dB values are never multiplied by \$0.707 \approx \sqrt{\frac{1}{2}}\$. Rather, a -3 dB drop occurs, relative to some maximum, when the signal voltage drop to 0.707 of that maximum. In other words, 70.7% voltage is -3dB.
What is special about -3dB is that it represents half the power. If the voltage drops to \$\sqrt{\frac{1}{2}}\$, then since power is proportional to the square of voltage, power has dropped to \$\frac{1}{2}\$.
Actually, half power is not exactly -3dB; this is an approximation. The true value is \$10\log_{10}\frac{1}{2} = -3.010299...\$.
The multiplier \$10\$ becomes \$20\$ when we calculate decibels using voltage, because of the squaring. We want the decibels corresponding to \$\sqrt{\frac{1}{2}}\$ voltage to be the same as those for \$\frac{1}{2}\$ power, thus:
$$20\log_{10}\sqrt{\frac{1}{2}} = 20\log_{10}\left(\frac{1}{2}\right)^{1/2} = \frac{1}{2}20\log_{10}\frac{1}{2} = 10\log_{10}\frac{1}{2}$$
If your graph shows dB values, then you can find the -3 dB points directly, relative to the amplitude across the amplifier's mid-band. If the graph shows linear signal level as a voltage, then you use 0.707. If the graph shows power, then use 0.5. You must know what units are on the dependent axis of the graph, of course.
