Effect of Rc in Transistor Amplifier

So why does the point with the 30 Ohm stay above 8 V while the other returns to 0?

Think of the transistor as another resistor in series with the 30 Ohm resistor; because it has a voltage drop, it has some resistance, so it's like a voltage divider!!! This voltage drop across the CE region is dependant upon the voltage present at the base (almost like an electrically controlled potentiometer), which is usually a minimum of 0.75V, as the base voltage increases, the Voltage drop across CE increases, thus leaving a voltage equal to Vcc-Vce across the resistor... which is why it functions as an inverter! Bare with me, it'll make sense in just minute. Now consider that both transistors have the same input voltage... this means the only other factor affecting this minimum 8V at the 30 Ohm resistor is the value of the resistor! Because both transistors have the same varying base voltage, then they should have the same voltage drop across Vce...... thus, THEY BOTH HAVE THE SAME RESISTANCE AT ALL TIMES! And if the other resistors with the transistors form voltage dividers whose value of R2 (the transistors) stay the same, but the R1's are different, then the only reason why the lower peak voltages differ is because the Rc resistors are different values, meaning they hold a different voltage that is divided at that point!!! There we have it, that's why! Did this help?

After rereading what I just wrote, I realized that was incorrect about the voltage drop, I meant to say Vbe, not Vce. Sorry if that was confusing

No you said Vce and you mean Vce not Vbe. Why the next sentence? "Thus leaving a voltage equal to Vcc-Vce...." so again Vce which you said before. Not easy to explain things if youself dont understand them. Issac you are funny, but must talk to anything on everycircuit, very funny.