what are the vertical and horizontal asymptotes for the function f(x)=(x^2 + x - 6)/(x^3 - 1)? vertical…

Explanation:

Step1: Find vertical asymptotes

Set the denominator equal to zero: $x^{3}-1 = 0$. Using the formula $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$, we have $(x - 1)(x^{2}+x + 1)=0$. The real - root of $x^{3}-1 = 0$ is $x = 1$ (since the discriminant of $x^{2}+x + 1$ is $\Delta=1^{2}-4\times1\times1=- 3<0$ and has no real roots). So the vertical asymptote is $x = 1$.

Step2: Find horizontal asymptotes

Degree of numerator $n = 2$ and degree of denominator $m=3$. When $n<m$, $\lim_{x\rightarrow\pm\infty}\frac{x^{2}+x - 6}{x^{3}-1}=0$. So the horizontal asymptote is $y = 0$.

Answer:

vertical asymptote: $x = 1$ horizontal asymptote: $y = 0$