Date: 10/16/2001 at 10:23:27 From: Amrita Subject: Slant asymptotes. I love the idea of your Web site - the archives are really helpful. I need to know how to find slant asymptotes. For example, does this problem have slant asymptotes? f(x)=(x^(4/3)+x^(1/3)-2)/(x^(4/3)-16). I'm also having trouble taking the derivative of this problem: f(x)=((x^2(x+1))-(1-x))^(1/2) I've started it, but it's so long I get confused. Thanks a million! -Amrita
Date: 10/16/2001 at 14:48:51 From: Doctor Rob Subject: Re: Slant asymptotes. Thanks for writing to Ask Dr. Math, Amrita. One way to find nonvertical asymptotes is to take the derivative of f(x) and see what lim f(x)/x, x->infinity is. If it exists and equals m, that is the slope of an asymptote. The y-intercept b of the asymptote is then found as lim [f(x)-m*x] = b. x->infinity Then the asymptote is y = m*x + b. For example, the function f(x) = (3*x^2+1)/x has m = lim (3*x^2+1)/x^2, x->infinity lim 3 + 1/x^2 = 3. x->infinity b = lim [(3*x^2+1)/x-3*x], x->infinity = lim 1/x = 0. x->infinity Thus the line y = 3*x is a nonvertical asymptote. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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