SOLUTION: Find the horizontal and vertical asymptotes of R(x)= x^3-1/x^3+1

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Question 138276: Find the horizontal and vertical asymptotes of R(x)= x^3-1/x^3+1
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) About Me  (Show Source):
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Find the horizontal and vertical asymptotes of R(x)= x^3-1/x^3+1
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Horizontal asymptotes: as x approaches inf, R(x) approaches x^3/x^3 = 1
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Vertical asymptote:
Occurs when x^3+1 = 0
x^3 = -1
x = -1
Vertical asymptote at x=-1
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Cheers,
Stan H.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
To find the vertical asymptote, you need to find values of x that would make the denominator of the rational expression go to zero. Use the sum of two cubes rules to factor the denominator. You will get a 1st degree binomial that has a zero, and a quadratic that has no real roots. The zero of the binomial is the only value, a, that will make the denominator zero, so the equation of the vertical asymptote is x=a. I'll let you figure out what value to substitute for a.

For any rational function there is either 1 horizontal asymptote or none.

If the degree of the denominator polynomial is greater than the degree of the numerator, then the horizontal asymptote is y=0.

If the degree of the denominator polynomial is equal to the degree of the numerator polynomial, then the horizontal asymptote is y=p%2Fq where p is the lead coefficient on the numerator and q is the lead coefficient on the denominator.

If the degree of the denominator is less than the degree of the numerator, then there is no horizontal asymptote. If the degrees differ by 1 in this case, there is a straight line slant or oblique asymptote equal to the polynomial long division quotient of the numerator divided by the denominator excluding any remainder.