Question

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

Explanation:

Step1: Recall Horizontal Asymptote Rules

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), where \( N(x) \) is the numerator polynomial and \( D(x) \) is the denominator polynomial with degrees \( n \) and \( d \) respectively:

• If \( n < d \), horizontal asymptote is \( y = 0 \).
• If \( n = d \), horizontal asymptote is \( y=\frac{\text{leading coefficient of }N(x)}{\text{leading coefficient of }D(x)} \).
• If \( n > d \), there is no horizontal asymptote (but there may be an oblique asymptote).

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

• Degree of numerator \( n = 2 \), degree of denominator \( d = 1 \).
• Since \( n > d \) (2 > 1), this function has no horizontal asymptote (we'll check others to confirm).

Step3: Analyze \( f(x)=\frac{3x^2}{x^2 - 1} \)

• Degree of numerator \( n = 2 \), degree of denominator \( d = 2 \).
• \( n = d \), so horizontal asymptote is \( y=\frac{3}{1}=3 \).

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

• Degree of numerator \( n = 1 \), degree of denominator \( d = 2 \).
• \( n < d \), so horizontal asymptote is \( y = 0 \).

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

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

Answer:

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