How do I find the asymptotes of $y=1/((x-1)(x-3))$ ?

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Answer 1 · 2018-03-29T04:37:52

Horizontal is when $limxto+-oo1/((x-3)(x-1))=0$

and vertical is when x is 1 or 3

The horizontal assymptotes are the assymptotes as x approaches infinity or negative infinity $limxtooo$ or $limxto-oo$

$limxtooo 1/(x^2-4x+3)$

Divide top and bottom by the highest power in the denominator
$limxtooo (1/x^2)/(1-4/x+3/x^2)$

$0/(1-0-0)=0/1=0$ so this is your horizontal assymptote negative infinty gives us the same result

For the vertical asymptote we are looking for when the denominator is equal to zero

$(x-1)(x-3)=0$ so you have a vertical asymptote when

$x=3 or 1$

How do I find the asymptotes of y=1/((x-1)(x-3))y=1/((x-1)(x-3))?