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

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

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

Answer

Answer:

B. vertical asymptote: $x = 1$; horizontal asymptote: $y = 0$

Explanation:

Step1: Find vertical asymptotes

Set denominator $x^{3}-1 = 0$. Since $x^{3}-1=(x - 1)(x^{2}+x + 1)$ and $x^{2}+x + 1$ has no real - roots ($\Delta=1^{2}-4\times1\times1=- 3<0$), the real root of $x^{3}-1 = 0$ is $x = 1$. 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 $y = 0$.