how do you find horizontal asymptotes of implicit equations?

Question

i need the answer to the equation:
2*y^3 + 6*x^2*y - 12*x^2 + 6*y =1

Answer 1

Just look for values of y that can't be substituted in.

Firstly, rearrange to solve for x:
x^2(6y - 12) + (2y^3 + 6y - 1) = 0
x^2(6y-12) = (1-6y-2y^3)
x^2 = (1-6y-3y^2)/(6y-12).

So you'll get an asymptote at y = 2, because you can't substitute in y=2 and solve for x, but you can put in y=1.9999, etc.

Note there are lots of other values of y which can't be substituted in, because they lead to a negative on the right hand side. However, if some y makes it negative, any y close to that will also make it negative, so the graph isn't getting closer and closer to that value. The only special y value is 2, which makes the right hand side undefined.