# How do you graph f(x)=x^2/(x-4) using holes, vertical and horizontal asymptotes, x and y intercepts?

Jan 27, 2018

No holes, one VA at $x = 4$, $x$- and $y$-intercept at the origin, and EBA is $y = x + 4$.

#### Explanation:

A vertical asymptote occurs in a rational function when there is $0$ value in the denominator. Let's take a look our function:

$f \left(x\right) = {x}^{2} / \left(x - 4\right)$

If we set the denominator equal to $0$, we can find out when our asymptote will occur:

$x - 4 = 0 \implies$

$x = 4$

Now we know that the function has a vertical asymptote at $x = 4$.

We also know that there are no holes in the function because holes occur when there are factors in common in the numerator and the denominator; this function doesn't have any common factors.

An $x$-intercept occurs when the numerator of the rational function is equal to $0$. We can solve for the x-intercept(s):

${x}^{2} = 0 \implies$

$x = 0$

There's an $x$-intercept at $\left(0 , 0\right)$. Since the $y$-value is $0$, this point is also the $y$-intercept. Additionally, we know that this $x$-intercept only "bounces" off the $x$-axis (and doesn't cross it) because the multiplicity of ${x}^{2}$ is $2$, which is even.

Lastly, find the EBA (end behavior asymptote). Since the power in the numerator is greater than the power in the denominator, we have to divide the two using synthetic division:

$4$ | $1 \text{ "0" } 0$
$\textcolor{w h i t e}{w}$|
$\text{ ---------------}$
$\text{ } |$

$4$ | $1 \text{ "0" } 0$
$\textcolor{w h i t e}{w}$|
$\text{ ---------------}$
$\text{ "1" } |$

$4$ | $1 \text{ "0" } 0$
$\textcolor{w h i t e}{w}$|$\text{ } 4$
$\text{ ---------------}$
$\text{ "1" "4" } |$

$4$ | $1 \text{ "0" } 0$
$\textcolor{w h i t e}{w}$|$\text{ "4" } 16$
$\text{ ---------------}$
$\text{ "1" "4" | } 16$

The quotient is $x + 4$, so the EBA is the line $y = x + 4$.

The final graph looks like this:

graph{x^2/(x-4) [-47.03, 56.97, -18.04, 34]}

As you can see, there are no holes, 1 $x$- and $y$-intercept, a vertical asymptote at $x = 4$ and a diagonal asymptote (EBA) of $y = x + 4$.