Horizontal asymptote HELP (1 Viewer)

Triage

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This may be an easy question but I just cannot understand the explanation.

Find the horizontal asymptote of these functions by dividing through by the highest power of x in the denominator.

F(x) = (5-x)/(4-2x)

Cheers
 

zeebobDD

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lol wtf was i thinking
 
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Timske

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so the highest power of x in the denominator is 1, since its just x

f(x) =[ (5/x)-1 ]/ [(4/x)-2]

then the limit is as x approaches infinity 5/x and 4/x become 0, so the asymptote will be -1/-2 = 2
^ vertical asymptote,

4-2x =/= 0
-2x =/= -4
x =/= 2
 

Sy123

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Ok since I cant find that long post on asymptotes I made, Ill give a brief explanation:

Take our function:



The horizontal asymptote will be

And this can be derived from mere observation as well. Take any very large values of x, k and m are really insignificant.
Now we have in the numerator a multiplied by our very large number, in denominator, b multiplied by very large number.
These very large numbers 'cancel' out and then we are left with a/b

For functions such as:



Horizontal asymptote will always be zero, since the denominator will largely dominate the numerator (in terms of value). Hence it comes to 0.

For the question you posted:


Our 'a' is -1, and our 'b' is -2

Hence our horizontal asymptote is y=0.5

You can find horizontal asymptotes by subbing in very large (or very small) values of x on your calculator, and see where they are coming close to
 

D94

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This may be an easy question but I just cannot understand the explanation.

Find the horizontal asymptote of these functions by dividing through by the highest power of x in the denominator.

F(x) = (5-x)/(4-2x)

Cheers
You divide by the highest power of x, then you take the limit as x approaches infinity.



The highest power of x is x1, so you divide each term by x1. If the highest power was a square (2), then you divide through each term by x2.

Then you recognise as x approaches infinity, 1/x approaches 0 (zero), so you can then simplify it like so.
 
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