Question

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

Answer 1

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]}