Find the slant asymptote for $x= \dfrac{3t}{1+t^3}$, $y= \dfrac{3t^2}{1+t^3}$

Question

I am geting $y=-x$, while the answer $y=-x-1$

What i did is to write $y = tx$ and as $t$ goes to $-1$. $x, y$ go to $\infty$, hence the asymptote is $y=-x$

Where did i go wrong.

Answer 1

I must confess that I really do not know how this topic has been teached to you. So, I give you my approach and forgive me if it does not fit in what you are supposed to do.

As you noticed, all the problem is around $t=-1$. So, let us look at the Taylor McLaurin expansions around this point. What we obtain is $$x= \dfrac{3t}{1+t^3}=-\frac{1}{t+1}+\frac{t+1}{3}+\frac{1}{3} (t+1)^2+O\left((t+1)^3\right)$$ $$y= \dfrac{3t^2}{1+t^3}=\frac{1}{t+1}-1-\frac{t+1}{3}+O\left((t+1)^3\right)$$ which confirms the answer.