SOLUTION: Asymptote questions (x+y)²(x+2y+2)-x-9y-2=0 x³+y³-3axy=0 y²(x²-a²)=x²(x²-4a²) x³-2y³+xy(2x-y) +y(x-1)+1=0 x²y²-a²(x²+y²)-a³(x+y)+a ki power 4 =0

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Question 1023991: Asymptote questions
(x+y)²(x+2y+2)-x-9y-2=0
x³+y³-3axy=0
y²(x²-a²)=x²(x²-4a²)
x³-2y³+xy(2x-y) +y(x-1)+1=0
x²y²-a²(x²+y²)-a³(x+y)+a ki power 4 =0
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
We want to know the behavior of the function as x goes to infinity and -infinity
a. Divide both sides of %28x%2By%29%5E2%28x%2B2y%2B2%29-x-9y-2=0+ by x%5E3. We get

Upon letting x-%3Einfinity or x-%3E-infinity, the equation becomes
%281%2By%2Fx%29%5E2%281%2B%282y%29%2Fx%29+=+0.
Now let w = y/x.==> %281%2Bw%29%5E2%281%2B2w%29+=+0. Solving for w, we get w = -1 or w = -1/2. Hence the asymptotes are y = -x and y = -x/2.
b. +x%5E3%2By%5E3-3axy=0. Divide both sides again by x%5E3.
==> 1%2B%28y%2Fx%29%5E3+-+3a%28y%2Fx%29%281%2Fx%29+=+0. As x goes to plus or minus infinity, the equation becomes 1%2B%28y%2Fx%29%5E3+=+0.
==> %28y%2Fx%29%5E3+=+-1 . Let w = y/x, and so w%5E3+=+-1.
==> w = -1, or y/x = -1. Hence the asymptote is y = -x.
d. x%5E3-2y%5E3%2Bxy%282x-y%29+%2By%28x-1%29%2B1=0
Again divide both sides by x%5E3.
The equation becomes . Letting x go to plus or minus infinity, the equation becomes
1+-+2%28y%2Fx%29%5E3%2B%28y%2Fx%29%282-y%2Fx%29+=+0. Let y/x = w.
==> 1+-+2w%5E3%2Bw%282-w%29+=+0 <==> 0+=+2w%5E3%2Bw%5E2+-+2w-1=0
Solving for w, we get w = 1, -1, or -1/2.
Hence the asymptotes are y = x, y = -x, and y = -x/2.
I leave parts c and d to you.