I need to find possible curves, with asymptotes given as $x=0 (x \to -\infty)$ and $y=mx \hspace{0.5cm} m>0$.
it is easy to find curves for individual lines, $y= \exp(-\lambda_1 x) + mx$ for
$y=mx$ and $y= \exp(\lambda_2 x)$ for $y=0$. And with piecewise definition of $y$ as,
\begin{eqnarray}
y= \exp(\lambda_2 x) \hspace{0.5cm} x<0 \\
y= \exp(-\lambda_1 x) + mx \hspace{0.5cm} x>0
\end{eqnarray}
Which is continuous at $x=0$ we have one family of curve. In this case however derivatives might not be continuous.
How can I proceed to find different types of curves for given asymptotes?
There are many many possible answers to your problem. But here's a general approach you could take:
Try looking at the derivative that you want. On the left ($x << 0$), it should be just about $0$; on the right ($x >> 0$), you want it to be just about $m$. So you're going to need a function that's flat on both sides of the origin, but rises as it passes through the origin. My personal favorite among these is $\arctan$, which goes from $-\pi/2$ to $\pi/2$. So if you let $$ g(x) = 1 + \frac{2}{\pi} \arctan(x) $$ then you get a function that's nearly 0 on the left half-axis, and nearly 1 on the right half-axis. You can then let $$ f(x) = \int_{-\infty}^x m g(t) dt $$ and you've got a function that has the derivatives you want. The only question is "What about the correct values?" Well, as $x \to -\infty$, the value of $f$ clearly goes to $0$, so we don't need to add a constant, and you've got a solution.
I want to add that there's no real merit in this solution except that it satisfies your requirements ... if your slopes come from some real data in the real world, there's no reason to believe that the interpolating function looks like this one. For instance, I could add any multiple of $$ k(x) = \frac{1}{1 + x^{2k}} $$ (for any positive integer $k$) to my answer and still have a perfectly good answer that satisfies all the requirements you gave.