What are the asymptote(s) and hole(s), if any, of # f(x) =(7x)/(x-3)^3 #?

1 Answer
Jul 24, 2018

Answer:

no holes
vertical asymptote at #x = 3#
horizontal asymptote is #y = 0#

Explanation:

Given: #f(x) = (7x)/(x-3)^3#

This type of equation is called a rational (fraction) function.

It has the form: #f(x) = (N(x))/(D(x)) = (a_nx^n + ...)/(b_m x^m + ...)#,

where #N(x))# is the numerator and #D(x)# is the denominator,

#n# = the degree of #N(x)# and #m# = the degree of #(D(x))#

and #a_n# is the leading coefficient of the #N(x)# and

#b_m# is the leading coefficient of the #D(x)#

Step 1, factor : The given function is already factored.

Step 2, cancel any factors that are both in #(N(x))# and #D(x))# (determines holes):

The given function has no holes #" "=> " no factors that cancel"#

Step 3, find vertical asymptotes: #D(x) = 0#

vertical asymptote at #x = 3#

Step 4, find horizontal asymptotes:
Compare the degrees:

If #n < m# the horizontal asymptote is #y = 0#

If #n = m# the horizontal asymptote is #y = a_n/b_m#

If #n > m# there are no horizontal asymptotes

In the given equation: #n = 1; m = 3 " "=> y = 0#

horizontal asymptote is #y = 0#

Graph of #(7x)/(x-3)^3#:
graph{(7x)/(x-3)^3 [-6, 10, -15, 15]}