Finding horizontal and vertical asymptotes

[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 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.

 

[HallsofIvy]

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.

 

[dnt]

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

and can you explain how you got those oblique asymptotes?