Question

which function has no horizontal asymptote?
○ $f(x)=\frac{2x - 1}{3x^2}$
○ $f(x)=\frac{x - 1}{3x}$
○ $f(x)=\frac{2x^2}{3x - 1}$
○ $f(x)=\frac{3x^2}{x^2 - 1}$

Explanation:

Step1: Recall horizontal asymptote rules

For a rational function \(f(x)=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}\), if \(n < m\), \(y = 0\) is the horizontal asymptote. If \(n=m\), \(y=\frac{a_n}{b_m}\) is the horizontal asymptote. If \(n>m\), there is no horizontal asymptote.

Step2: Analyze \(f(x)=\frac{2x - 1}{3x^2}\)

Degree of numerator \(n = 1\), degree of denominator \(m=2\). Since \(n

Step3: Analyze \(f(x)=\frac{x - 1}{3x}\)

Degree of numerator \(n = 1\), degree of denominator \(m = 1\). \(y=\frac{1}{3}\) is the horizontal asymptote.

Step4: Analyze \(f(x)=\frac{2x^2}{3x - 1}\)

Degree of numerator \(n = 2\), degree of denominator \(m = 1\). Since \(n>m\), there is no horizontal asymptote.

Step5: Analyze \(f(x)=\frac{3x^2}{x^2-1}\)

Degree of numerator \(n = 2\), degree of denominator \(m = 2\). \(y = 3\) is the horizontal asymptote.

Answer:

\(f(x)=\frac{2x^2}{3x - 1}\)