How do you find the slant asymptote of $\frac{x^{4} + 1}{x^{2} + 2}$ $( x^4 + 1 ) / ( x^2 + 2 )$ ?

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How do you find the slant asymptote of $\frac{x^{4} + 1}{x^{2} + 2}$ $( x^4 + 1 ) / ( x^2 + 2 )$ ?

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Answer 1 · 2016-01-15T21:18:33

This rational function is asymptotic to a parabola, not a line.

It has no slant asymptote.

Explanation:

The degree of the numerator is $4$ $4$ and the degree of the numerator is $2$ $2$ .

As a result this rational function is asymptotic to a parabola, not a line.

More explicitly:

$f (x) = \frac{x^{4} + 1}{x^{2} + 2}$ $f(x) = (x^4+1)/(x^2+2)$

$= \frac{x^{4} + 2 x^{2} - 2 x^{2} - 4 + 5}{x^{2} + 2}$ $=(x^4+2x^2-2x^2-4+5)/(x^2+2)$

$= \frac{x^{2} (x^{2} + 2) - 2 (x^{2} + 2) + 5}{x^{2} + 2}$ $=(x^2(x^2+2)-2(x^2+2)+5)/(x^2+2)$

$= x^{2} - 2 + \frac{5}{x^{2} + 2}$ $=x^2-2 + 5/(x^2+2)$

So as $x \to \pm \infty$ $x->+-oo$ we find $(f (x) - (x^{2} - 2)) \to 0$ $(f(x) - (x^2-2)) -> 0$

That is $f (x)$ $f(x)$ is asymptotic to $x^{2} - 2$ $x^2-2$

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How do you find the slant asymptote of $\frac{x^{4} + 1}{x^{2} + 2}$ $( x^4 + 1 ) / ( x^2 + 2 )$ ?

1 Answer

Explanation:

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