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avatar+22 

Howdy, been spending a lot of time trying to figure this one out.

 

The equation of an asymptote can be a constant, which makes it a horizontal asymptote. An asymptote can also be linear, which makes it an oblique asymptote. But asymptotes can have even higher degrees. For example, an asymptote that is quadratic is called a parabolic asymptote.


Find the equation of the parabolic asymptote of the graph of

 

y= (2x+4-3)/(x^2-4x+1)


Enter your answer in the form "y=ax^2+bx+C".

 

Thank you.  Any help would be helpful

- Chad

 May 24, 2024
 #1
avatar+129487 
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Do you mean                     ( 2x^4 -3)

                           y =          ____________         ????

                                          (x^2 - 4x + 1)

 

cool cool cool

 May 24, 2024
 #2
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Yes I'm sorry.  This is my first time using this platform.  Ain't sure how to format things.

 

- Chad

Chadly  May 24, 2024
 #3
avatar+129487 
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No problem

 

Just some polynomial division

 

                        x^2  + 4x  + 15

x^2 - 4x + 1   [  x^4  + 0x^3 + 0x^2 + 0x - 3 ] 

                         x^4 -  4x^3  + 1x^2

                       ___________________

                                   4x^3 - 1x^2 + 0x

                                   4x^3 -16x^2 + 4x

                                  ________________

                                           15x^2  - 4x - 3

                                           15x^2 - 60x +15

                                           ______________

                                                    56x - 18        (this remainder is  ignored)

 

The parabolis asymtote is    y =   x^2 + 4x + 15       

 

cool cool cool           

CPhill  May 24, 2024

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