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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: enter image description here

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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up vote 4 down vote accepted

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 $y\to 1^-$.

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So, the graph of the parametric equations will be the same as the rectangular equations, even down to the asymptotes? – Mack Nov 23 '12 at 23:26
But parametric equations allow us to see the orientation of the graph, isn't that correct? – Mack Nov 23 '12 at 23:30
Thank you Will! – amWhy Nov 24 '12 at 19:24

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