Question

what are the vertical and horizontal asymptotes for the function \\(f(x) = \frac{x^2 + x - 6}{x^3 - 1}\\)?

vertical asymptote: \\(x = 1\\)
horizontal asymptote: none

vertical asymptote: \\(x = 1\\)
horizontal asymptote: \\(y = 0\\)

vertical asymptote: \\(x = -2, x = 3\\)
horizontal asymptote: \\(y = 0\\)

vertical asymptote: \\(x = -2, x = -3\\)
horizontal asymptote: none

Explanation:

Find vertical asymptotes

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Since \(x = 1\) does not make the numerator \(x^2 + x - 6\) zero:

$$ \text{vertical asymptote: } x = 1 $$

Find horizontal asymptotes

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$$ \text{horizontal asymptote: } y = 0 $$

Answer:

• vertical asymptote: \(x = 1\)

horizontal asymptote: none

• **vertical asymptote: \(x = 1\)

horizontal asymptote: \(y = 0\) (Correct answer)**

• vertical asymptote: \(x = -2, x = 3\)

horizontal asymptote: \(y = 0\)

• vertical asymptote: \(x = -2, x = -3\)

horizontal asymptote: none