How do you find the zeroes for $R( x) = (3x)/(x^2 - 2x)$ ?

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Answer 1 · 2015-06-05T23:58:23

$R(x)=(3x)/(x^2-2x) = (3cancel(x))/((x-2)cancel(x)) = 3/(x-2)$

with restrictions $x!=0$ and $x!= 2$ .

When $x < 0$ or $0 < x < 2$ we have $x-2 < 0$

so $3/(x-2) < 0$

In fact $3/(x-2) -> 0_-$ as $x -> -oo$

When $x > 2$ we have $x-2 > 0$

so $3/(x-2) > 0$

In fact $3/(x-2) -> 0_+$ as $x -> oo$

So $R(x) != 0$ for all $x in RR$

graph{3x/(x^2-2x) [-10, 10, -5, 5]}

How do you find the zeroes for R( x) = (3x)/(x^2 - 2x)R( x) = (3x)/(x^2 - 2x)?