Oblique asymptote
- Thread starter alyren
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alyren said:
Hint:
G(x)= (2x[sup:ci25v7bt]3[/sup:ci25v7bt]+11x[sup:ci25v7bt]2[/sup:ci25v7bt]+5x-1)/(x[sup:ci25v7bt]2[/sup:ci25v7bt]+6x+5) = 2x - 1 +[(x+4)/(x[sup:ci25v7bt]2[/sup:ci25v7bt]+6x+5)]
.
BigGlenntheHeavy
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\(\displaystyle f(x) \ = \ \frac{2x^3+11x^2+5x-1}{x^2+6x+5} \ = \ \frac{2x^3+11x^2+5x-1}{(x+5)(x+1)} \ = \ (2x-1)+\frac{x+4}{x^2+6x+5}.\)
\(\displaystyle Now \ \lim_{x\to\pm\infty}\frac{x+4}{x^2+6x+5} \ = \ 0, \ this \ is \ the \ remainder.\)
\(\displaystyle Ergo, \ we \ have \ y \ = \ 2x-1 \ as \ a \ oblique \ asymptote \ and \ x \ = \ -1,-5 \ as \ vertical \ asymptotes.\)
\(\displaystyle See \ graph \below.\)
\(\displaystyle Post \ Script: \ What \ is \ the \ domain \ and \ range \ of \ f(x)?\)
[attachment=0:leqekim2]aaa.jpg[/attachment:leqekim2]
\(\displaystyle Now \ \lim_{x\to\pm\infty}\frac{x+4}{x^2+6x+5} \ = \ 0, \ this \ is \ the \ remainder.\)
\(\displaystyle Ergo, \ we \ have \ y \ = \ 2x-1 \ as \ a \ oblique \ asymptote \ and \ x \ = \ -1,-5 \ as \ vertical \ asymptotes.\)
\(\displaystyle See \ graph \below.\)
\(\displaystyle Post \ Script: \ What \ is \ the \ domain \ and \ range \ of \ f(x)?\)
[attachment=0:leqekim2]aaa.jpg[/attachment:leqekim2]
mmm4444bot
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alyren said:How to find oblique asymptote?
I use polynomial longhand division or synthetic division to write:
Quotient + Remainder/Divisor
When x goes toward ±infinity, the Divisor becomes infinitely larger than the remainder, so the ratio Remainder/Divisor goes toward zero, leaving the Quotient (a linear function) as the oblique asymptote.
Cheers