Finding horizontal and vertical asymptotes

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SUMMARY

The discussion centers on finding horizontal and vertical asymptotes for the equation x² + xy + y² = 25. The user employs implicit differentiation, yielding the derivative y' = (-2x - y)/(x + 2y). They correctly identify that setting the numerator to zero results in y = -2x, but struggle to determine the existence of horizontal and vertical asymptotes. Ultimately, it is concluded that the only asymptotes present are oblique asymptotes, specifically y = x and y = -3x.

PREREQUISITES
  • Understanding of implicit differentiation
  • Knowledge of asymptotes in calculus
  • Familiarity with polynomial equations
  • Ability to analyze limits and behavior of functions
NEXT STEPS
  • Study the concept of oblique asymptotes in detail
  • Learn how to determine horizontal asymptotes for rational functions
  • Explore the implications of implicit differentiation on curve behavior
  • Investigate the graphical representation of asymptotes using graphing software
USEFUL FOR

Students and educators in calculus, mathematicians analyzing polynomial behavior, and anyone interested in understanding asymptotic analysis in functions.

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the equation is:

x^2 + xy + y^2 = 25

my method was to implicitely differentiate and then set the numerator = 0 to get points of horizontal asymptotes, and then set the denominator = 0 to get the points for the veritcal asymptotes.

when i use implicit differentiation i get:

y' = (-2x - y)/(x + 2y)

am i correct so far? now I am stuck because if i set the top equal to 0 i just get y = -2x. what can i do from here to solve for the points where there are both horizontal and vertical asymptotes? thanks.
 
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Do you have any reason to think the graph of that expression has horizontal and vertical asymptotes? I get that the only asymptotes are the oblique asymptotes y= x and y= -3x.
 
how can i know if it does have any to begin with?

and can you explain how you got those oblique asymptotes?
 

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