Answer:
Option B - Vertical asymptote: x = 1, horizontal asymptote: y = 0
Step-by-step explanation:
Given : Function [tex]\frac{x^2+x-6}{x^3-1}[/tex]
To find : What are the vertical and horizontal asymptotes for the function ?
Solution :
Function [tex]\frac{x^2+x-6}{x^3-1}[/tex]
For vertical asymptote,
We equate the denominator to zero,
[tex]x^3-1=0[/tex]
[tex]\Rightarrow x^3=1[/tex]
[tex]\Rightarrow x=1[/tex]
So, x=1 is the vertical asymptote.
For horizontal asymptote,
We compare the degree of numerator and denominator.
Degree of numerator is 2 and degree of denominator is 3.
When degree of denominator is greater than degree of numerator then y=0 is the horizontal asymptote.
So, y=0 is the horizontal asymptote.
Therefore, Option B is correct.
