Example 6 x + 2 </mrow> x

starbright49ly

starbright49ly

Answered question

2022-05-28

Example
6 x + 2 x 2 9 = 6 x + 2 ( x + 3 ) ( x 3 )
I know how to find the vertical and horizontal asypmtotes and everything, I just don't know how to find end behavior for a RATIONAL function without plugging in a bunch of numbers.

Answer & Explanation

odczepneyv

odczepneyv

Beginner2022-05-29Added 10 answers

If you are concerned by the behavior of the function when x starts to be large, just perform the long division of polynomials.
For
f ( x ) = 6 x + 2 x 2 9
this will give
f ( x ) 6 x + 2 x 2
and then the asymptote would be function 6 x .
Changing to
g ( x ) = 6 x 2 + 2 x 2 9
this will give
g ( x ) 6 + 56 x 2
and then the asymptote would be function 6, an horizontal asymptote.
Changing to
h ( x ) = 6 x 3 + 2 x 2 9
this will give
h ( x ) 6 x + 54 x + 2 x 2
and then the asymptote would be function 6 x, an oblique asymptote.
You could notice that this simple division gives you the asymptote as well as the manner the function appoaches it.
qtbabe9876a9

qtbabe9876a9

Beginner2022-05-30Added 4 answers

Horizontal asymptotes (if they exist) are the end behavior. However horizontal asymptotes are really just a special case of slant asymptotes (slope=0).
The slant asymptote is found by using polynomial division to write a rational function F ( x ) G ( x ) in the form
F ( x ) G ( x ) = Q ( x ) + R ( x ) G ( x )
Where Q ( x ) (the quotient) is the line of your slant asymptote, and the degree of R ( x ) (the remainder) is strictly less than the degree of G ( x ).
In your example, the degree of the numerator is already strictly less than the degree of the denominator, so what is your slant asymptote? This will tell you the end behavior.

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