Question

what is the flutterign asymptote of (x^4- 4 ) / (2x^2 - 3x + 2 )
it's supposed to be (x^2)/2 + 3x/4 + 5/8 according to wolframalpha but I don't see how, since the limit of f(x)/x would be infinite

Answer 1

Do you mean the oblique asymptote? If that the case, that function doesn't have any. You can check it "by eye", because the degree of the denominator is two orders smaller than the degree of the numerator.

Answer 2

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Answer 3

I don't even know what "oblique" means q.q

all I have is : http://www.wolframalpha.com/input/?i=%28x^4+-+4+%29+%2F+%282x^2+-+3x+%2B+2%29+asymptote

it's a curve & I only know how to calculate straight lines, NOT curves
according to wolfram alpha it would be x^2/2 + 3x/4 + 5/8 but that would require an x left in the previous equation
( the formula we have to use is g(x)= ax + b with a= f(x)/x lim x -> infinity
and b = f(x) - ax lim x -> infinity
that would mean a must contain another x but that's impossible due to the limit )

Answer 4

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Answer 5

2 is not a scary number. The degree differs by 2. This means only that anything that might be called an asymptote will Quadratic in shape - degree 2. The precise shape is obtained by long division at worst. There may be other clues along the way. There is no oblique (linear but not horizontal) asymptote.

\(\dfrac{x^{4}-4}{2x^{2}-3x+2} = \dfrac{1}{8}\left(4x^{2}+6x+5\right)+\dfrac{3}{4}\cdot\dfrac{x-7}{2x^{2}-3x+2}\)

And we see that the last term, if taken alone, has an horizontal asymptote at y = 0 and thus can be ignored for purposes of determining the other "asymptote", which is simply,
\(y = \dfrac{1}{8}\left(4x^{2}+6x+5\right)\)

Answer 6

* will be Quadratic

Answer 7

Note: I may have missed something in that particularly tedious long division. It's early and my head hurts, so I'm not doing it over.

Okay, fine, the last term should be \(\dfrac{3}{8}\cdot\dfrac{x-14}{2x^{2}-3x+2}\).

I should get more sleep.