What are the vertical and horizontal asymptotes for the function f(x)= x^2+x-6/x^3-1?
a) vertical asymptote: x = 1
horizontal asymptote: none
b) vertical asymptote: x = 1
horizontal asymptote: y = 0
c) vertical asymptote: x = –2, x = 3
horizontal asymptote: y = 0
d) vertical asymptote: x = –2, x = –3
horizontal asymptote: none

Option B

Step-by-step explanation:

Given

f(x)=  (x^2+x-2)/(x^3-1)

For vertical asymptotes we have to put the denominator of the function equal to zero:

x^3-1=0

x^3=1

So,

x=1

As the degree of denominator is greater than the degree of numerator, there will be only horizontal asymptote which will be y=0 because all the y values will be dragged down to the x-axis.  

So the horizontal asymptote: y=0

And Vertical asymptote: x =1

i belive those are the ones that are under 90 degreese.

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point