Question

find the horizontal asymptote, if any, of the graph of the rational function.
$g(x) = \frac{12x^2}{3x^2 + 1}$
$\bigcirc$ $y = \frac{1}{4}$
$\bigcirc$ $y = 0$
$\bigcirc$ $y = 4$
$\bigcirc$ no horizontal asymptote

Explanation:

Step 1: Recall the rule for horizontal asymptotes of rational functions

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), where \( N(x) \) is the numerator and \( D(x) \) is the denominator, we consider the degrees of \( N(x) \) and \( D(x) \). Let the degree of \( N(x) \) be \( n \) and the degree of \( D(x) \) be \( m \).

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

Step 2: Determine the degrees and leading coefficients

For the function \( g(x)=\frac{12x^{2}}{3x^{2}+1} \), the numerator \( N(x)=12x^{2} \) has degree \( n = 2 \) and leading coefficient \( 12 \). The denominator \( D(x)=3x^{2}+1 \) has degree \( m = 2 \) and leading coefficient \( 3 \).

Step 3: Calculate the horizontal asymptote

Since \( n = m \), we use the formula \( y=\frac{\text{leading coefficient of } N(x)}{\text{leading coefficient of } D(x)} \). Substituting the values, we get \( y=\frac{12}{3}=4 \).

Answer:

C. \( y = 4 \) (assuming the options are labeled A, B, C, D with A: \( y=\frac{1}{4} \), B: \( y = 0 \), C: \( y = 4 \), D: no horizontal asymptote)