Answer:

Step-by-step explanation:
It is given that the graph corresponds to a natural logarithmic function.
That means, the function
has a natural log (Log with base
) of some terms of x.
It is given that asymptote of given curve is at
. i.e. when we put value
, the function will have a value
.
We know that natural log of 0 is not defined.
So, we can say the following:
is not defined at

i.e.
is the point where 
a = 3
Hence, the function becomes:

Also, given that the graph crosses x axis at x = -2
When we put x = -2 in the function:

And y axis at 1.
Put x = 0, we should get y = 1

So, the function is: 