+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
+22 Yes I'm sorry. This is my first time using this platform. Ain't sure how to format things.
- Chad
+129567 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
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4x^3 - 1x^2 + 0x
4x^3 -16x^2 + 4x
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15x^2 - 4x - 3
15x^2 - 60x +15
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56x - 18 (this remainder is ignored)
The parabolis asymtote is y = x^2 + 4x + 15
