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

1 Answer
Mar 29, 2018

Answer:

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

and vertical is when x is 1 or 3

Explanation:

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#