MAT 142: Rational Functions and Equations

Reminders

At our last meeting, we encountered and utilized two tools for working with polynomials in expanded form:

The Rational Roots Theorem narrowed down candidates for rational roots.
Polynomial long division was used to factor out a known root \(r\) of \(p\left(x\right)\), rewriting \(p\left(x\right) = \left(x - r\right)q\left(x\right)\) where \(q\left(x\right)\) is of lower degree than \(p\left(x\right)\).
Together, these tools let us factor any polynomial into linear and quadratic factors, ultimately permitting us to find all of its roots.

Try the following warm-up problems.

Problem 1: Use polynomial long division to factor \(p\left(x\right) = x^3 + x^2 - 21x - 45\) and sketch a graph. (Hint: \(\left(x + 3\right)\) is a factor.)

Problem 2: Find the domain of \(\displaystyle{f\left(x\right) = \frac{x^2 - 5x}{x^2 - 25}}\).

Problem 3: Find the domain of \(\displaystyle{f\left(x\right) = \frac{x + 8}{\sqrt{\left(x^2 - 36\right)}}}\).

Note: Problems 2 and 3 both involve domain restrictions, but for different reasons. Think about what makes each one different.

Objectives

Today we’ll leverage those tools from polynomial functions to analyse rational functions, which are ratios of polynomials.

Rational functions introduce genuinely new geometric features that our previous classes of functions don’t have.

Domain restrictions
Asymptotes

After today’s class meeting, you should be able to:

Define rational functions and identify their domains.
Find the roots and \(y\)-intercept of a rational function.
Identify vertical and horizontal asymptotes and explain what they represent.
Use asymptotes, roots, the \(y\)-intercept, and additional reference points to sketch the graph of a rational function.

What Is a Rational Function?

Definition (Rational Function): A function \(f\left(x\right)\) is a rational function if it can be written as

\[r\left(x\right) = \frac{p\left(x\right)}{q\left(x\right)}\]

where \(p\left(x\right)\) and \(q\left(x\right)\) are polynomials.

A few notes:

Polynomial functions are rational functions with \(q\left(x\right) = 1\).
Rational functions introduce a new concern in that the denominator can possibly evaluate to zero, creating domain restrictions and asymptotes.
Finding roots of a rational function means solving \(\displaystyle{\frac{p(x)}{q(x)} = 0}\), which is a rational equation.

Domain, Roots, and \(y\)-Intercept

Consider the general rational function \(\displaystyle{r\left(x\right) = \frac{p\left(x\right)}{q\left(x\right)}}\)

Domain: Since division by zero is undefined, the domain of \(r\left(x\right)\) excludes all \(x\) where \(q\left(x\right) = 0\).

Roots: A fraction equals zero only when its numerator is zero (and its denominator is not). So roots occur where \(p\left(x\right) = 0\), provided \(q\left(x\right) \neq 0\) at those points.

\(y\)-Intercept: Evaluate \(r\left(0\right)\), provided \(0\) is in the domain. If \(x = 0\) is not a permissible input, then there is no \(y\)-intercept.

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0 \end{align}\]

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0 \end{align}\]

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0\\ \implies x \neq 5 &\text{ or } x \neq -4 \end{align}\]

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0\\ \implies x \neq 5 &\text{ or } x \neq -4 \end{align}\]

So the domain is \(x \neq -4\) and \(x \neq 5\), or \(\boxed{~\left(-\infty, -4\right)\cup\left(-4, 5\right)\cup\left(5, \infty\right)~}\).

Roots: Solve \(\displaystyle{\frac{2x^2 - 8x + 6}{x^2 - x - 20} = 0}\)

\[\begin{align} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= 0 \end{align}\]

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0\\ \implies x \neq 5 &\text{ or } x \neq -4 \end{align}\]

So the domain is \(x \neq -4\) and \(x \neq 5\), or \(\boxed{~\left(-\infty, -4\right)\cup\left(-4, 5\right)\cup\left(5, \infty\right)~}\).

Roots: Solve \(\displaystyle{\frac{2x^2 - 8x + 6}{x^2 - x - 20} = 0}\)

\[\begin{align} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= 0\\ \implies 2x^2 - 8x + 6 &= 0 \end{align}\]

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0\\ \implies x \neq 5 &\text{ or } x \neq -4 \end{align}\]

So the domain is \(x \neq -4\) and \(x \neq 5\), or \(\boxed{~\left(-\infty, -4\right)\cup\left(-4, 5\right)\cup\left(5, \infty\right)~}\).

Roots: Solve \(\displaystyle{\frac{2x^2 - 8x + 6}{x^2 - x - 20} = 0}\)

\[\begin{align} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= 0\\ \implies 2x^2 - 8x + 6 &= 0\\ \implies 2\left(x^2 - 4x + 3\right)&= 0 \end{align}\]

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0\\ \implies x \neq 5 &\text{ or } x \neq -4 \end{align}\]

So the domain is \(x \neq -4\) and \(x \neq 5\), or \(\boxed{~\left(-\infty, -4\right)\cup\left(-4, 5\right)\cup\left(5, \infty\right)~}\).

Roots: Solve \(\displaystyle{\frac{2x^2 - 8x + 6}{x^2 - x - 20} = 0}\)

\[\begin{align} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= 0\\ \implies 2x^2 - 8x + 6 &= 0\\ \implies 2\left(x^2 - 4x + 3\right) &= 0\\ \implies x^2 - 4x + 3 &= 0 \end{align}\]

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0\\ \implies x \neq 5 &\text{ or } x \neq -4 \end{align}\]

So the domain is \(x \neq -4\) and \(x \neq 5\), or \(\boxed{~\left(-\infty, -4\right)\cup\left(-4, 5\right)\cup\left(5, \infty\right)~}\).

Roots: Solve \(\displaystyle{\frac{2x^2 - 8x + 6}{x^2 - x - 20} = 0}\)

\[\begin{align} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= 0\\ \implies 2x^2 - 8x + 6 &= 0\\ \implies 2\left(x^2 - 4x + 3\right) &= 0\\ \implies x^2 - 4x + 3 &= 0\\ \implies \left(x - 3\right)\left(x - 1\right) &= 0 \end{align}\]

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0\\ \implies x \neq 5 &\text{ or } x \neq -4 \end{align}\]

So the domain is \(x \neq -4\) and \(x \neq 5\), or \(\boxed{~\left(-\infty, -4\right)\cup\left(-4, 5\right)\cup\left(5, \infty\right)~}\).

Roots: Solve \(\displaystyle{\frac{2x^2 - 8x + 6}{x^2 - x - 20} = 0}\)

\[\begin{align} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= 0\\ \implies 2x^2 - 8x + 6 &= 0\\ \implies 2\left(x^2 - 4x + 3\right) &= 0\\ \implies x^2 - 4x + 3 &= 0\\ \implies \left(x - 3\right)\left(x - 1\right) &= 0\\ \implies x - 3 = 0 &\text{ or } x - 1 = 0 \end{align}\]

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0\\ \implies x \neq 5 &\text{ or } x \neq -4 \end{align}\]

So the domain is \(x \neq -4\) and \(x \neq 5\), or \(\boxed{~\left(-\infty, -4\right)\cup\left(-4, 5\right)\cup\left(5, \infty\right)~}\).

Roots: Solve \(\displaystyle{\frac{2x^2 - 8x + 6}{x^2 - x - 20} = 0}\)

\[\begin{align} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= 0\\ \implies 2x^2 - 8x + 6 &= 0\\ \implies 2\left(x^2 - 4x + 3\right) &= 0\\ \implies x^2 - 4x + 3 &= 0\\ \implies \left(x - 3\right)\left(x - 1\right) &= 0\\ \implies x - 3 = 0 &\text{ or } x - 1 = 0\\ \implies x = 3 &\text{ or } x = 1 \end{align}\]

So the function \(r\left(x\right)\) has roots at \(\boxed{~\left(1, 0\right)~}\) and \(\boxed{~\left(3, 0\right)~}\)

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0\\ \implies x \neq 5 &\text{ or } x \neq -4 \end{align}\]

So the domain is \(x \neq -4\) and \(x \neq 5\), or \(\boxed{~\left(-\infty, -4\right)\cup\left(-4, 5\right)\cup\left(5, \infty\right)~}\).

\(y\)-intecept: Since \(x = 0\) is in the domain of \(r\left(x\right)\), the function has a \(y\)-intercept.

We’ll find it by evaluating \(r\left(0\right)\).

\(\displaystyle{r\left(0\right) = \frac{2\left(0\right)^2 - 8\left(0\right) + 6}{\left(0\right)^2 - \left(0\right) - 20}}\)

Roots: Solve \(\displaystyle{\frac{2x^2 - 8x + 6}{x^2 - x - 20} = 0}\)

\[\begin{align} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= 0\\ \implies 2x^2 - 8x + 6 &= 0\\ \implies 2\left(x^2 - 4x + 3\right) &= 0\\ \implies x^2 - 4x + 3 &= 0\\ \implies \left(x - 3\right)\left(x - 1\right) &= 0\\ \implies x - 3 = 0 &\text{ or } x - 1 = 0\\ \implies x = 3 &\text{ or } x = 1 \end{align}\]

So the function \(r\left(x\right)\) has roots at \(\boxed{~\left(1, 0\right)~}\) and \(\boxed{~\left(3, 0\right)~}\)

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0\\ \implies x \neq 5 &\text{ or } x \neq -4 \end{align}\]

So the domain is \(x \neq -4\) and \(x \neq 5\), or \(\boxed{~\left(-\infty, -4\right)\cup\left(-4, 5\right)\cup\left(5, \infty\right)~}\).

\(y\)-intecept: Since \(x = 0\) is in the domain of \(r\left(x\right)\), the function has a \(y\)-intercept.

We’ll find it by evaluating \(r\left(0\right)\).

\(\displaystyle{r\left(0\right) = \frac{2\left(0\right)^2 - 8\left(0\right) + 6}{\left(0\right)^2 - \left(0\right) - 20} = \frac{6}{-20}}\)

Roots: Solve \(\displaystyle{\frac{2x^2 - 8x + 6}{x^2 - x - 20} = 0}\)

\[\begin{align} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= 0\\ \implies 2x^2 - 8x + 6 &= 0\\ \implies 2\left(x^2 - 4x + 3\right) &= 0\\ \implies x^2 - 4x + 3 &= 0\\ \implies \left(x - 3\right)\left(x - 1\right) &= 0\\ \implies x - 3 = 0 &\text{ or } x - 1 = 0\\ \implies x = 3 &\text{ or } x = 1 \end{align}\]

So the function \(r\left(x\right)\) has roots at \(\boxed{~\left(1, 0\right)~}\) and \(\boxed{~\left(3, 0\right)~}\)

Analysing Rational Functions I

Example 1: For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), find the domain, roots, and \(y\)-intercept.

Solution.

Domain: Solve \(x^2 - x - 20 \neq 0\).

\[\begin{align} x^2 - x - 20 &\neq 0\\ \implies \left(x - 5\right)\left(x + 4\right) &\neq 0\\ \implies x \neq 5 &\text{ or } x \neq -4 \end{align}\]

So the domain is \(x \neq -4\) and \(x \neq 5\), or \(\boxed{~\left(-\infty, -4\right)\cup\left(-4, 5\right)\cup\left(5, \infty\right)~}\).

\(y\)-intecept: Since \(x = 0\) is in the domain of \(r\left(x\right)\), the function has a \(y\)-intercept.

We’ll find it by evaluating \(r\left(0\right)\).

\(\displaystyle{r\left(0\right) = \frac{2\left(0\right)^2 - 8\left(0\right) + 6}{\left(0\right)^2 - \left(0\right) - 20} = \frac{6}{-20} = \boxed{~\frac{-3}{10}~}}\)

Roots: Solve \(\displaystyle{\frac{2x^2 - 8x + 6}{x^2 - x - 20} = 0}\)

\[\begin{align} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= 0\\ \implies 2x^2 - 8x + 6 &= 0\\ \implies 2\left(x^2 - 4x + 3\right) &= 0\\ \implies x^2 - 4x + 3 &= 0\\ \implies \left(x - 3\right)\left(x - 1\right) &= 0\\ \implies x - 3 = 0 &\text{ or } x - 1 = 0\\ \implies x = 3 &\text{ or } x = 1 \end{align}\]

So the function \(r\left(x\right)\) has roots at \(\boxed{~\left(1, 0\right)~}\) and \(\boxed{~\left(3, 0\right)~}\)

Graphing Rational Functions I

We can now draw a partial graph of the function \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\).

As a reminder, \(r\left(x\right)\) has

Graphing Rational Functions I

We can now draw a partial graph of the function \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\).

As a reminder, \(r\left(x\right)\) has

domain restrictions at \(x = -4\) and \(x = 5\)

Graphing Rational Functions I

We can now draw a partial graph of the function \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\).

As a reminder, \(r\left(x\right)\) has

domain restrictions at \(x = -4\) and \(x = 5\)
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\)

Graphing Rational Functions I

We can now draw a partial graph of the function \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\).

As a reminder, \(r\left(x\right)\) has

domain restrictions at \(x = -4\) and \(x = 5\)
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\)
a \(y\)-intercept at \(\left(0, -3/10\right)\)

Analysis of Rational Function Practice

Try It! 1: Identify the domain, roots, and \(y\)-intercept for the function \(\displaystyle{g\left(x\right) = \frac{-4x^2 - 4x + 24}{5x^2 + 5x - 100}}\). Draw a partial graph of the function by identifying locations for discontinuities, the \(y\)-intercept, and any roots.

Asymptotes

Definition (Asymptote): An asymptote is a line (or curve) that a function approaches arbitrarily closely. Rational functions can have vertical, horizontal, or slant asymptotes.

A Common Misconception

You may have previously heard that “A function can never cross its asymptote”, but this is not generally true. A function can never cross its vertical asymptotes, but it can cross its horizontal or slant asymptote.

Definition (Vertical Asymptote): The line \(x = a\) is a vertical asymptote of \(\displaystyle{\frac{p(x)}{q(x)}}\) if \(q\left(a\right) = 0\) and \(p\left(a\right) \neq 0\).

Types of Discontinuities

There are two types of discontinuity that a rational function \(\displaystyle{r\left(x\right) = \frac{p\left(x\right)}{q\left(x\right)}}\) may have.

A hole occurs at \(x = a\) if \(\left(x - a\right)\) is a factor of both \(p\left(x\right)\) and \(q\left(x\right)\).
A vertical asymptote occurs at \(x = a\) when \(\left(x - a\right)\) is a factor of \(q\left(x\right)\), but not \(p\left(x\right)\).

Vertical Asymptotes and Holes

Definition (Vertical Asymptote): The line \(x = a\) is a vertical asymptote of \(\displaystyle{\frac{p(x)}{q(x)}}\) if \(q\left(a\right) = 0\) and \(p\left(a\right) \neq 0\).

Types of Discontinuities

There are two types of discontinuity that a rational function \(\displaystyle{r\left(x\right) = \frac{p\left(x\right)}{q\left(x\right)}}\) may have.

A hole occurs at \(x = a\) if \(\left(x - a\right)\) is a factor of both \(p\left(x\right)\) and \(q\left(x\right)\).
A vertical asymptote occurs at \(x = a\) when \(\left(x - a\right)\) is a factor of \(q\left(x\right)\), but not \(p\left(x\right)\).

Strategy: Solve \(q\left(x\right) = 0\). Each solution is either the location of a vertical asymptote or of a hole.

For \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\), we already found that \(q\left(x\right) = 0\) at \(x = -4\) and \(x = 5\).

Since neither of these are also roots of \(p\left(x\right)\), the discontinuities don’t cancel out – they aren’t holes.

Both of these discontinuities result in vertical asymptotes, represented by the dashed lines in the partial graph we produced earlier.

Vertical Asymptotes Practice

Try It! 2: Determine whether the domain restrictions of \(\displaystyle{g\left(x\right) = \frac{-4x^2 - 4x + 24}{5x^2 + 5x - 100}}\) correspond to vertical asymptotes or holes.

Horizontal Asymptotes

Definition (Horizontal Asymptote): The line \(y = b\) is a horizontal asymptote of \(r\left(x\right)\) if \(\displaystyle{\lim_{x\to\infty}{f\left(x\right)} = b}\) or \(\displaystyle{\lim_{x\to-\infty}{f\left(x\right)} = b}\).

Strategy: Divide numerator and denominator by \(x^n\), where \(n\) is the highest degree in the entire expression. Then take the limit — terms with \(x\) in the denominator vanish.

Example 2: Find any horizontal asmyptotes of the function \(\displaystyle{f\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\).

*Solution.

\[\begin{align} \lim_{x\to -\infty} \frac{2x^2 - 8x + 6}{x^2 - x - 20} \end{align}\]

Horizontal Asymptotes

Definition (Horizontal Asymptote): The line \(y = b\) is a horizontal asymptote of \(r\left(x\right)\) if \(\displaystyle{\lim_{x\to\infty}{f\left(x\right)} = b}\) or \(\displaystyle{\lim_{x\to-\infty}{f\left(x\right)} = b}\).

Strategy: Divide numerator and denominator by \(x^n\), where \(n\) is the highest degree in the entire expression. Then take the limit — terms with \(x\) in the denominator vanish.

Example 2: Find any horizontal asmyptotes of the function \(\displaystyle{f\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\).

*Solution.

\[\begin{align} \lim_{x\to -\infty} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= \lim_{x\to -\infty}{\frac{\frac{2x^2}{x^2} - \frac{8x}{x^2} + \frac{6}{x^2}}{\frac{x^2}{x^2} - \frac{x}{x^2} - \frac{20}{x^2}}} \end{align}\]

Horizontal Asymptotes

Definition (Horizontal Asymptote): The line \(y = b\) is a horizontal asymptote of \(r\left(x\right)\) if \(\displaystyle{\lim_{x\to\infty}{f\left(x\right)} = b}\) or \(\displaystyle{\lim_{x\to-\infty}{f\left(x\right)} = b}\).

Strategy: Divide numerator and denominator by \(x^n\), where \(n\) is the highest degree in the entire expression. Then take the limit — terms with \(x\) in the denominator vanish.

Example 2: Find any horizontal asmyptotes of the function \(\displaystyle{f\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\).

Solution.

\[\begin{align} \lim_{x\to -\infty} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= \lim_{x\to -\infty}{\frac{\frac{2x^2}{x^2} - \frac{8x}{x^2} + \frac{6}{x^2}}{\frac{x^2}{x^2} - \frac{x}{x^2} - \frac{20}{x^2}}}\\ &= \lim_{x\to -\infty}{\frac{2 - \frac{8}{x} + \frac{6}{x^2}}{1 - \frac{1}{x} - \frac{20}{x^2}}} \end{align}\]

Horizontal Asymptotes

Definition (Horizontal Asymptote): The line \(y = b\) is a horizontal asymptote of \(r\left(x\right)\) if \(\displaystyle{\lim_{x\to\infty}{f\left(x\right)} = b}\) or \(\displaystyle{\lim_{x\to-\infty}{f\left(x\right)} = b}\).

Strategy: Divide numerator and denominator by \(x^n\), where \(n\) is the highest degree in the entire expression. Then take the limit — terms with \(x\) in the denominator vanish.

Example 2: Find any horizontal asmyptotes of the function \(\displaystyle{f\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\).

Solution.

\[\begin{align} \lim_{x\to -\infty} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= \lim_{x\to -\infty}{\frac{\frac{2x^2}{x^2} - \frac{8x}{x^2} + \frac{6}{x^2}}{\frac{x^2}{x^2} - \frac{x}{x^2} - \frac{20}{x^2}}}\\ &= \lim_{x\to -\infty}{\frac{2 - \frac{8}{x} + \frac{6}{x^2}}{1 - \frac{1}{x} - \frac{20}{x^2}}}\\ &= \frac{2}{1} \end{align}\]

Horizontal Asymptotes

Definition (Horizontal Asymptote): The line \(y = b\) is a horizontal asymptote of \(r\left(x\right)\) if \(\displaystyle{\lim_{x\to\infty}{f\left(x\right)} = b}\) or \(\displaystyle{\lim_{x\to-\infty}{f\left(x\right)} = b}\).

Strategy: Divide numerator and denominator by \(x^n\), where \(n\) is the highest degree in the entire expression. Then take the limit — terms with \(x\) in the denominator vanish.

Example 2: Find any horizontal asmyptotes of the function \(\displaystyle{f\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\).

Solution.

\[\begin{align} \lim_{x\to \infty} \frac{2x^2 - 8x + 6}{x^2 - x - 20} &= \lim_{x\to \infty}{\frac{\frac{2x^2}{x^2} - \frac{8x}{x^2} + \frac{6}{x^2}}{\frac{x^2}{x^2} - \frac{x}{x^2} - \frac{20}{x^2}}}\\ &= \lim_{x\to \infty}{\frac{2 - \frac{8}{x} + \frac{6}{x^2}}{1 - \frac{1}{x} - \frac{20}{x^2}}}\\ &= \frac{2}{1}\\ &= 2 \end{align}\]

So \(r\left(x\right)\) has a horizontal asymptote at \(y = 2\).

Note that the exact same analysis would show that \(\displaystyle{\lim_{x\to -\infty}}{r\left(x\right)} = 2\).

If the end behavior of a rational function approaches a finite limit, then it will always be the case that \(\displaystyle{\lim_{x\to -\infty}{r\left(x\right)} = \lim_{x\to \infty}{r\left(x\right)}}\).

This is not the case for all function classes though.

Horizontal Asymptotes Practice

Try It! 3: Determine whether the function \(\displaystyle{g\left(x\right) = \frac{-4x^2 - 4x + 24}{5x^2 + 5x - 100}}\) has a horizontal asymptote. Write the equation of the horizontal asymptote, if so.

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Reference Point at \(x = -5\):

\(\displaystyle{r\left(-5\right) = \frac{2\left(-5\right)^2 - 8\left(-5\right) + 6}{\left(-5\right)^2 - \left(-5\right) - 20}}\)

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Reference Point at \(x = -5\):

\(\displaystyle{r\left(-5\right) = \frac{2\left(-5\right)^2 - 8\left(-5\right) + 6}{\left(-5\right)^2 - \left(-5\right) - 20} = \frac{50 + 40 + 6}{25 + 5 - 20}}\)

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Reference Point at \(x = -5\):

\(\displaystyle{r\left(-5\right) = \frac{2\left(-5\right)^2 - 8\left(-5\right) + 6}{\left(-5\right)^2 - \left(-5\right) - 20} = \frac{96}{10}}\)

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Reference Point at \(x = -5\):

\(\displaystyle{r\left(-5\right) = \frac{2\left(-5\right)^2 - 8\left(-5\right) + 6}{\left(-5\right)^2 - \left(-5\right) - 20} = \frac{48}{5}}\)

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Reference Point at \(x = -5\):

\(\displaystyle{r\left(-5\right) = \frac{2\left(-5\right)^2 - 8\left(-5\right) + 6}{\left(-5\right)^2 - \left(-5\right) - 20} = \frac{48}{5} > 2}\)

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Reference Point at \(x = -5\):

\(\displaystyle{r\left(-5\right) = \frac{2\left(-5\right)^2 - 8\left(-5\right) + 6}{\left(-5\right)^2 - \left(-5\right) - 20} = \frac{48}{5} > 2}\)

Reference Point at \(x = 6\):

\(\displaystyle{r\left(6\right) = \frac{2\left(6\right)^2 - 8\left(6\right) + 6}{\left(6\right)^2 - \left(6\right) - 20}}\)

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Reference Point at \(x = -5\):

\(\displaystyle{r\left(-5\right) = \frac{2\left(-5\right)^2 - 8\left(-5\right) + 6}{\left(-5\right)^2 - \left(-5\right) - 20} = \frac{48}{5} > 2}\)

Reference Point at \(x = 6\):

\(\displaystyle{r\left(6\right) = \frac{2\left(6\right)^2 - 8\left(6\right) + 6}{\left(6\right)^2 - \left(6\right) - 20} = \frac{72 - 48 + 6}{36 - 6 - 20}}\)

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Reference Point at \(x = -5\):

\(\displaystyle{r\left(-5\right) = \frac{2\left(-5\right)^2 - 8\left(-5\right) + 6}{\left(-5\right)^2 - \left(-5\right) - 20} = \frac{48}{5} > 2}\)

Reference Point at \(x = 6\):

\(\displaystyle{r\left(6\right) = \frac{2\left(6\right)^2 - 8\left(6\right) + 6}{\left(6\right)^2 - \left(6\right) - 20} = \frac{30}{10}}\)

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Reference Point at \(x = -5\):

\(\displaystyle{r\left(-5\right) = \frac{2\left(-5\right)^2 - 8\left(-5\right) + 6}{\left(-5\right)^2 - \left(-5\right) - 20} = \frac{48}{5} > 2}\)

Reference Point at \(x = 6\):

\(\displaystyle{r\left(6\right) = \frac{2\left(6\right)^2 - 8\left(6\right) + 6}{\left(6\right)^2 - \left(6\right) - 20} = 3}\)

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Reference Point at \(x = -5\):

\(\displaystyle{r\left(-5\right) = \frac{2\left(-5\right)^2 - 8\left(-5\right) + 6}{\left(-5\right)^2 - \left(-5\right) - 20} = \frac{48}{5} > 2}\)

Reference Point at \(x = 6\):

\(\displaystyle{r\left(6\right) = \frac{2\left(6\right)^2 - 8\left(6\right) + 6}{\left(6\right)^2 - \left(6\right) - 20} = 3 > 2}\)

Now we draw an approximate graph respecting all asymptotes and reference points.

Sketching the Complete Graph

We now know a great deal about \(\displaystyle{r\left(x\right) = \frac{2x^2 - 8x + 6}{x^2 - x - 20}}\). It has…

vertical asymptotes at \(x = -4\) and \(x = 5\),
a horizontal asymptote at \(y = 2\),
roots at \(\left(1, 0\right)\) and \(\left(3, 0\right)\), and
a \(y\)-intercept at \(\left(0, -3/10\right)\)

To complete the sketch, we’ll need reference points from each of the domain regions. We’ll find one to the left of \(x = -4\) and one to the right of \(x = 5\) because we already have reference points in the middle section.

Reference Point at \(x = -5\):

\(\displaystyle{r\left(-5\right) = \frac{2\left(-5\right)^2 - 8\left(-5\right) + 6}{\left(-5\right)^2 - \left(-5\right) - 20} = \frac{48}{5} > 2}\)

Reference Point at \(x = 6\):

\(\displaystyle{r\left(6\right) = \frac{2\left(6\right)^2 - 8\left(6\right) + 6}{\left(6\right)^2 - \left(6\right) - 20} = 3 > 2}\)

Now we draw an approximate graph respecting all asymptotes and reference points.

Sketching the Graph Practice

Try It! 4: Use your findings from the earlier Try It! activities to sketch the graph of \(\displaystyle{g\left(x\right) = \frac{-4x^2 - 4x + 24}{5x^2 + 5x - 100}}\). Be sure to include any holes, vertical asymptotes, horizontal asymptote, roots, the \(y\)-intercept, and at least one reference point from each region.

Before sketching: Based on your analysis of the function \(g\left(x\right)\) so far, how many separate “pieces” will the graph have? In which region will each root appear?

The Full Strategy

Strategy (Analyzing and Graphing a Rational Function): Consider the rational function \(\displaystyle{r\left(x\right) = \frac{p\left(x\right)}{q\left(x\right)}}\).

Start by factoring both \(p\left(x\right)\) and \(q\left(x\right)\) completely.
Domain: find all \(x\) where \(q\left(x\right) = 0\) and exclude them.
Vertical asymptotes and holes: each excluded \(x\) value corresponds to either a hole or a vertical asymptote.
- A hole occurs at \(x = a\) if \(\left(x - a\right)\) is a factor of both \(p\left(x\right)\) and \(q\left(x\right)\).
- A vertical asymptote occurs at \(x = a\) if \(\left(x - a\right)\) is a factor of \(q\left(x\right)\), but not \(p\left(x\right)\).
Roots: solve \(p\left(x\right) = 0\), and verify that the denominator is not also \(0\) there (otherwise there is a hole instead of a root).
\(y\)-intercept: evaluate \(r\left(0\right)\), again checking that \(0\) is a permissible input. Otherwise, the function \(r\left(x\right)\) has no \(y\)-intercept.
Horizontal asymptote: evaluate \(\displaystyle{\lim_{x\to\infty}{\frac{p\left(x\right)}{q\left(x\right)}}}\) by dividing every term in the numerator and denominator by the highest power of \(x\) in the entire function.
- Terms with a remaining \(x\) in the denominator after simplifying will vanish as \(x\to\infty\).
Reference points: evaluate \(r\left(x\right)\) for at least one \(x\)-value in each region beyond and between asymptotes.
- Compare the height (\(y\)-coordinate) of the reference point to any horizontal asymptote (if one exists) and be sure to plot the location of the reference point appropriately.
Sketch: connect the dots, respecting asymptotes and root behavior.

The Full Strategy Practice

Try It! 5: Conduct a full analysis of the rational function \(\displaystyle{h\left(x\right) = \frac{x^4 - 7x^3 - 9x^2 + 115x - 100}{x^4 - 7x^3 - 25x^2 + 151x - 120}}\)

Hints. The expression \(x^4 - 7x^3 - 9x^2 + 115x -100\) factors into \(\left(x + 4\right)\left(x - 1\right)\left(x - 5\right)^2\). Similarly, the expression \(x^4 - 7x^3 - 25x^2 + 151x - 120\) factors into \(\left(x+ 5\right)\left(x - 1\right)\left(x - 3\right)\left(x - 8\right)\).

Exit Ticket Task

Navigate to our MAT142 Exit Ticket Form, answer the questions, and complete the task below.

Note. Today’s discussion is listed as 14. Rational Functions and Equations

Task: Consider \(\displaystyle{j\left(x\right) = \frac{x - 4}{x^2 - x - 6}}\).

\(\left(a\right)\) Find the domain of \(j\left(x\right)\).

\(\left(b\right)\) Find the roots of \(j\left(x\right)\), if any exist.

\(\left(c\right)\) Find any vertical and horizontal asymptotes.

\(\left(d\right)\) Sketch an approximate graph of \(j\left(x\right)\).

Summary and Next Time…

Ideas From Today

A rational function \(\displaystyle{r\left(x\right) = \frac{p(x)}{q(x)}}\) is undefined where \(q\left(x\right) = 0\) — those are domain restrictions.
Vertical asymptotes occur at \(x = a\) where \(q\left(a\right) = 0\) but \(p\left(a\right)\neq 0\).
Holes occur at \(x = a\), where \(q\left(a\right) = 0\) and \(p\left(a\right) = 0\).
Horizontal asymptotes describe end behavior. A rational function may have zero or one horizontal asymptote. Find it by evaluating \(\displaystyle{\lim_{x\to \infty}{\frac{p\left(x\right)}{q\left(x\right)}}}\).
Roots come from the numerator: solve \(p\left(x\right) = 0\).
Use reference points between asymptotes and roots to determine the location of the graph of \(r\left(x\right)\) in each domain region, then sketch.

Looking Ahead

We’ll continue our exploration of different function classes with radical functions.
This class of function introduces us to a new type of domain restriction.
Our analysis will include more of the same – domain, \(y\)-intercept, roots, end behavior, etc.

Next Time:
Radical Functions

Homework:
Start Homework 10 on MyOpenMath