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

the slope of the line is 60

one point on the line is (40,10)

step-by-step explanation:

in the equation outside the parenthesis the equation always tells you the slope. in this case it is 60.

in the parenthesis is also says 40 which is x because it is near the x and 10 is y because it is near y.

what "following statement"?

step-by-step explanation: