Question

What are the vertical and horizontal asymptotes for the function f(x)= x^2+x-6/x^3-1?

Answer 1

The horizontal asymptotes will occur when x approaches ±oo

The easy way to see this is to divide all terms by the highest order variable...

(x^2/x^3 +x/x^3-6/x^3)/(x^3/x^3-1/x^3) and all of that mess boils down to

0/1 or just 0 as x approaches ±oo

So the horizontal asymptote is the horizontal line y=0

The vertical asymptote will occur when there is division by zero, which is undefined because it is not a real value.

x^3-1=0

(x-1)(x^2+x+1)=0

So this only occurs as x approaches -1, so the vertical asymptote is the vertical line x=-1.

Answer 2

Hence, x=1 is the vertical asymptote

Hence, y=0 is the horizontal asymptote

Explanation:

We have been given an expression

`(x^2+x-6)/(x^3-1)`

For vertical asymptote we equate the denominator to zero that means

`x^3-1=0`

`\Rightarrow x^3=1`

`\Rightarrow x=1`

Hence, x=1 is the vertical asymptote.

Now, for horizontal asymptote we compare the degree of numerator and denominator

Here degree of numerator is 2 and degree of denominator is 3

When degree of denominator is greater than degree of numerator y=0 is the horizontal asymptote

Hence, y=0 is the horizontal asymptote.