How do you find all the asymptotes for # (x-2)/(x^2-12x+12) #?

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Answer 1 · 2016-03-12T13:18:39

Two vertical asymptotes #x=10.899# or #x=1.101#

Using quadratic formulahe denominator #x^2-12x+12# will be zero at

#x=(12+-sqrt(12^2-4xx1xx12))/2=(12+-sqrt96)/2#or

#x=(12+-4sqrt6)/2=6+-2xx2.4495# i.e.

#x=10.899# or #x=1.101#

Hence we have two vertical asymptotes #x=10.899# or #x=1.101#.

As the degree of numerator is less than that of denominator, there is no horizontal or slanting asymptote.

graph{(x-2)/(x^2-12x+12) [-5, 20, -5, 5]}