What is the amplification factor for a 6J5 vacuum tube under given conditions?

[Acuben]

Homework Statement

What is amplification factor?

In case you need to know the context to answer this question, I'll post the homework problem here
a) Plate characteristics for the 6J5 vacuum-tube are shown in ___ http://img301.imageshack.us/f/6j5loadline60klu9.png/
Find the plate current in the triode shown here
(btw this isn't my picture, I just googled it, but it seems very similar to the one I had on the problem)
b) Find the amplification factor of the tube under the conditions in part a)I'll post the picture later if needed,. It's a triode vacuume tube with battery on the left side of 8v making counterclockwise current
and on the right side,
there is an resistor with 25k Ohms, and 350v battery making counterclockwise current as well
(if it's not enough information, let me know)

answer for part a) is 5.6mA under 210 Potential different (it does not have to be exact, but an estimate)
I have no problems for part a
answer for part b) approx: 20

Homework Equations

Load line equation: E-IR = 350(volts)-I*25000(Ohms)

The Attempt at a Solution

I don't know what amplification factor is o.oedit: nvm just got it.
it was
(let u be mu)
u= delta V(b) / delta V

 

[lippyka]

(a)where delta V(b) is the change in the output voltage and delta V(a) is the change in the input voltageu = Vout/Vinu(amplification factor) = 210V/8V = 26.25