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

Jul 24, 2018

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

#### Explanation:

Given: $f \left(x\right) = \frac{7 x}{x - 3} ^ 3$

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

It has the form: $f \left(x\right) = \frac{N \left(x\right)}{D \left(x\right)} = \frac{{a}_{n} {x}^{n} + \ldots}{{b}_{m} {x}^{m} + \ldots}$,

where N(x)) is the numerator and $D \left(x\right)$ is the denominator,

$n$ = the degree of $N \left(x\right)$ and $m$ = the degree of $\left(D \left(x\right)\right)$

and ${a}_{n}$ is the leading coefficient of the $N \left(x\right)$ and

${b}_{m}$ is the leading coefficient of the $D \left(x\right)$

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

Step 2, cancel any factors that are both in $\left(N \left(x\right)\right)$ and D(x)) (determines holes):

The given function has no holes $\text{ "=> " no factors that cancel}$

Step 3, find vertical asymptotes: $D \left(x\right) = 0$

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 $\frac{7 x}{x - 3} ^ 3$:
graph{(7x)/(x-3)^3 [-6, 10, -15, 15]}