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Sagot :
To determine which rational function [tex]\( h \)[/tex] has a horizontal asymptote at [tex]\( y = 1 \)[/tex], we analyze the horizontal asymptotes of each option. A horizontal asymptote describes the behavior of the function as [tex]\( x \)[/tex] approaches infinity or negative infinity. The horizontal asymptote depends on the degrees of the polynomial in the numerator and the denominator.
### Option A: [tex]\( h(x) = \frac{x^2 - 16}{x^2 + 16} \)[/tex]
For this function:
- The degrees of the numerator and the denominator are both 2.
- When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.
- The leading coefficients of [tex]\( x^2 \)[/tex] in both the numerator and the denominator are 1.
Thus, the horizontal asymptote is:
[tex]\[ y = \frac{1}{1} = 1 \][/tex]
### Option B: [tex]\( h(x) = \frac{x^2 + 16}{x^2 - 16} \)[/tex]
For this function:
- Again, the degrees of the numerator and the denominator are both 2.
- When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.
- The leading coefficients of [tex]\( x^2 \)[/tex] in both the numerator and the denominator are still 1.
Thus, the horizontal asymptote is:
[tex]\[ y = \frac{1}{1} = 1 \][/tex]
### Option C: [tex]\( h(z) = \frac{z^2 - 16}{z - 4} \)[/tex]
For this function:
- The degree of the numerator is 2, and the degree of the denominator is 1.
- When the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Instead, the behavior at infinity will be a slant (oblique) asymptote or, in certain cases, another form of asymptotic behavior (specifically, this will approach infinity linearly rather than having a horizontal asymptote).
Thus, there is no horizontal asymptote at [tex]\( y = 1 \)[/tex].
### Option D: [tex]\( h(x) = \frac{x + 4}{x^2 + 16} \)[/tex]
For this function:
- The degree of the numerator is 1, and the degree of the denominator is 2.
- When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
Thus, the horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
### Conclusion
The only options where the horizontal asymptote is [tex]\( y = 1 \)[/tex] are Options A and B. Thus, the correct answers are:
A. [tex]\( h(x) = \frac{x^2 - 16}{x^2 + 16} \)[/tex]
B. [tex]\( h(x) = \frac{x^2 + 16}{x^2 - 16} \)[/tex]
Therefore, since the question asks to select the correct answer:
1 (which matches Option A): [tex]\( h(x) = \frac{x^2 - 16}{x^2 + 16} \)[/tex].
### Option A: [tex]\( h(x) = \frac{x^2 - 16}{x^2 + 16} \)[/tex]
For this function:
- The degrees of the numerator and the denominator are both 2.
- When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.
- The leading coefficients of [tex]\( x^2 \)[/tex] in both the numerator and the denominator are 1.
Thus, the horizontal asymptote is:
[tex]\[ y = \frac{1}{1} = 1 \][/tex]
### Option B: [tex]\( h(x) = \frac{x^2 + 16}{x^2 - 16} \)[/tex]
For this function:
- Again, the degrees of the numerator and the denominator are both 2.
- When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.
- The leading coefficients of [tex]\( x^2 \)[/tex] in both the numerator and the denominator are still 1.
Thus, the horizontal asymptote is:
[tex]\[ y = \frac{1}{1} = 1 \][/tex]
### Option C: [tex]\( h(z) = \frac{z^2 - 16}{z - 4} \)[/tex]
For this function:
- The degree of the numerator is 2, and the degree of the denominator is 1.
- When the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Instead, the behavior at infinity will be a slant (oblique) asymptote or, in certain cases, another form of asymptotic behavior (specifically, this will approach infinity linearly rather than having a horizontal asymptote).
Thus, there is no horizontal asymptote at [tex]\( y = 1 \)[/tex].
### Option D: [tex]\( h(x) = \frac{x + 4}{x^2 + 16} \)[/tex]
For this function:
- The degree of the numerator is 1, and the degree of the denominator is 2.
- When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
Thus, the horizontal asymptote is:
[tex]\[ y = 0 \][/tex]
### Conclusion
The only options where the horizontal asymptote is [tex]\( y = 1 \)[/tex] are Options A and B. Thus, the correct answers are:
A. [tex]\( h(x) = \frac{x^2 - 16}{x^2 + 16} \)[/tex]
B. [tex]\( h(x) = \frac{x^2 + 16}{x^2 - 16} \)[/tex]
Therefore, since the question asks to select the correct answer:
1 (which matches Option A): [tex]\( h(x) = \frac{x^2 - 16}{x^2 + 16} \)[/tex].
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