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Let's go through each function one by one, finding the equations of vertical asymptotes, identifying any holes, and determining if there are horizontal or oblique asymptotes.

### Analyzing Each Function

#### (a) \( y = \frac{x}{x - 2} \)

1. **Vertical Asymptotes**: Set the denominator \( x - 2 = 0 \). Thus, there is a vertical asymptote at \( x = 2 \).
2. **Holes**: There are no common factors in the numerator and denominator, so there are no holes.
3. **Horizontal Asymptotes**: Since the degree of the numerator (1) is equal to the degree of the denominator (1), the horizontal asymptote is given by the ratio of the leading coefficients. So, \( y = \frac{1}{1} = 1 \).

#### (b) \( y = \frac{x - 1}{3x - 3} \)

1. **Vertical Asymptotes**: Set \( 3x - 3 = 0 \Rightarrow x = 1 \). So, there is a vertical asymptote at \( x = 1 \).
2. **Holes**: The numerator \( x - 1 \) and the denominator \( 3x - 3 \) share a common factor of \( x - 1 \). After canceling, the function simplifies to \( y = \frac{1}{3} \). Thus, there is a hole at \( x = 1 \).
3. **Horizontal Asymptotes**: After simplification, the function is constant \( y = \frac{1}{3} \), so this line acts as a horizontal asymptote.

#### (c) \( y = \frac{-7x}{4x + 2} \)

1. **Vertical Asymptotes**: Set \( 4x + 2 = 0 \Rightarrow x = -\frac{1}{2} \). So, there is a vertical asymptote at \( x = -\frac{1}{2} \).
2. **Holes**: There are no common factors in the numerator and denominator, so there are no holes.
3. **Horizontal Asymptotes**: Since the degree of the numerator (1) is equal to the degree of the denominator (1), the horizontal asymptote is given by the ratio of the leading coefficients. Therefore, \( y = \frac{-7}{4} \).

#### (d) \( y = \frac{x^2 + 2}{x - 6} \)

1. **Vertical Asymptotes**: Set \( x - 6 = 0 \Rightarrow x = 6 \). So, there is a vertical asymptote at \( x = 6 \).
2. **Holes**: There are no common factors in the numerator and denominator, so there are no holes.
3. **Oblique Asymptotes**: Since the degree of the numerator (2) is greater than the degree of the denominator (1), there is an oblique asymptote. Perform polynomial division on \( \frac{x^2 + 2}{x - 6} \), yielding \( y = x + 6 + \frac{38}{x - 6} \). Thus, the oblique asymptote is \( y = x + 6 \).

#### (e) \( y = \frac{1}{x^2 + 2x - 15} \)

1. **Vertical Asymptotes**: Factor the denominator: \( x^2 + 2x - 15 = (x + 5)(x - 3) \). Setting each factor to zero gives vertical asymptotes at \( x = -5 \) and \( x = 3 \).
2. **Holes**: There are no common factors between the numerator and the denominator, so there are no holes.
3. **Horizontal Asymptotes**: Since the degree of the numerator (0) is less than the degree of the denominator (2), there is a horizontal asymptote at \( y = 0 \).

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Let me know if you would like a deeper explanation for any of these solutions.

Here are some related questions for further exploration:

1. How do you determine whether a rational function has an oblique asymptote?
2. What steps are involved in finding holes in a rational function?
3. How can you verify the horizontal asymptote for a rational function?
4. Why do holes occur only when the numerator and denominator share a factor?
5. How can polynomial division be used to find oblique asymptotes?

**Tip**: Remember that a vertical asymptote occurs when the denominator is zero and cannot be canceled out, while a hole occurs when a factor in both the numerator and denominator is canceled.

For each function, determine the equations of any vertical asymptotes, the locations of any holes, and the existence of any horizontal asymptotes (other than the x-axis) or oblique asymptotes.

Explore step-by-step solutions for finding vertical, horizontal, and oblique asymptotes, as well as identifying holes in rational functions. Learn how to analyze rational functions like y = x/(x - 2) and y = 1/(x^2 + 2x - 15) with techniques such as polynomial division and degree comparison. Suitable for students in Grades 10-12.

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