Finding horizontal and vertical asymptotes

1. Oct 26, 2006

dnt

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 im 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.

2. Oct 26, 2006

HallsofIvy

Staff Emeritus
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.

3. Oct 28, 2006

dnt

how can i know if it does have any to begin with?

and can you explain how you got those oblique asymptotes?