Question

find the equation of all horizontal asymptotes (if any) of the rational function.
$f(x) = \frac{6}{2x^3 + 8x^2 - 6}$
select the correct choice below and fill in any answer boxes within your choice.
a. the function has one horizontal asymptote.
(simplify your answer. type an equation. use integers or fractions for any numbers in the equation.)
b. the function has two horizontal asymptotes. the top asymptote is \\(\square\\), and the bottom asymptote is \\(\square\\).
(simplify your answers. type equations. use integers or fractions for any numbers in the equations.)
c. the function has no horizontal asymptotes.

Explanation:

Step1: Analyze degrees of numerator and denominator

The function is \( f(x)=\frac{6}{2x^{3}+8x^{2}-6} \). The degree of the numerator (highest power of \( x \)) is \( 0 \) (since \( 6 = 6x^{0} \)), and the degree of the denominator is \( 3 \) (from \( 2x^{3} \)).

Step2: Apply horizontal asymptote rules

For a rational function \( \frac{N(x)}{D(x)} \):

• If \( \text{deg}(N) < \text{deg}(D) \), the horizontal asymptote is \( y = 0 \).
• If \( \text{deg}(N)=\text{deg}(D) \), it's \( y=\frac{\text{leading coefficient of }N}{\text{leading coefficient of }D} \).
• If \( \text{deg}(N) > \text{deg}(D) \), there's no horizontal asymptote (or an oblique one).

Here, \( \text{deg}(N)=0 < \text{deg}(D)=3 \), so horizontal asymptote is \( y = 0 \). Also, since \( \text{deg}(N) < \text{deg}(D) \), there's one horizontal asymptote.

Answer:

A. The function has one horizontal asymptote. \( y = 0 \)