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

User Neetu
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2 Answers

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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.
User Chaine
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8.8k points
5 votes

Answer:

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.

User Daniel Vogelnest
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