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?
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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
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?
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