How do you identify all asymptotes or holes for #f(x)=(x^2-5x+6)/(x^2-4x+3)#?

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Answer 1 · 2016-12-10T13:37:42

There is a hole at #x=3#
The vertical asymptote is #x=1#
The horizontal asymptote is #y=1#

Let's factorise the numerator and denominator

#x^2-5x+6=(x-3)(x-2)#

#x^2-4x+3=(x-1)(x-3)#

So,

#f(x)=(x^2-5x+6)/(x^2-4x+3)=(cancel(x-3)(x-2))/((x-1)cancel(x-3))#

#=(x-2)/(x-1)#

So, there is a hole at #x=3#

The domain of #f(x)# is #D_f(x)=RR-{1} #

As we cannot divide by #0#, #x!=1#

So, the vertical asymptote is #x=1#

To calculate the limits as #x->+-oo#, we take the terms of highest degree in the numerator and the denominator.

#lim_(x->+-oo)f(x)=lim_(x->+-oo)x/x=1#

So, the horizontal asymptote is #y=1#

graph{(y-(x-2)/(x-1))(y-1)=0 [-11.25, 11.25, -5.625, 5.625]}