This equation: f(x)= (x-2)2^(x-1)
Says to state the equation of the horizontal asymptote.

How do I do I do this without using a calculator?

Well the asymptote is horizontal, so it's of the form y = n, where you need to find n.

You need to find the point where x --> infinity

And... how is that done... algebraically?

we need to take limits when x goes to + and - infinity

\lim_{x\to\infty}f(x)=(\infty -2)2^{\infty -1}=\infty
\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}x2^x=0 \;\text{because 2^x shrinks faster than x grows}
Therefore, the horizontal asymptote is y=0.

A plot of this function confirms this result.

I see... Thanks :)
I don't understand the lim x2^x=0 bit however... :S

you don't understand limits or you don't understand why x2^x is going to 0?

Don't understand why x2^x is going to 0...

The basic idea is that as x becomes negative 2^x is going to be getting closer and closer to 0 and x is going to be getting more and more negative. However 2^x is going to be going to 0 faster than x is going to - infinity so 2^x wins.

Try a few examples:
-1 * 2^-1 = -1 * 1/2 = -1/2 = -0.5
-2 * 2^-2 = -2 * 1/4 = -1/2 = -0.5
-3 * 2^-3 = -3 * 1/8 = -3/8 = -0.375
-4 * 2^-4 = -4 * 1/16 = -1/4 = -0.25
-5 * 2^-5 = -5 * 1/32 = -5/32 = -0.15625
-6 * 2^-6 = -6 * 1/64 = -3/32 = -0.09375

you can see its getting smaller and smaller. The magnitude of x is increasing by 1 each time but the magnitude of 2^x is halving each time.

do you follow now?