To find the horizontal asymptote of the rational function [tex]\( f(x) = \frac{11x}{3x^2 + 1} \)[/tex], follow these steps:
1. Identify the degrees of the numerator and the denominator:
- The degree of the numerator (the highest power of [tex]\( x \)[/tex] in the numerator) is 1 because [tex]\( 11x \)[/tex] has [tex]\( x^1 \)[/tex].
- The degree of the denominator (the highest power of [tex]\( x \)[/tex] in the denominator) is 2 because [tex]\( 3x^2 \)[/tex] has [tex]\( x^2 \)[/tex].
2. Compare the degrees:
- 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 (there may be an oblique asymptote).
In this case, the degree of the numerator (1) is less than the degree of the denominator (2). Therefore, the horizontal asymptote is:
[tex]\[
\text{The horizontal asymptote is } y = 0
\][/tex]
Hence, the correct choice is:
A. The horizontal asymptote is [tex]\( y = 0 \)[/tex].
