How do you find the asymptotes for $f (x) = \frac{2 x}{x^{2} + 16}$ $f(x)= (2x)/(x^2+16)$ ?

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Answer 1 · 2016-01-17T02:14:29

Since the denominator will never equal zero, there are no vertical asymptotes .

There is one horizontal asymptote:

$_{x \to \pm \infty}^{lim} \frac{2 x}{x^{2} + 16} = 0$ $""_(xrarr+-oo)^"lim"(2x)/(x^2+16)=0$

Horizontal : $y = 0$ $y=0$

Hope that helped

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How do you find the asymptotes for f(x)= (2x)/(x^2+16)f(x)=2xx2+16f(x)= (2x)/(x^2+16)?