Question

find the horizontal asymptote, if any, of the graph of the rational function.

(f(x) = \frac{-6x}{3x^3 + x^2 + 1})

a. (y = 0)
b. (y = -\frac{1}{2})
c. (y = -2)
d. no horizontal asymptote

Explanation:

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<reasoning>

Identify the degrees of the numerator and denominator

Using the Horizontal Asymptotes and Rational Function End Behavior knowledge points
\[

$$\begin{aligned} &p(x) = -6x \implies \text{degree } n = 1\\ &q(x) = 3x^3 + x^2 + 1 \implies \text{degree } m = 3 \end{aligned}$$

\]

Compare the degrees to find the horizontal asymptote

Using the Horizontal Asymptotes and Rational Function End Behavior knowledge points
\[

$$\begin{aligned} &n < m \quad (1 < 3)\\ &\lim_{x \to \pm\infty} f(x) = 0 \implies y = 0 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) \(y = 0\)</mcq-correct>
<mcq-option>(B) \(y = -\frac{1}{2}\)</mcq-option>
<mcq-option>(C) \(y = -2\)</mcq-option>
<mcq-option>(D) no horizontal asymptote</mcq-option>
</answer>

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Answer:

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<reasoning>

Identify the degrees of the numerator and denominator

Using the Horizontal Asymptotes and Rational Function End Behavior knowledge points
\[

$$\begin{aligned} &p(x) = -6x \implies \text{degree } n = 1\\ &q(x) = 3x^3 + x^2 + 1 \implies \text{degree } m = 3 \end{aligned}$$

\]

Compare the degrees to find the horizontal asymptote

Using the Horizontal Asymptotes and Rational Function End Behavior knowledge points
\[

$$\begin{aligned} &n < m \quad (1 < 3)\\ &\lim_{x \to \pm\infty} f(x) = 0 \implies y = 0 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) \(y = 0\)</mcq-correct>
<mcq-option>(B) \(y = -\frac{1}{2}\)</mcq-option>
<mcq-option>(C) \(y = -2\)</mcq-option>
<mcq-option>(D) no horizontal asymptote</mcq-option>
</answer>

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