Bode plot of Transfer Function
To determine slopes:

Starting from the left, only the transfer function zero is active as the two poles start contributing at their respective break frequencies (100/1000 rad/s).  Slope of a system with a single zero is +20 db/decade.  The first pole (lower frequency, 100 rad/s) begins to contribute at its break frequency (100 rad/s) and  reduces the overall gain slope to 0 dB/decade [+20 (zero) -20 (pole)].  The second pole at 1,000 rad/s then changes the gain slope to +20-20-20 = -20 dB/decade.

Each pole reduces the slope by 20 dB/decade and each zero increases the slope by 20 dB/decade.

To figure out the vertical position of the plot, one can look at 1/10th of the frequency of the lowest frequency pole/zero (but not located at the origin), calculate the gain at the point, and go from there using the slope criteria.

In this particular case, the lowest frequency pole is at 100 rad/s.  So choose 10 rad/s (1/10th of 100 rad/s) and plug the value in the transfer function:

\$H(10) = 100*10 = 1,000\$

In [dB] : \$20 * \log_{10}(1,000) = 60 \textrm{dB}\$

Note that the contribution of the two poles can be neglected:

\$H(10) = 100*10 / [(1+10/100)*(1+10/1000)] \approx 1,000\$

This particular graph starts 2 decades below the lowest break frequency:

\$ H(1) = 100*1 = 100\$

In [dB] : \$20 * \log_{10}(100) = 40 \textrm{dB}\$

Further help: Example 2 in this link shows a very similar transfer function and the way to develop its Bode plot:  http://lpsa.swarthmore.edu/Bode/BodeExamples.html