How do you identify all asymptotes or holes and intercepts for #f(x)=1/(x-2)^2#?

1 Answer
Dec 16, 2017

Answer:

The #y#-intercept is #(0,1/4)#; #y=0# is the horizontal asymptote; #x=2# is the vertical asymptote.

Explanation:

When #x=0# we have #f(0)=1/(0-2)^2=1/4#, so the #y#-intercept is #(0,1/4)#.

The equation #f(x)=0# has no solutions so there are no #x#-intercepts.

Since the degree of the numerator is less than the degree of the denominator we know that #y=0# is the horizontal asymptote. (Or you could justify it by saying #lim_(x\to oo)f(x)=0# if you've gotten to that concept yet.)

There are no holes because there are no common factors in the numerator and denominator.

#x=2# is the vertical asymptote because it is the zero of the denominator. (Or you could use limits to justify that as well.)