# What are the asymptote(s) and hole(s), if any, of  f(x) = (x*(x-2))/(x^2-2x+1)?

Aug 19, 2017

$x = 1 \text{ }$ is the vertical asymptote of $f \left(x\right)$.
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$y = 1 \text{ }$ is the horizantal asymptote of $f \left(x\right)$

#### Explanation:

This rational equation has a vertical and horizantal asymptote .
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Vertical asymptote is determined by factorizing the denominator :
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${x}^{2} - 2 x + 1$
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$= {x}^{2} - 2 \left(1\right) \left(x\right) + {1}^{2}$
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$= {\left(x - 1\right)}^{2}$
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Then,$\text{ "x=1" }$is a vertical asymptote.
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Let us find the horizantal asymptote :
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As it is known we have To check both degrees of the
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numerator and denominator .
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Here , the degree of the numerator is $2$ and that of the
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denominator is $2$ too .
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If $\frac{a {x}^{2} + b x + c}{{a}_{1} {x}^{2} + {b}_{1} x + {c}_{1}}$then the horizantal asymptote is $\textcolor{b l u e}{\frac{a}{{a}_{1}}}$
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In $f \left(x\right) = \frac{x . \left(x - 2\right)}{{x}^{2} - 2 x + 1} = \frac{{x}^{2} - 2 x}{{x}^{2} - 2 x + 1}$
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Same degree in the numerator and denominator then horizantal
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asymptote is $y = \textcolor{b l u e}{\frac{1}{1}} = 1$
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$\therefore x = 1 \mathmr{and} y = 1 \text{ }$ are the asymptotes of $f \left(x\right)$.