Domain of Rational Functions

Section 11.1 Domain of Rational Functions

This section is mostly review, but with a focus on rational functions (functions in a fraction). We will look at one new vocabulary word for a thing we have seen before.

Remember what we talked about in Section 1.2 about intercepts:

Fact 11.1.

Since the \(x\)-intercepts are where the graph crosses the \(x\)-axis, and every point there has \(y=0\text{,}\) to find \(x\)-intercepts algebraically, plug \(y=0\) into the equation and solve for \(x\).
Since the \(y\)-intercepts are where the graph crosses the \(y\)-axis, and every point there has \(x=0\text{,}\) to find \(y\)-intercepts algebraically, plug \(x=0\) into the equation and solve for \(y\).

The same still applies for rational functions. Let's look at an example, just to make sure we remember.

Example 11.2.

Suppose \(f(x)=\dfrac{3x-7}{x^2-9}\text{,}\) and we want to find its x- and y-intercepts.

Let's start with the x-intercepts. Remember, that is where the graph crosses the \(x\)-axis, which is where \(y=0\text{.}\) So, our strategy is to set \(y=0\) and solve for \(x\text{:}\)

\begin{align*} 0 \amp= \dfrac{3x-7}{x^2-9}\\ {\color{red}{(x^2-9)}}0 \amp= \dfrac{3x-7}{x^2-9}{\color{red}{(x^2-9)}}\\ 0 \amp= 3x-7\\ 0 {\color{red}{+7}} \amp= 3x-7{\color{red}{+7}} \\ 7\amp=3x\\ \dfrac{7}{\color{red}{3}}\amp=\dfrac{3x}{\color{red}{3}}\\ \dfrac{7}{3}\amp=x \end{align*}

Since we always write our intercepts as ordered pairs, we know that our x-intercept is \(\left(\frac{7}{3},0\right)\text{.}\)

Now, let's find our y-intercept. Remember, that is where the graph crosses the \(y\)-axis, which is where \(x=0\text{.}\) So, our strategy is to set \(x=0\) and solve for \(y\text{:}\)

\begin{equation*} y= \dfrac{3(0)-7}{(0)^2-9} = \dfrac{-7}{-9}=\dfrac{7}{9} \end{equation*}

Since we always write our intercepts as ordered pairs, we know that our y-intercept is \(\left(0,\frac{7}{9}\right)\text{.}\)

Checkpoint 11.3.

Find the x- and y-intercepts of \(f(x)=\dfrac{6x+5}{x+3}\text{.}\)

Answer.

The x-intercept is \(\left(\frac{-5}{6},0\right)\text{.}\) The y-intercept is \(\left(0,\frac{5}{3}\right)\text{.}\)

Solution.

Let's start with the x-intercepts. Remember, that is where the graph crosses the \(x\)-axis, which is where \(y=0\text{.}\) So, our strategy is to set \(y=0\) and solve for \(x\text{:}\)

\begin{align*} 0 \amp= \dfrac{6x+5}{x+3}\\ {\color{red}{(x+3)}}0 \amp= \dfrac{6x+5}{x+3}{\color{red}{(x+3)}}\\ 0 \amp= 6x+5\\ 0 {\color{red}{-5}} \amp= 6x+5{\color{red}{-5}} \\ -5\amp=6x\\ \dfrac{-5}{\color{red}{6}}\amp=\dfrac{6x}{\color{red}{6}}\\ \dfrac{-5}{6}\amp=x \end{align*}

Since we always write our intercepts as ordered pairs, we know that our x-intercept is \(\left(\frac{-5}{6},0\right)\text{.}\)

Now, let's find our y-intercept. Remember, that is where the graph crosses the \(y\)-axis, which is where \(x=0\text{.}\) So, our strategy is to set \(x=0\) and solve for \(y\text{:}\)

\begin{equation*} y= \dfrac{6(0)+5}{(0)+3} = \dfrac{5}{3} \end{equation*}

Since we always write our intercepts as ordered pairs, we know that our y-intercept is \(\left(0,\frac{5}{3}\right)\text{.}\)

Now, let's review what we learned in Section 8.4 about finding domain from a formula. In that case, we had two possible problems we had to contend with: variables in the denominator, and variables in a square root. In this section, we are only focused on functinos that are fractions, no square roots in sight. So, the only problem we need to deal with is when there is a variable in the denominator, so we need to make sure it is never 0.

Fact 11.4. Domain of a Function with Variables in the Denominator.

The domain of a function that has variable in the denominator of a fraction is everything except the \(x\)-values that make the denominator 0.

Example 11.5.

Suppose \(f(x)=\dfrac{4-x}{3x+2}\text{,}\) and we want to find its domain. Well, the denominator is \(3x+2\text{,}\) and the domain is everywhere that that isn't 0. So, let's find where it's zero and then our answer will be everything else:

\begin{align*} 3x+2\amp=0\\ 3x \amp=-2\\ x \amp= -\dfrac{2}{3} \end{align*}

So, our domain is everything except \(x = -\dfrac{2}{3}\text{.}\) Therefore, our domain is \(\left(-\infty, -\frac{2}{3}\right)\cup\left(-\frac{2}{3},\infty\right)\text{.}\)

Definition 11.6.

The vertical asymptote of a rational function is a vertical line where the denominator is 0.

Given the example we just did above, we can see that the vertical asymptote is just a vertical line at the place where the domain is a problem. So, we solve it the same way we did the example before, but our answer will be the problem value, not "everything except the problem".

Example 11.7.

Suppose \(f(x)=\dfrac{4x+1}{5-x}\text{,}\) and we want to find its vertical asymptote. Remember that the vertical asymptote is where the denominator is 0. So, we will set the denominator equal to 0 and solve.

\begin{align*} 5-x \amp=0\\ 5 \amp=x \end{align*}

So, we have our answer. Remember that the vertical asymptote is specifically a vertical line, so our answer is the entire equation \(x=5\text{,}\) not just the number \(5\text{.}\)

Checkpoint 11.8.

Suppose \(f(x)=\dfrac{3x-5}{x-8}\text{.}\) What is the vertical asymptote of \(f(x)\text{?.}\)

Answer.

The vertical asymptote is \(x=8\text{.}\) Remember that the vertical asymptote is specifically a vertical line, so our answer is the entire equation \(x=8\text{,}\) not just the number \(8\text{.}\)

Solution.

Remember that the vertical asymptote is where the denominator is 0. So, we will set the denominator equal to 0 and solve.

\begin{align*} x-8 \amp=0\\ x \amp=8 \end{align*}

So, we have our answer. Remember that the vertical asymptote is specifically a vertical line, so our answer is the entire equation \(x=8\text{,}\) not just the number \(8\text{.}\)