  - Canceling \( (x - 1) \), we get \( f(x) = -(x - 1) \) for \( x \neq 1 \).
  - At \( x = 1 \), \( f(x) \approx -(1 - 1) = 0 \).

So, the hole is at \( (1, 0) \).

### Summary of Features
- **Horizontal Asymptote**: None
- **Vertical Asymptote**: None
- **x-Intercept**: None
- **y-Intercept**: \( (0, 1) \)
- **Hole**: \( (1, 0) \)

Would you like a further breakdown of any of these steps, or have any questions?

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### Related Questions
1. How do you determine if a function has an oblique asymptote?
2. What is the difference between a hole and a vertical asymptote in a rational function?
3. How do you find the x- and y-intercepts for other types of functions?
4. Why does a rational function with equal degrees in numerator and denominator have a horizontal asymptote?
5. Can a rational function have both a vertical asymptote and a hole at the same x-value?

### Tip
For rational functions, always check for common factors in the numerator and denominator to identify holes.

Determine each feature of the graph of the function \( f(x) = \frac{-x^2 + 2x - 1}{x - 1} \) including horizontal asymptote, vertical asymptote, x-intercept, y-intercept, and hole.

Explore the features of the rational function \( f(x) = \frac{-x^2 + 2x - 1}{x - 1} \), including horizontal and vertical asymptotes, x- and y-intercepts, and identifying any holes in the graph. This detailed solution is suitable for high school students learning about rational functions.

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