How many vertical asymptotes does the function #f(x)=(x-2)/(x^2-4x-5)# have?

Question

1246 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-22T18:01:46

The vertical asymptotes are #x=-1# and #x=5#

Let's factorise the denominator.

#x^2-4x-5=(x+1)(x-5)#

The domain of #f(x)# is #D_(f(x))=RR-{-1,5} #

As you cannot divide by #0#, #x!=-1# and #x!=5#

Threfore there are 2 vertical asymptotes.

The vertical asymptotes are #x=-1# and #x=5#

graph{(x-2)/(x^2-4x-5) [-10, 10, -5, 5]}