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

Explanation:

Step1: Recall Horizontal Asymptote Rules

For a rational function ( f(x) = \frac{N(x)}{D(x)} ), where ( N(x) ) is the numerator and ( D(x) ) is the denominator (both polynomials):

• If the degree of ( N(x) ) (( n )) is less than the degree of ( D(x) ) (( 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 ): No horizontal asymptote (may have an oblique/slant asymptote).

Step2: Analyze Each Function

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

• Degree of numerator (( n )): 1 (highest power of ( x ) in numerator is ( x^1 )).
• Degree of denominator (( d )): 2 (highest power of ( x ) in denominator is ( x^2 )).
• Since ( n = 1 < d = 2 ), horizontal asymptote is ( y = 0 ).

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

• Degree of numerator (( n )): 1.
• Degree of denominator (( d )): 1.
• Since ( n = d ), horizontal asymptote is ( y = \frac{1}{3} ) (leading coefficients: 1 (numerator) and 3 (denominator)).

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

• Degree of numerator (( n )): 2 (highest power ( x^2 )).
• Degree of denominator (( d )): 1 (highest power ( x^1 )).
• Since ( n = 2 > d = 1 ), no horizontal asymptote (check next function to confirm).

Function 4: ( f(x) = \frac{3x^2}{x^2 - 1} )

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

Answer:

( f(x) = \frac{2x^2}{3x - 1} ) (the third option: ( \boldsymbol{f(x)=\frac{2x^2}{3x - 1}} ))