# Section 3.5

:::{prf:definition} Rational Function
:label: ratFunction
Let $p(x)$ and $q(x)$ be any polynomial function where $q(x)$ is not the zero function. A **rational function** is 

$$f(x)=\frac{p(x)}{q(x)}$$

The domain of $f$ is the set of all $x$ such that $q(x)\ne0$.
:::

When finding the domain of a rational function **immediately** solve $q(x)=0$. Do not simplify $f$ and then set the denominator equal to zero.

If $f(x)=\frac{p(x)}{q(x)}$, then the list of possible zeros for $f(x)$ is the zeros for $p(x)$.

:::{prf:property} Graph of $f(x)=\frac{1}{x}$.
:label: oneoverxGraphProp
The graph of $f(x)=\frac{1}{x}$ is:

![The graph of y equals one over x](images/oneoverxgraph.png)

* The domain of $f$ is the set of all $x$ such that $x\ne0$.
* The range of $f$ is the set of all $y$ such that $y\ne0$.
* The function is continuous for all $x$, not zero.
* The function is decreasing in its domain.
* The function has a vertical asymptote $x=0$.
* The function has a horizontal asymptote $y=0$.
:::

:::{prf:property} Graph of $f(x)=\frac{1}{x^2}$.
:label: oneoverxsqrGraphProp
The graph of $f(x)=\frac{1}{x^2}$ is:

![The graph of y equals one over x squared](images/oneoverxsqrgraph.png)

* The domain $f$ is the set of all $x$ such that $x\ne0$.
* The range of $f$ is the set of all $y$ such that $y\ne0$.
* The function is continuous for all $x$, not zero.
* The function is increasing on the open interval $(-\infty,0)$.
* The function is decreasing on the open interval $(0,\infty)$.
* The function has a vertical asymptote $x=0$.
* The function has a horizontal asymptote $y=0$.
:::

:::{prf:definition} Vertical/Horizontal Asymptote
:label: vertHorAsy
Let $f(x)$ be a rational function.
* If the size of $f(x)$ approaches infinity as $x$ approaches a fixed number (say $a$). Then we say $f$ has a vertical asymptote at $x=a$.
* If $f(x)$ approaches a number $L$ as $x$ approaches infinity. Then $f$ has a horizontal asymptote at $y=L$.
* If $f(x)$ approaches a number $M$ as $x$ approaches negative infinity. The $f$ has a horizontal asymptote at $y=M$.
:::

When finding the domain of a rational function do not simplify. When finding vertical or horizontal asymptote(s) first simplify.

Important facts when dealing with a horizontal asymptote. The function $\frac{1}{x}$ approaches zero when $x$ approaches $\infty$.

::::{prf:example}
:label: ratFunctionExam1
Let $f(x)=\frac{x+1}{(2x-1)(x+3)}$. 

Find the domain of the function.
:::{dropdown} Solution:
Immediately setting the denominator equal to zero we say $x$ cannot be $\frac{1}{2}$ or $-3$. Therefore, the domain of $f$ is the set of all $x$ such that $x\ne\frac{1}{2}$ or $x\ne-3$ (interval notation would be $(-\infty,-3)\cup(-3,\frac{1}{2})\cup(\frac{1}{2},\infty)$.
:::

Find the vertical asymptote.
:::{dropdown} Solution:
Since $f$ is already simplified and the denominator is zero when $x$ is $\frac{1}{2}$ or $-3$ we say $f$ has vertical asymptote $y=\frac{1}{2}$ and $y=-3$.
:::

Find the horizontal asymptote.
:::{dropdown} Solution:
First, we will write the function in the following way:

\begin{align*}
    f(x) & = \frac{x+1}{(2x-1)(x+3)}\\
    & = \frac{x+1}{2x^2+5x-3}
\end{align*}

Next, we divide the top and bottom by $x^2$.

\begin{align*}
    \frac{x+1}{2x^2+5x-3} & = \frac{\frac{1}{x}+\frac{1}{x^2}}{2+\frac{5}{x}-\frac{3}{x^2}}\\
    & \to \frac{0+0}{2+0-0} = 0
\end{align*}

Therefore, the function has a horizontal asymptote $x=0$.
:::
::::

::::{prf:example}
:label: ratFunctionExam2
Let $f(x)=\frac{2x+1}{x-3}$. Find horizontal and vertical asymptotes.

Horizontal Asymptote:
:::{dropdown} Solution:
First, divide the top and bottom by $x$. 

\begin{align*)
    \frac{2x+1}{x-3} & = \frac{2+\frac{1}{x}}{1-\frac{3}{x}}\\
    & \to \frac{2+0}{1-0} = 2
\end{align*}

As $x\to0$ we see that $f(x)\to0$. Therefore, $f$ has horizontal asymptote $y=2$.
:::

Vertical Asymptote:
:::{dropdown} Solution:
Since $f$ is fully simplified and $x-3=0$ when $x=3$ we say the line $x=3$ is a vertical asymptote.
:::
::::

::::{prf:example}
:label: ratFuncExam3
Let $f(x)=\frac{x^2+x-6}{x^2-x-12}$. Find horizontal and vertical asymptotes.

Before moving forwards, we should first simplify $f$.

:::{dropdown} Solution:
\begin{align*}
    \frac{x^2+x-6}{x^2-x-12} & = \frac{(x+3)(x-2)}{(x-4)(x+3)}\\
    & = \frac{x-2}{x-4}
\end{align*}
:::

Horizontal asymptote:
:::{dropdown} Solution:
First, we will divide the top and bottom by $x$.

\begin{align*}
    \frac{x-2}{x-4} & = \frac{1-\frac{2}{x}}{1-\frac{4}{x}}\\
    & \to \frac{1-0}{1-0}=1
\end{align*}

That is, $f\to 1$ as $x\to \infty$. Therefore, $f$ has a horizontal asymptote $y=1$.
:::

Vertical asymptote:

:::{dropdown} Solution:
Since $f$ simplified is $\frac{x-2}{x-4}$ and $x-4=0$ whenever $x=4$; we say, $x=4$ is a vertical asymptote.
:::
::::

In the last example it is important to note that $f(x)\ne\frac{x-2}{x-4}$ and $f(-3)$ is undefined (point-wise). If you graph $y=\frac{x^2+x-6}{x^2-x-12}$ and $y=\frac{x-2}{x-4}$ you will think they are the same; however, they are not. The graph $y=\frac{x^2+x-6}{x^2-x-12}$ will have a hole when $x=-3$. The graph $y=\frac{x-2}{x-4}$ will not have a hole when $x=-3$.

::::{prf:example}
:label: ratFuncExam4
Find the oblique asymptote for the function $f(x)=\frac{x^2+1}{x-2}$.
:::{dropdown} Solution:
After polynomial long division we have:

$$\frac{x^2+1}{x-2} = x+2+\frac{5}{x-2}$$

Therefore, the oblique asymptote is $y=x+2$. That is, the graph $y=\frac{x^2+1}{x-2}$ will always approach the line $y=x+2$ but never touch it.
:::
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