Question
A rational function has the given information. Choose the equation that could represent the same rational function.
horizontal asymptote: y= 3/2
vertical asymptote: x= −2
hole at x= 5
(1 point)
Responses
y=3x² − 15x / 2x² − 10x
y=3x² − 12x − 15/2x² − 6x − 20
y=3x⁴ − 12x − 15/2x⁶ − 6x³ − 20x
y=x² − 4x − 5 x² − 3x − 10
horizontal asymptote: y= 3/2
vertical asymptote: x= −2
hole at x= 5
(1 point)
Responses
y=3x² − 15x / 2x² − 10x
y=3x² − 12x − 15/2x² − 6x − 20
y=3x⁴ − 12x − 15/2x⁶ − 6x³ − 20x
y=x² − 4x − 5 x² − 3x − 10
Answers
GPT-4o mini
To determine which equation represents a rational function that has the specified characteristics (horizontal asymptote \( y = \frac{3}{2} \), vertical asymptote \( x = -2 \), and a hole at \( x = 5 \)), let's analyze these conditions:
1. **Horizontal Asymptote**:
- The horizontal asymptote of a rational function depends on the degrees and leading coefficients of the numerator and denominator.
- For horizontal asymptote \( y = \frac{3}{2} \), the degrees of the numerator and denominator must be the same (let's denote them as \( n \)). The leading coefficient of the numerator divided by the leading coefficient of the denominator should equal \( \frac{3}{2} \).
2. **Vertical Asymptote**:
- A vertical asymptote occurs when the denominator is zero and the numerator is not zero at that point. For \( x = -2 \) to be a vertical asymptote, \( (x + 2) \) must be a factor of the denominator.
3. **Hole**:
- A hole in the graph occurs when both the numerator and denominator share a common factor that cancels out. For a hole at \( x = 5 \), both the numerator and denominator must have \( (x - 5) \) as a factor.
Let's analyze the given options one by one for these conditions:
1. **Option**: \( y=\frac{3x^2-15x}{2x^2-10x} \)
- **Asymptotes**: The denominator factors to \( 2x(x - 5) \), so the vertical asymptote is at \( x = 0 \) (not the required \( x = -2 \)). Also, the numerator has a factor of \( 3x(x - 5) \), resulting in a hole at \( x = 5 \).
- **Horizontal Asymptote**: The leading coefficients give \( \frac{3}{2} \) (correct).
- **Conclusion**: Incorrect due to vertical asymptote.
2. **Option**: \( y=\frac{3x^2-12x-15}{2x^2-6x-20} \)
- **Asymptotes**: The denominator factors to \( 2(x + 2)(x - 5) \), so has a vertical asymptote at \( x = -2 \) (correct) but also has a hole at \( x = 5 \).
- **Horizontal Asymptote**: The leading coefficients give \( \frac{3}{2} \) (correct).
- **Conclusion**: This option is valid.
3. **Option**: \( y=\frac{3x^4 - 12x - 15}{2x^6 - 6x^3 - 20x} \)
- **Analysis**: The degrees of denominator and numerator are different, so it cannot have \( y = \frac{3}{2} \) as a horizontal asymptote.
- **Conclusion**: Incorrect due to degrees of numerator and denominator.
4. **Option**: \( y=\frac{x^2 - 4x - 5}{x^2 - 3x - 10} \)
- **Asymptotes**: The denominator does not factor to include \( (x + 2) \), thus the vertical asymptote is not \( x = -2 \).
- **Conclusion**: Incorrect due to vertical asymptote.
Based on the analysis, the equation that could represent the same rational function, given the conditions of horizontal asymptote \( y = \frac{3}{2} \), vertical asymptote \( x = -2 \), and a hole at \( x = 5 \), is:
\[
\boxed{y=\frac{3x^2-12x-15}{2x^2-6x-20}}
\]
1. **Horizontal Asymptote**:
- The horizontal asymptote of a rational function depends on the degrees and leading coefficients of the numerator and denominator.
- For horizontal asymptote \( y = \frac{3}{2} \), the degrees of the numerator and denominator must be the same (let's denote them as \( n \)). The leading coefficient of the numerator divided by the leading coefficient of the denominator should equal \( \frac{3}{2} \).
2. **Vertical Asymptote**:
- A vertical asymptote occurs when the denominator is zero and the numerator is not zero at that point. For \( x = -2 \) to be a vertical asymptote, \( (x + 2) \) must be a factor of the denominator.
3. **Hole**:
- A hole in the graph occurs when both the numerator and denominator share a common factor that cancels out. For a hole at \( x = 5 \), both the numerator and denominator must have \( (x - 5) \) as a factor.
Let's analyze the given options one by one for these conditions:
1. **Option**: \( y=\frac{3x^2-15x}{2x^2-10x} \)
- **Asymptotes**: The denominator factors to \( 2x(x - 5) \), so the vertical asymptote is at \( x = 0 \) (not the required \( x = -2 \)). Also, the numerator has a factor of \( 3x(x - 5) \), resulting in a hole at \( x = 5 \).
- **Horizontal Asymptote**: The leading coefficients give \( \frac{3}{2} \) (correct).
- **Conclusion**: Incorrect due to vertical asymptote.
2. **Option**: \( y=\frac{3x^2-12x-15}{2x^2-6x-20} \)
- **Asymptotes**: The denominator factors to \( 2(x + 2)(x - 5) \), so has a vertical asymptote at \( x = -2 \) (correct) but also has a hole at \( x = 5 \).
- **Horizontal Asymptote**: The leading coefficients give \( \frac{3}{2} \) (correct).
- **Conclusion**: This option is valid.
3. **Option**: \( y=\frac{3x^4 - 12x - 15}{2x^6 - 6x^3 - 20x} \)
- **Analysis**: The degrees of denominator and numerator are different, so it cannot have \( y = \frac{3}{2} \) as a horizontal asymptote.
- **Conclusion**: Incorrect due to degrees of numerator and denominator.
4. **Option**: \( y=\frac{x^2 - 4x - 5}{x^2 - 3x - 10} \)
- **Asymptotes**: The denominator does not factor to include \( (x + 2) \), thus the vertical asymptote is not \( x = -2 \).
- **Conclusion**: Incorrect due to vertical asymptote.
Based on the analysis, the equation that could represent the same rational function, given the conditions of horizontal asymptote \( y = \frac{3}{2} \), vertical asymptote \( x = -2 \), and a hole at \( x = 5 \), is:
\[
\boxed{y=\frac{3x^2-12x-15}{2x^2-6x-20}}
\]