Answer :
To determine which rational function has a horizontal asymptote at [tex]\( y = 3 \)[/tex] and vertical asymptotes at [tex]\( x = 4 \)[/tex] and [tex]\( x = -3 \)[/tex], let’s analyze each of the functions step-by-step.
### Horizontal Asymptote Analysis
The horizontal asymptote of a rational function is determined by the degrees of the numerator and denominator:
1. If the degrees are the same, the horizontal asymptote [tex]\( y = \frac{a}{b} \)[/tex] is found by dividing the leading coefficients.
2. If the degree of the numerator is less than the denominator, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
3. If the degree of the numerator is greater than the denominator, there is no horizontal asymptote.
For [tex]\( y = 3 \)[/tex], the degrees must be the same and the quotient of the leading coefficients must be 3.
### Vertical Asymptote Analysis
Vertical asymptotes occur where the denominator is zero, and the numerator is non-zero. We analyze the denominator to find these points.
### Checking the Given Functions
#### Function 1: [tex]\( y = \frac{x^2}{x^2 - x - 12} \)[/tex]
1. Factor the denominator: [tex]\( x^2 - x - 12 \)[/tex]
[tex]\[ x^2 - x - 12 = (x - 4)(x + 3) \][/tex]
2. Vertical asymptotes at [tex]\( x = 4 \)[/tex] and [tex]\( x = -3 \)[/tex].
3. Horizontal asymptote:
[tex]\[ \text{Leading coefficients ratio} = \frac{1}{1} = 1 \][/tex]
- Horizontal asymptote at [tex]\( y = 1 \)[/tex].
This does not satisfy [tex]\( y = 3 \)[/tex] as the horizontal asymptote.
#### Function 2: [tex]\( y = \frac{x^2}{x^2 + x - 12} \)[/tex]
1. Factor the denominator: [tex]\( x^2 + x - 12 \)[/tex]
[tex]\[ x^2 + x - 12 = (x + 4)(x - 3) \][/tex]
2. Vertical asymptotes at [tex]\( x = -4 \)[/tex] and [tex]\( x = 3 \)[/tex] (different from given).
3. Horizontal asymptote:
[tex]\[ \text{Leading coefficients ratio} = \frac{1}{1} = 1 \][/tex]
- Horizontal asymptote at [tex]\( y = 1 \)[/tex].
This does not match vertical asymptotes or horizontal asymptote conditions.
#### Function 3: [tex]\( y = \frac{3x^2}{x^2 - x - 12} \)[/tex]
1. Factor the denominator: [tex]\( x^2 - x - 12 \)[/tex]
[tex]\[ x^2 - x - 12 = (x - 4)(x + 3) \][/tex]
2. Vertical asymptotes at [tex]\( x = 4 \)[/tex] and [tex]\( x = -3 \)[/tex].
3. Horizontal asymptote:
[tex]\[ \text{Leading coefficients ratio} = \frac{3}{1} = 3 \][/tex]
- Horizontal asymptote at [tex]\( y = 3 \)[/tex].
This matches both vertical and horizontal asymptote conditions.
#### Function 4: [tex]\( y = \frac{3x^2}{x^2 + x - 12} \)[/tex]
1. Factor the denominator: [tex]\( x^2 + x - 12 \)[/tex]
[tex]\[ x^2 + x - 12 = (x + 4)(x - 3) \][/tex]
2. Vertical asymptotes at [tex]\( x = -4 \)[/tex] and [tex]\( x = 3 \)[/tex] (different from given).
3. Horizontal asymptote:
[tex]\[ \text{Leading coefficients ratio} = \frac{3}{1} = 3 \][/tex]
- Horizontal asymptote at [tex]\( y = 3 \)[/tex].
This does not match vertical asymptote conditions.
### Conclusion
Only Function 3 [tex]\( y = \frac{3x^2}{x^2 - x - 12} \)[/tex] meets the required horizontal asymptote at [tex]\( y = 3 \)[/tex] and vertical asymptotes at [tex]\( x = 4 \)[/tex] and [tex]\( x = -3 \)[/tex].
The correct function is:
[tex]\[ \boxed{y = \frac{3x^2}{x^2 - x - 12}} \][/tex] and the corresponding number is [tex]\( \boxed{3} \)[/tex].
### Horizontal Asymptote Analysis
The horizontal asymptote of a rational function is determined by the degrees of the numerator and denominator:
1. If the degrees are the same, the horizontal asymptote [tex]\( y = \frac{a}{b} \)[/tex] is found by dividing the leading coefficients.
2. If the degree of the numerator is less than the denominator, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
3. If the degree of the numerator is greater than the denominator, there is no horizontal asymptote.
For [tex]\( y = 3 \)[/tex], the degrees must be the same and the quotient of the leading coefficients must be 3.
### Vertical Asymptote Analysis
Vertical asymptotes occur where the denominator is zero, and the numerator is non-zero. We analyze the denominator to find these points.
### Checking the Given Functions
#### Function 1: [tex]\( y = \frac{x^2}{x^2 - x - 12} \)[/tex]
1. Factor the denominator: [tex]\( x^2 - x - 12 \)[/tex]
[tex]\[ x^2 - x - 12 = (x - 4)(x + 3) \][/tex]
2. Vertical asymptotes at [tex]\( x = 4 \)[/tex] and [tex]\( x = -3 \)[/tex].
3. Horizontal asymptote:
[tex]\[ \text{Leading coefficients ratio} = \frac{1}{1} = 1 \][/tex]
- Horizontal asymptote at [tex]\( y = 1 \)[/tex].
This does not satisfy [tex]\( y = 3 \)[/tex] as the horizontal asymptote.
#### Function 2: [tex]\( y = \frac{x^2}{x^2 + x - 12} \)[/tex]
1. Factor the denominator: [tex]\( x^2 + x - 12 \)[/tex]
[tex]\[ x^2 + x - 12 = (x + 4)(x - 3) \][/tex]
2. Vertical asymptotes at [tex]\( x = -4 \)[/tex] and [tex]\( x = 3 \)[/tex] (different from given).
3. Horizontal asymptote:
[tex]\[ \text{Leading coefficients ratio} = \frac{1}{1} = 1 \][/tex]
- Horizontal asymptote at [tex]\( y = 1 \)[/tex].
This does not match vertical asymptotes or horizontal asymptote conditions.
#### Function 3: [tex]\( y = \frac{3x^2}{x^2 - x - 12} \)[/tex]
1. Factor the denominator: [tex]\( x^2 - x - 12 \)[/tex]
[tex]\[ x^2 - x - 12 = (x - 4)(x + 3) \][/tex]
2. Vertical asymptotes at [tex]\( x = 4 \)[/tex] and [tex]\( x = -3 \)[/tex].
3. Horizontal asymptote:
[tex]\[ \text{Leading coefficients ratio} = \frac{3}{1} = 3 \][/tex]
- Horizontal asymptote at [tex]\( y = 3 \)[/tex].
This matches both vertical and horizontal asymptote conditions.
#### Function 4: [tex]\( y = \frac{3x^2}{x^2 + x - 12} \)[/tex]
1. Factor the denominator: [tex]\( x^2 + x - 12 \)[/tex]
[tex]\[ x^2 + x - 12 = (x + 4)(x - 3) \][/tex]
2. Vertical asymptotes at [tex]\( x = -4 \)[/tex] and [tex]\( x = 3 \)[/tex] (different from given).
3. Horizontal asymptote:
[tex]\[ \text{Leading coefficients ratio} = \frac{3}{1} = 3 \][/tex]
- Horizontal asymptote at [tex]\( y = 3 \)[/tex].
This does not match vertical asymptote conditions.
### Conclusion
Only Function 3 [tex]\( y = \frac{3x^2}{x^2 - x - 12} \)[/tex] meets the required horizontal asymptote at [tex]\( y = 3 \)[/tex] and vertical asymptotes at [tex]\( x = 4 \)[/tex] and [tex]\( x = -3 \)[/tex].
The correct function is:
[tex]\[ \boxed{y = \frac{3x^2}{x^2 - x - 12}} \][/tex] and the corresponding number is [tex]\( \boxed{3} \)[/tex].
