Plotting bode plot, how the starting gain is in negative?
Your transfer function needs to be rewritten in a low-entropy format as recommended by the fast analytical circuits techniques or FACTs. Factor \$10s\$ in the numerator and \$\frac{s}{2}\$ in the denominator and simplify by \$s\$:

\$H(s)=\frac{10s}{(1+\frac{s}{2})(1+\frac{s}{10})}=\frac{10}{0.5}\frac{1}{(1+\frac{1}{0.5s})(1+\frac{s}{10})}=H_0\frac{1}{(1+\frac{\omega_{p1}}{s})(1+\frac{s}{\omega_{p2}})}\$ with:

\$H_0=20\$ or 26 dB, \$\omega_{p1}=2\;rad/s\$ and \$\omega_{p2}=10\;rad/s\$

This way, you see a plateau gain of 26 dB while the zero being at the origin (for \$s=0\$) is merged into the inverted pole \$\omega_{p1}\$. This is the best way to write the transfer function and see the gain in between the two poles as this might very likely be your design target in a filter design. If you do not rewrite the function in this format, you do not reveal it immediately.

The zero lies in the origin and for \$s=0\$, the magnitude is 0. Therefore, as you approach the origin on the x-axis, you see an attenuation indicated by the negative sign which appears when extracting the log of the transfer function magnitude at this point. The 40-dB attenuation or -40-dB magnitude read at 0.001 rad, simply indicates that the stimulus (your input signal), is divided by 100 at this point.