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What is the slant asymptote of f(x) = (x^2-1)/x?
Can you give me the process of how to do this? I was taught that you divide the numerator by the denominator and the answer would be the slant asymptote and the remainder would be the vertical asymptote, but it's not really working here since the answer I got from using the process was different from what my calculator graphed. Thanks!
I forgot to say that I know that the slant asymptote is y=x, but I want to know how to get it. Thanks!
3 Answers
- ?Lv 79 years agoFavourite answer
You are correct on how to obtain a slant asymptote, but to obtain vertical asymptotes we need to find when division by zero occurs.
By splitting up the fraction, we have:
f(x) = (x^2 - 1)/x = x^2/x - 1/x = x - 1/x,
so dividing x^2 - 1 by x leaves a quotient of x and, hence, the slant asymptote is:
y = x.
Then, since division by zero occurs when x = 0, the vertical asymptote is x = 0.
I hope this helps!
- Anonymous9 years ago
The slant asymptote would be the line y=x.
You were right about dividing the x's. All you do is divide the x's with the highest exponents on the numerator and denominator. So it would look like this...
Y=x^2/x
Y=x
- ?Lv 44 years ago
(a million - x + x^2)(a million - x + x^2) = (x^2 - x + a million)(x^2 - x + a million) = x^4 - x^3 + x^2 - x^3 + x^2 - x + x^2 - x + a million = x^4 - x^3 - x^3 + x^2 + x^2 + x^2 - x - x + a million = x^4 - 2x^3 + 3x^2 - 2x + a million
