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

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Answer 1 · 2015-06-18T03:58:47

The vertical asymptotes are $x=0 and x=4 and x=-4$
The horizontal asymptote is the line $y=0$
.

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

$16x-x^3=0$

Factor out $x$ .

$x(16-x^2)=0$

$x=0$

$16-x^2=0$

Factor to get

$(4+x)(4-x) =0$

$4+x =0$ 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

$(x-4)/(16x-x^3) = (cancel(x^3) (1/x^2 -4/x^3))/(cancel(x^3)(16/x^2 - 1))$ gets closer and closer to $0/1$ .

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