Answer:
vertical asymptote at x = 1
horizontal asymptote at y = 0
Step-by-step explanation:
[tex]f(x)=\frac{x^2+x-6}{x^3-1}[/tex]
WE factor both top and bottom
x^2 + x-6 can be factored as (x+3)(x-2)
Now we factor x^3 -1
x^3-y^3= (x-y)(x^2+xy+y^2)
x^3 - 1^3=(x-1)(x^2+x+1)
replace the factors
[tex]f(x)=\frac{(x+3)(x-2)}{(x-1)(x^2+x+1)}[/tex]
To find vertical asymptote , we set the denominator =0 and solve for x
x^3 -1 =0
x^3 = 1
Now take cube root
x = 1
So vertical asymptote at x=1
To find horizontal asymptote we look at the degree of numerator and denominator
Degree of numerator =2 and degree of denominator = 3
Degree of numerator is smaller than the degree of denominator so horizontal asymptote at y=0
