Question
Describe the vertical asymptote(s) and hole(s) for the graph of y=(x−3)(x+8)/(x+8)(x+2) (1 point) Responses asymptote: x= -8 and hole: x= -2 asymptote: x= -8 and hole: x= -2 asymptote: x=2 and hole: x=8 asymptote: x=2 and hole: x=8 asymptote x=-2 and hole: none asymptote x=-2 and hole: none asymptote: x= -2 and hole: x=-8
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GPT-4o mini
To analyze the function \( y = \frac{(x - 3)(x + 8)}{(x + 8)(x + 2)} \), we need to identify its vertical asymptotes and holes.
1. **Holes**: A hole occurs where a factor cancels out. Here, the factor \( (x + 8) \) is present in both the numerator and the denominator. This means there is a hole at \( x = -8 \).
2. **Vertical Asymptotes**: A vertical asymptote occurs where the denominator is zero but that factor does not cancel with the numerator. After canceling out the \( (x + 8) \) from both numerator and denominator, the function simplifies to:
\[
y = \frac{x - 3}{x + 2}
\]
The denominator \( x + 2 = 0 \) gives \( x = -2 \). Since this factor does not cancel, there is a vertical asymptote at \( x = -2 \).
Summarizing:
- Hole: \( x = -8 \)
- Vertical asymptote: \( x = -2 \)
Therefore, the correct response based on these observations would be:
**asymptote: x = -2 and hole: x = -8**.
1. **Holes**: A hole occurs where a factor cancels out. Here, the factor \( (x + 8) \) is present in both the numerator and the denominator. This means there is a hole at \( x = -8 \).
2. **Vertical Asymptotes**: A vertical asymptote occurs where the denominator is zero but that factor does not cancel with the numerator. After canceling out the \( (x + 8) \) from both numerator and denominator, the function simplifies to:
\[
y = \frac{x - 3}{x + 2}
\]
The denominator \( x + 2 = 0 \) gives \( x = -2 \). Since this factor does not cancel, there is a vertical asymptote at \( x = -2 \).
Summarizing:
- Hole: \( x = -8 \)
- Vertical asymptote: \( x = -2 \)
Therefore, the correct response based on these observations would be:
**asymptote: x = -2 and hole: x = -8**.