Originally Posted by
Resilient ...
what can we say about the asymptotes of $\displaystyle f(x) = \frac{x^4+1}{x^3-x}$? I see that it has vertical asymptote at x = 1, x = -1 and x = 0, but I'm not sure what else
Originally Posted by
Resilient Sorry, I meant
$\displaystyle f(x) = \frac{x^4+1}{x^3-x}$
1. As you stated correctly the vertical asymptotes correspond to the zeros of the denominator.
2. To get the horizontal or slanted asymptotes transform the term of the function into a complete polynomial and a fraction: Code:
x^2 + 1
(x^4 + 1) ÷ (x^3 - x) = x + -----------
-(x^4 - x^2) x^3 - x
------------
x^2 + 1
The none-fractional part of the term is the term of the slanted asymptote. That means in your case the slanted asymptote has the equation:
$\displaystyle y = x$