How do you identify all horizontal and slant asymptote for #f(x)=4/(x-2)^3#?

1 Answer
Nov 7, 2016

Answer:

The vertical asymptote is #x=2#
The horizontal asymptote is #y=0#

Explanation:

As you cannot divide by #0#, you have a vertical asymptote #x=2#
There is no slant asymptote as the degree of the numerator is smaller than the degree of the denominator.
#lim_(n rarr -oo )f(x)=lim_(narr-oo)4/x^3=0^-#

#lim_(n rarr +oo )f(x)=lim_(narr+oo)4/x^3=0^+#

So there is a horizontal asymptote #y=0#
graph{4/(x-2)^3 [-16.02, 16.01, -8.01, 8.01]}