How do you find vertical, horizontal and oblique asymptotes for $\frac{x}{{(1 - x)}^{2}}$ $x/(1-x)^2$ ?

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How do you find vertical, horizontal and oblique asymptotes for $\frac{x}{{(1 - x)}^{2}}$ $x/(1-x)^2$ ?

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Answer 1 · 2017-06-13T21:30:53

$y = 0$ $y=0$

$x = 1$ $x=1$

Explanation:

The degree of the bottom is 2, and the degree of the top is 1. Therefore, there will be a horizontal asymptote at $y = 0$ $y=0$ since the function will tend towards 0 as x goes to $\infty$ $oo$ . This also means there will be no oblique asymptote.

As far as vertical asymptotes, they will occur when the denominator is 0 and the numerator is non-zero. For this function, this will happen when $x = 1$ $x=1$ since this makes the bottom equal to 0 and the top equal to 1.

So our asymptotes are $y = 0$ $y=0$ and $x = 1$ $x=1$

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How do you find vertical, horizontal and oblique asymptotes for $\frac{x}{{(1 - x)}^{2}}$ $x/(1-x)^2$ ?

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Explanation:

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How do you find vertical, horizontal and oblique asymptotes for $\frac{x}{{(1 - x)}^{2}}$ $x/(1-x)^2$ ?