# What are the asymptotes for (x-4)/(16x-x^3)?

Jun 18, 2015

The vertical asymptotes are $x = 0 \mathmr{and} x = 4 \mathmr{and} x = - 4$
The horizontal asymptote is the line $y = 0$
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#### Explanation:

Set the denominator $16 x - {x}^{3}$ equal to zero, and solve for $x$.

$16 x - {x}^{3} = 0$

Factor out $x$.

$x \left(16 - {x}^{2}\right) = 0$

$x = 0$

$x = 0$

$16 - {x}^{2} = 0$

Factor to get

$\left(4 + x\right) \left(4 - x\right) = 0$

$4 + x = 0$ so $x = - 4$

$4 - x = 0$ so $x = 4$

The horizontal asymptote is $y = 0$ (The $x$-axis)

because as $x$ increases without bound, the value of

$\frac{x - 4}{16 x - {x}^{3}} = \frac{\cancel{{x}^{3}} \left(\frac{1}{x} ^ 2 - \frac{4}{x} ^ 3\right)}{\cancel{{x}^{3}} \left(\frac{16}{x} ^ 2 - 1\right)}$ gets closer and closer to $\frac{0}{1}$.