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Consider $$ x(t) = 2 e^{-t} + 3e^{2t}$$

$$y(t) = 5 e^{-t} + 2 e^{2t}$$

which represents a non rectilinear paths

Horizontal and Verical Asymptotes :

If $t \rightarrow +\infty \ \ or \ \ -\infty$, then $x(t) \ \ and \ \ y(t) \ \ \rightarrow \infty$, So there are no asymptotes parallel to coordinate axis

oblique Asymptotes:

Please tell me how to find the Oblique asymptotes

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There are no horizontal asymptotes: this would mean $x\to\infty$ and $y\to$ some finite value. For obligue asymptotes look at the limit when $t\to\pm\infty$ of $y/x$. This is a plot of the curve.

enter image description here

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  • $\begingroup$ you are finding the slope of the oblique asymptotes two different ways which one is correct or both correct . oblique asymptote is $y = mx + c$ and how to find the value of c. $\endgroup$ – user120386 Feb 15 '15 at 10:40
  • $\begingroup$ There is one oblique asymptote at $+\infty$ and another at $-\infty$. You find $c$ as $\lim_{t\to\pm\infty}y-m\,x$. $\endgroup$ – Julián Aguirre Feb 15 '15 at 12:07
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There are two asymptotes by inspection which are at an angle to x-axis.

We need to find out not $ \frac{y(t)}{x(t)} $ tendency but tendency of limits of oblique asymptotes.

$ \dfrac{dy/dt}{ dx/dt}$ when $t\to +\infty $

and also

$ \dfrac{dy/dt}{ dx/dt}$ when $t\to -\infty $

separately.

These limits evaluate to $5/2$ and $2/3$ for each asymptote as coefficients for positive and negative exponents.

Better to make parametric plot of the curve and an ordinary plot of the slope.

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