Question

what are the vertical and horizontal asymptotes for the function $f(x)=\frac{x^{2}+x - 6}{x^{3}-1}$? vertical asymptote: $x=-2,x = 3$ horizontal asymptote: $y = 0$ vertical asymptote: $x = 1$ horizontal asymptote: $y = 0$ vertical asymptote: $x=-2,x=-3$ horizontal asymptote: none vertical asymptote: $x = 1$ horizontal asymptote: none

Explanation:

Step1: Find the vertical asymptote

Set the denominator equal to zero: \(x^{3}-1 = 0\).
Factor using the formula \(a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})\), so \(x^{3}-1=(x - 1)(x^{2}+x + 1)\).
The quadratic \(x^{2}+x + 1\) has discriminant \(\Delta=1^{2}-4\times1\times1=-3<0\).
The root of \(x^{3}-1 = 0\) is \(x = 1\). So the vertical asymptote is \(x = 1\).

Step2: Find the horizontal asymptote

For a rational function \(y=\frac{f(x)}{g(x)}\) where \(f(x)=x^{2}+x - 6\) (degree \(n = 2\)) and \(g(x)=x^{3}-1\) (degree \(m=3\)).
Since \(n

Answer:

vertical asymptote: \(x = 1\); horizontal asymptote: \(y = 0\) (the second option)