Monday

April 21, 2014

April 21, 2014

Posted by **Anonymous** on Monday, October 12, 2009 at 7:07pm.

r(x)= 2x-1/ x^2-x-2 = 2x-1/ (x-2)(x+1)

s(x)= x^3+27/x^2+4 = (x+3)(x^2-3x+9)/(x-2)(x+2)

t(x)=x^3-9x= x(x-3)(x+3)/x+2

u(x)=x^2+x-6/x^2-25= (x+3)(x-2)/(x+5)(x-5)

I do not know how to determine the asymptotes. Please help me. I would sincerely appreciate it. Thanks!

- Calculus -
**MathMate**, Monday, October 12, 2009 at 8:28pmVertical asymptotes are typically present in functions with a polynomial denominator. The vertical asymptote is located where the value of x is such that the denominator is zero. If the denominator is not a polynomial, the same applies whenever the denominator becomes zero.

To find horizontal asymptotes, you would divide the polynomials by long division.

If the result is of the form p+q/P(x), where p and q are integers and P(x) is a polynomial, the horizontal asymptote is y=p, since the second term vanishes when P(x) becomes infinity.

If the result of the long division is px+q+r/P(x), then the line y=px+q is the oblique asymptote, again, the last term vanishes as x becomes infinity.

You can conclude that the function has no vertical asymptote when the denominator does not vanish for all real values of x.

If you need a check on the answers, feel free to post.

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