What are the vertical asymptotes of the graph $y = \frac{2 x - 6}{x^{3} - x}$ $y=(2x-6)/(x^3-x)$ ?

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What are the vertical asymptotes of the graph $y = \frac{2 x - 6}{x^{3} - x}$ $y=(2x-6)/(x^3-x)$ ?

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Answer 1 · 2014-12-18T12:45:26

$x = 0, x = 1$ $x=0,x=1$ and $x = - 1$ $x=-1$

The vertical asymptotes are where the denominator is equal to zero, because you can't divide by that.
So $x^{3} - x = 0$ $x^3-x=0$ is equivalent with $x (x^{2} - 1) = 0$ $x(x^2-1)=0$ , so $x = 0$ $x=0$ or $x^{2} = 1$ $x^2=1$ . By taking the square root of the second one, we find $x = 1$ $x=1$ and $x = - 1$ $x=-1$ . So all the answers are $x = 0, x = 1$ $x=0,x=1$ and $x = - 1$ $x=-1$ .

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What are the vertical asymptotes of the graph $y = \frac{2 x - 6}{x^{3} - x}$ $y=(2x-6)/(x^3-x)$ ?

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What are the vertical asymptotes of the graph $y = \frac{2 x - 6}{x^{3} - x}$ $y=(2x-6)/(x^3-x)$ ?