How can I guess a function just by looking at the graph of it? 
I have been asked to guess function of this graph.
Can anyone please tell me how can I say this is a graph of an exponential function ?
Is there any other function which looks like this ?
 A: There really isn't enough information in that graph to know whether it is the graph of an exponential function. The best that we could do is to try to decide, based on the shape of the graph, and on estimation of a few values of the function, whether this graph is consistent with an exponential function; one might also say that $f$ is approximately an exponential function or that $f$ is modelled by an exponential function.
First let's look at the shape of the graph: It is rapidly decreasing until it gets close to the positive $x$-axis, and from there onward it looks like it is asymptotic to the positive $x$-axis. So far that would be consistent with an exponentially decreasing function of the form
$$(*) \qquad f(x) = C b^{-x}
$$
for some $C>0$ and $b>1$.
Now let's test at some values of the function. There's not that many values to test.
To start with, we can certainly estimate $f(0) \approx 5$, which is perhaps accurate to within $.1$. Using formula $(*)$ we would get $5 = f(0) = C b^0 = C$ and so we would get
$$f(x) = 5 b^{-x}
$$
Next, perhaps we can estimate $f(1) \approx 1.5$, although now the accuracy does not look any better than $.5$.
Finally, perhaps we can also estimate $f(-1) \approx 16$, although now the accuracy is even worse, not better than $1$ or $2$.
Okay, what can we conclude from this data?
\begin{align*}
f(-1) &\approx 16 \\
f(0) &\approx 5 \\
f(1) &\approx 1.5
\end{align*}
In particular, we see that for each increment of $x$ by a single unit, first from $-1$ to $0$, and then from $0$ to $1$, the value of $f$ decreases by a factor of about $3$. That is consistent with $b=3$, and so it looks like $f$ can be modelled by the exponential function
$$f(x) \approx 5 \cdot 3^{-x}
$$
A: 
How can I guess a function just by looking at the graph of it?

In general, you can't. There are usually multiple different functions that would be equally plausible for a graph.
However, in very simple cases like yours it may be possible to say one hypothesis is the only sensible candidate and any other explanation is weeded out by Occam's razor.
Particularly telling for this graph is that it is for most of the positive range very close to 0. Now, there are a couple of different functions that have 0-asymptotes; broadly they can be sorted into inverse-polynomial and exponential asymptotes. Your function approaches the x-axis fast, that could either be an inverse-like function with reasonably high power, or just about any exponential function, which always converge quickly to 0 unless the basis is very small.
$$
  y = 5 \cdot 3^{-x}
$$

A rational function would need extra care to avoid producing poles, i.e. discontinuities where the function then jumps back. The example graph does of course shoot over the scale, which might also happen for a rational function, but a rational function always jumps then:
$$
  y = \frac{405}{(x+3)^4}
$$

Avoiding the discontinuity itself is easy enough in case of an even exponent: just add an offset to keep the denominator always positive
$$
  y = \frac{405}{(x+3)^4 + 1}
$$

...but that still doesn't stop the function from “returning” for large negative values. In fact, that is not possible to prevent with a rational function alone, though it can be easily enough prevented with a piecewise-definition:
$$
  y = \begin{cases}
          \frac{405}{(x+3)^4} & \text{for $x < -2$}
       \\ -1620\cdot x -2835  & \text{else}
      \end{cases}
$$

There are also other functions with similar shape, e.g.
$$
  y = \frac58 \cdot \left(\sqrt{x^2 + 4} - x\right)^3
$$

So, there we have actually some pretty convincing alternative explanations, but both are quite a bit more verbose to write than the exponential one. And if you look really closely, you'll see that they matches the graph worse too, namely they don't converge quickly enough but have a too steep derivative at $x=0$. That could be improved still, but would make the definitions even more complicated.
