Find the horizontal asymptote of f(x)= x^2+3x-2/x-2

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Find the horizontal asymptote of f(x)= x^2+3x-2/x-2

Mathematics
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I appreciate any help
You sure that is 3x - 2 in numerator and not 3x + 2 ??
Yes it's in the numerator

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As you can see, degree of numerator is greater than degree of denominator, you will get Slant Asymptote...
Ok
Do you know the Long Division Method??
No :(
You have to just divide numerator by denominator.. As you do simple division and then find out the quotient..
Ok so divide x^2+3x-2 by x-2?
yep...
Would it be the same as using synthetic division?
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yes, you can do synthetic division too..!!!
So, \(y = x + 5\), this is the Slant Asymptote in this case..
Where degree of numerator is greater than degree of denominator, there is no horizontal asymptote but there is slant asympote, as you can find that by \(y = quotient\), for that just use Long Division and find the quotient..
So I got x+1 - (4/x+2) :/
Wait so the answer is none?
yes None is the answer for this.. :P
There is no horizontal asymptote? What a trick question 😅
Thanks
I just told how to find slant asymptote equation if you do not have Horizontal asymptote..
Oh on :) thanks
Ok :D
Thanks again bye~
See, horizontal asymptote means you will get line parallel to x-axis or where y = 0, but if asymptote is not horizontal, then it must be inclined and y \(\ne\) 0 there, so how to find y there, I just told you that.. :)
\(\dagger\) ..

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