# Sketching A Plane Curve

The problem I am working on is, "Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter."

The parametric equations are: $x=t-3$ and $y=\Large\frac{t}{t-3}$

The rectangular function: $y=1+\Large\frac{3}{x}$

Here is the graph from the solution manual:

In my graph, I didn't include the horizontal asymptote. How would I deduce the horizontal from the parametric equations? Would I just take the limit as t went to $\pm \infty$?

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Using only the parametric equations, we know that as $x\to +\infty$, we have $t\to +\infty$ so that $y\to 1^+$. Similarly, as $x\to -\infty$, we have $t\to -\infty$ so that $y\to 1^-$. At the same time from $y=\frac{t}{t-3}$ we see that $y$ cannot take the value $1$.
Of course, one can just use the cartesian equation $y=1+\frac{3}{x}$ to deduce this too. Since $\frac{3}{x}$ cannot take the value $0$, $y$ cannot take the value $1$. As $x\to +\infty$, we have $y\to 1^+$. Similarly, as $x\to -\infty$, we have $y\to 1^-$.