An exponential question (1 Viewer)

[Aerlinn]

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?
:wave:

 

[ssglain]

Take limits?
as x -> infinity, f(x) -> infinity
as x -> - infinity, f(x) -> 0 from below
So x=0 is the horiz. asymptote.

 

[haque]

y=0 is the horizontal asymptote which is seen from ssglain's reasoning.

 

[Aerlinn]

Ah... ok... can that be done algebraically?

 

[haque]

yeah it can be shown algebraically-but in the hsc all we learn is tat exponential functions increase or decrease at a greater rate than linear functions.
However we could use L'Hopital's rule for indeterminate forms, in this case we can write it as -infinity/infinity so let
y=lim(xgoes to -infinity) (x-2)/2^(1-x)=lim(x to -infinity) 1/-ln2*2^(1-x) and as x goes to infinity the bottom expression's absolute value approaches infinity and the fraction is effectively zero. L'Hopital's rule isn't in the hsc thouhg.

 

[ssglain]

y=0 is the horizontal asymptote which is seen from ssglain's reasoning.

Oops. Y=0. Thanks for picking out the typo.

 

[Aerlinn]

y=lim(xgoes to -infinity) (x-2)/2^1-x

Isn't that meant to be y=lim(xgoes to -infinity) (x-2)2^(1-x) ?
Multiplication, not division, and if you don't put brackets in it changes the equation... ^^

 

[haque]

For L'Hopitals rule we express in the form of -infinity on infinty in this case so i've written (x-2) divided by 2^(1-x ) and then using the rule we differentiate "top and bottom individually". The expression (x-2)2^(1-x )isn't the same as (x-2)2^(x-1). I should have put brackets but forgot to in my haste-but my purpose was to convey the main idea on how it could have been done.