# Horizontal asympototes

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• Nov 28th 2006, 09:25 AM
dcfi6052
Horizontal asympototes
I was going thorough a horizontal asympototes question and got stuck. This is the queston.

Determine th H.A for the following function. Draw a sketch to show whether the function is above or below the asymptote as x approches infinity.

y=8x2-3/2x2+x-1 , (8x2-3 means 8x squared-3)

I determined that the H.A is y=4, but do not know how to determine if the function is above or below the asymptote as x approches infinity. Can someone explain how to figure this out with an example.
• Nov 28th 2006, 10:16 AM
earboth
Quote:

Originally Posted by dcfi6052
I was going thorough a horizontal asympototes question and got stuck. This is the queston.

Determine th H.A for the following function. Draw a sketch to show whether the function is above or below the asymptote as x approches infinity.

y=8x2-3/2x2+x-1 , (8x2-3 means 8x squared-3)

I determined that the H.A is y=4, but do not know how to determine if the function is above or below the asymptote as x approches infinity. Can someone explain how to figure this out with an example.

hello,

your horizontal asymptote is correct.

If you do synthetic division, you'll get:

$\displaystyle (8x^2-3)\div(2x^2+x-1)=4+\frac{-4x+1}{2x^2+x-1}$.

If x is positive you substract a number from 4 that means the graph is below y = 4.
If x is negative you add a number to 4 that means the graph is over (higher?) than y = 4.

EB
• Nov 28th 2006, 10:22 AM
earboth
Quote:

Originally Posted by dcfi6052
...
I determined that the H.A is y=4, but do not know how to determine if the function is above or below the asymptote as x approches infinity. Can someone explain how to figure this out with an example.

Hello,

I've attached the graph of your function. The blue line is the horizontal asymptote. I've added too the vertical asymtotes in red.

EB