To determine which function among the given has no horizontal asymptote, we analyze the degrees of the numerator and denominator of each given function.
1. Function [tex]\( f(x) = \frac{2x - 1}{3x^2} \)[/tex]
- The degree of the numerator ([tex]\(2x - 1\)[/tex]) is 1.
- The degree of the denominator ([tex]\(3x^2\)[/tex]) is 2.
- The degree of the numerator (1) is less than the degree of the denominator (2), so this function has a horizontal asymptote.
2. Function [tex]\( f(x) = \frac{x - 1}{3x} \)[/tex]
- The degree of the numerator ([tex]\(x - 1\)[/tex]) is 1.
- The degree of the denominator ([tex]\(3x\)[/tex]) is 1.
- The degree of the numerator (1) is equal to the degree of the denominator (1), so this function has a horizontal asymptote determined by the leading coefficients.
3. Function [tex]\( f(x) = \frac{2x^2}{3x - 1} \)[/tex]
- The degree of the numerator ([tex]\(2x^2\)[/tex]) is 2.
- The degree of the denominator ([tex]\(3x - 1\)[/tex]) is 1.
- The degree of the numerator (2) is greater than the degree of the denominator (1), so this function does not have a horizontal asymptote. Instead, it has an oblique or slant asymptote.
4. Function [tex]\( f(x) = \frac{3x^2}{x^2 - 1} \)[/tex]
- The degree of the numerator ([tex]\(3x^2\)[/tex]) is 2.
- The degree of the denominator ([tex]\(x^2 - 1\)[/tex]) is 2.
- The degree of the numerator (2) is equal to the degree of the denominator (2), so this function has a horizontal asymptote determined by the leading coefficients.
By comparing the degrees of the numerator and the denominator for each function, we conclude that the function without a horizontal asymptote is:
[tex]\[ f(x) = \frac{2x^2}{3x - 1} \][/tex]
Thus, the answer is the third function.
