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Old 20th May 2012, 11:51 PM   #1 (permalink)
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horizontal and vertical asymptote of graph



How do I do this question?

I used a software and the graph looks like this:


Apparently both the horizontal and verticle asymptote seem to be 2. How do I prove that value on paper?

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Old 20th May 2012, 11:56 PM   #2 (permalink)
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Re: horizontal and vertical asymptote of graph

have you learned l'hopital rule?

set x to infinity and use the rule to find value of y.
set y to infinity and find out what value of x would make that happen, focus on the denominator.

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Last edited by workingtoohard; 20th May 2012 at 11:58 PM.
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Old 21st May 2012, 01:10 AM   #3 (permalink)
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Re: horizontal and vertical asymptote of graph

Shoot, my brain is too dumb for this

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Old 21st May 2012, 10:02 AM   #4 (permalink)
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Re: horizontal and vertical asymptote of graph

Originally Posted by Crazyman View Post


How do I do this question?

I used a software and the graph looks like this:


Apparently both the horizontal and verticle asymptote seem to be 2. How do I prove that value on paper?
The vertical asymptote comes from the denominator. So denominator cannot be zero, vertical asymptote is obtain when we let the denominator to be zero. In this case,
x - 2 = 0
x = 2.

We can get the horizontal asymptote (or diagonal asymptote) when we make the fraction into proper fraction.
y = 2 + [5/(x-2)]
Ignoring the proper fraction, we have y = 2.

Another example,

y = 3x+1 + [x/(x-1)(x+2)]

Vertical asymptote are x = 1 and x = -2.
Diagonal asymptote (horizontal asymptote) is y = 3x+1


The reason behind the diagonal asymptote is that as x tends to negative or positive infinity, the proper fraction will tends to zero as the denominator gets larger and larger. Thus y will tends to the 3x + 1 in this case.

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Old 21st May 2012, 06:18 PM   #5 (permalink)
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Re: horizontal and vertical asymptote of graph

Has elaborated by namelessname... Just for your pleasure.

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Old 22nd May 2012, 12:09 AM   #6 (permalink)
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Re: horizontal and vertical asymptote of graph

Thanks guys. I understand how to do it now.

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Old 22nd May 2012, 12:21 AM   #7 (permalink)
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Re: horizontal and vertical asymptote of graph

Originally Posted by .Memo View Post
Has elaborated by namelessname... Just for your pleasure.
May I know the steps for (2x+1)/(x-2) = 2+ 5/(x-2)?

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Old 22nd May 2012, 01:52 AM   #8 (permalink)
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Re: horizontal and vertical asymptote of graph

2x + 1
= 2x - 4 + 4 + 1
= 2(x-2) + 5

divide that by (x-2) and you get

2 + 5/(x-2)

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Old 22nd May 2012, 07:50 PM   #9 (permalink)
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Re: horizontal and vertical asymptote of graph

workingtoohard method is correct too... but if you want the all-use-method, just do long division.

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