The extreme slope of the approach to the vertical asymptote suggests a log function. The plain log function will have x=0 as its vertical asymptote and will go through the point (1, 0). The base of the logarithm (b) can be determined by the fact that a log function will go through the point (b, 1). The graph has negative slope, also suggesting the function is reflected across the x-axis.
After we vertically reflect across y=0 and shift your graph to the right one place, it goes through the points (1, 0) and (2, 1). This tells us it is ...
[tex]\displaystyle y=-\log_2{(x+1)}[/tex]
