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Mathematics 11 Online
OpenStudy (anonymous):

Find vertical and horizontal asymptotes

OpenStudy (anonymous):

OpenStudy (anonymous):

Vertical where ever denominator is zero.

OpenStudy (anonymous):

what about horizontal?

OpenStudy (anonymous):

You have to compare the x on top and x on bottom. Some how, remove brackets or square root to compare.

OpenStudy (anonymous):

i don;t get it?

OpenStudy (anonymous):

Try to conjugate. Multiply by \[\sqrt{\frac{ 4+x }{ 4+x }}\]That would remove square root signs

OpenStudy (anonymous):

y= -1?

OpenStudy (anonymous):

Right, you're pretty good.

OpenStudy (anonymous):

Well, y is not equal to -1. When you compare the coefficient of top x and bottom x, you get -1.

OpenStudy (anonymous):

yes?

OpenStudy (anonymous):

Yes, what?

OpenStudy (anonymous):

why it doesnt equal to -1?

OpenStudy (anonymous):

Well, may be it does. But that is not relevant to trying to find the horizontal asymptote. What does your notes say about finding HA?

OpenStudy (anonymous):

you mean there is no horizontal asymtotes?>

OpenStudy (anonymous):

?

OpenStudy (anonymous):

You might need to brush up on finding horizontal asymptotes. It's really easy, read this attached half-page document.

OpenStudy (anonymous):

yeah ` horizontal asymtotes is y=-1

OpenStudy (anonymous):

am I right?>

OpenStudy (anonymous):

Just follow instructions, what does the doc say?

OpenStudy (anonymous):

if n/d is bigger than 1, then there is no horizontal asymptote

OpenStudy (anonymous):

there is no horzontal asymtotes

OpenStudy (anonymous):

It is \[\frac{ -x }{ x }\]or\[\frac{ -x ^{\frac{ 1 }{ 2 }} }{ x ^{\frac{ 1 }{ 2 }} }\]

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