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Sagot :
To find the horizontal asymptote(s) of the function [tex]\( f(x) = \frac{5x - 2}{x^2 + 3x - 18} \)[/tex], we analyze the behavior of the function as [tex]\( x \)[/tex] approaches both positive and negative infinity.
Step-by-Step Solution:
1. Understand the Function:
The given function is [tex]\( f(x) = \frac{5x - 2}{x^2 + 3x - 18} \)[/tex]. Here, the degree of the polynomial in the numerator (which is 1, because of [tex]\( 5x - 2 \)[/tex]) is less than the degree of the polynomial in the denominator (which is 2, because of [tex]\( x^2 + 3x - 18 \)[/tex]).
2. Horizontal Asymptote Rules:
For rational functions:
- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
- If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is [tex]\( y = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} \)[/tex].
- If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote (though there might be an oblique asymptote).
3. Applying the Rule:
In this case, since the degree of the numerator ([tex]\( 5x - 2 \)[/tex] which has degree 1) is less than the degree of the denominator ([tex]\( x^2 + 3x - 18 \)[/tex] which has degree 2), we apply the first rule. Therefore, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
4. Conclusion:
As [tex]\( x \)[/tex] approaches positive infinity ([tex]\( +\infty \)[/tex]) or negative infinity ([tex]\( -\infty \)[/tex]), the value of [tex]\( f(x) \)[/tex] approaches 0. Hence, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
The function [tex]\( f(x) = \frac{5x - 2}{x^2 + 3x - 18} \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex].
Step-by-Step Solution:
1. Understand the Function:
The given function is [tex]\( f(x) = \frac{5x - 2}{x^2 + 3x - 18} \)[/tex]. Here, the degree of the polynomial in the numerator (which is 1, because of [tex]\( 5x - 2 \)[/tex]) is less than the degree of the polynomial in the denominator (which is 2, because of [tex]\( x^2 + 3x - 18 \)[/tex]).
2. Horizontal Asymptote Rules:
For rational functions:
- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
- If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is [tex]\( y = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} \)[/tex].
- If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote (though there might be an oblique asymptote).
3. Applying the Rule:
In this case, since the degree of the numerator ([tex]\( 5x - 2 \)[/tex] which has degree 1) is less than the degree of the denominator ([tex]\( x^2 + 3x - 18 \)[/tex] which has degree 2), we apply the first rule. Therefore, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
4. Conclusion:
As [tex]\( x \)[/tex] approaches positive infinity ([tex]\( +\infty \)[/tex]) or negative infinity ([tex]\( -\infty \)[/tex]), the value of [tex]\( f(x) \)[/tex] approaches 0. Hence, the horizontal asymptote is [tex]\( y = 0 \)[/tex].
The function [tex]\( f(x) = \frac{5x - 2}{x^2 + 3x - 18} \)[/tex] has a horizontal asymptote at [tex]\( y = 0 \)[/tex].
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