5.6 Rational Functions

Rational functions are ratios of two polynomials, and they behave very differently from the polynomials you've studied so far. Because division by zero is undefined, these functions can have breaks in their graphs, vertical and horizontal asymptotes, and holes. Knowing how to find and interpret these features is essential for graphing rational functions and applying them to real-world problems.

Rational Function Fundamentals

Arrow notation for rational functions

A rational function is written as the ratio of two polynomial functions:

$f(x) = \frac{P(x)}{Q(x)}$

where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \neq 0$.

Arrow notation is just a compact way to express the same thing using an arrow instead of function notation:

• $f(x) = \frac{x+1}{x-2}$ becomes $x \rightarrow \frac{x+1}{x-2}$
• $g(x) = \frac{3x^2-4x+1}{2x+5}$ becomes $x \rightarrow \frac{3x^2-4x+1}{2x+5}$

You're describing the same function either way. Arrow notation just emphasizes the input-output relationship: you feed in an $x$-value, and the arrow shows what comes out.

Domain of rational functions

The domain of a rational function includes all real numbers except those that make the denominator equal to zero. To find the domain:

1. Set the denominator equal to zero.
2. Solve for $x$.
3. Exclude those values from the domain.

Example 1: For $f(x) = \frac{x+1}{x-2}$, set $x - 2 = 0$, which gives $x = 2$. The domain is all real numbers except 2, written as $\mathbb{R} \setminus \{2\}$.

Example 2: For $g(x) = \frac{3x^2-4x+1}{2x+5}$, set $2x + 5 = 0$, which gives $x = -\frac{5}{2}$. The domain is $\mathbb{R} \setminus \{-\frac{5}{2}\}$.

If the denominator is a quadratic or higher-degree polynomial, you may get multiple excluded values. Factor or use the quadratic formula as needed.

Graphing Rational Functions

Asymptotes of rational functions

Vertical asymptotes occur at $x$-values where the denominator equals zero and the factor does not cancel with the numerator. The graph shoots toward $+\infty$ or $-\infty$ near these values.

Horizontal asymptotes depend on comparing the degree of the numerator to the degree of the denominator. There are three cases:

• Degree of numerator < degree of denominator: The horizontal asymptote is $y = 0$.
• Degree of numerator = degree of denominator: The horizontal asymptote is $y = \frac{a_n}{b_n}$, where $a_n$ and $b_n$ are the leading coefficients of the numerator and denominator.
• Degree of numerator > degree of denominator: There is no horizontal asymptote. Instead, the function has oblique (slant) asymptote behavior, which you can find using polynomial long division.

Example 1: For $f(x) = \frac{2x+1}{x-3}$, the vertical asymptote is at $x = 3$. Since numerator and denominator are both degree 1, the horizontal asymptote is $y = \frac{2}{1} = 2$.

Example 2: For $g(x) = \frac{x^2-4}{x+2}$, notice that $x^2 - 4$ factors as $(x-2)(x+2)$. The $(x+2)$ cancels, so $x = -2$ is actually a hole, not a vertical asymptote. After canceling, the simplified function is $g(x) = x - 2$ (with a hole at $x = -2$). There is no horizontal or oblique asymptote for the simplified linear function.

Graphing rational functions

Follow these steps to graph a rational function:

1. Find the domain by setting the denominator equal to zero. These excluded values are candidates for vertical asymptotes or holes.

2. Factor numerator and denominator. Cancel any common factors to identify holes versus vertical asymptotes.

3. Determine horizontal (or oblique) asymptotes using the degree comparison rules above.

4. Find intercepts:

   • x-intercepts: Set the numerator of the simplified function equal to zero and solve.
   • y-intercept: Evaluate $f(0)$ (plug in $x = 0$).

5. Plot asymptotes, intercepts, and holes. Then test a few additional points on each side of the vertical asymptotes to determine the shape of the curve.

Example: Graph $f(x) = \frac{x+1}{x-2}$:

1. Domain: $\mathbb{R} \setminus \{2\}$
2. No common factors, so vertical asymptote at $x = 2$ (no holes).
3. Degrees are equal (both 1), so horizontal asymptote at $y = \frac{1}{1} = 1$.
4. x-intercept: Set $x + 1 = 0$, giving $(-1, 0)$. y-intercept: $f(0) = \frac{0+1}{0-2} = -\frac{1}{2}$, giving $(0, -\frac{1}{2})$.
5. Draw the asymptotes, plot the intercepts, and sketch the curve approaching the asymptotes.

Note: The original guide stated there was no y-intercept for this function, but there is one at $(0, -\frac{1}{2})$.

Features of rational function graphs

Holes occur when a factor appears in both the numerator and denominator and cancels out. The graph looks continuous except for a single missing point.

To find a hole:

1. Factor the numerator and denominator completely.
2. Identify and cancel common factors.
3. The $x$-value of the canceled factor is the hole's location.
4. Plug that $x$-value into the simplified function to get the $y$-coordinate of the hole.

Example: For $f(x) = \frac{(x-1)(x+2)}{(x-1)(x-3)}$, the $(x-1)$ cancels. There's a hole at $x = 1$. Plugging into the simplified function $\frac{x+2}{x-3}$: $f(1) = \frac{3}{-2} = -\frac{3}{2}$, so the hole is at $(1, -\frac{3}{2})$. The vertical asymptote is at $x = 3$ (the remaining denominator factor).

End behavior describes what happens to $f(x)$ as $x \rightarrow \infty$ or $x \rightarrow -\infty$:

• If the degree of the numerator is less than or equal to the degree of the denominator, the graph approaches the horizontal asymptote.
• If the degree of the numerator is exactly one more than the degree of the denominator, the graph approaches an oblique asymptote (found by polynomial long division).
• If the degree of the numerator exceeds the denominator by two or more, the end behavior grows without bound (no horizontal or oblique asymptote).

Example: For $g(x) = \frac{x^2+1}{x-2}$, the numerator has degree 2 and the denominator has degree 1. Performing long division gives $g(x) = x + 2 + \frac{5}{x-2}$, so the oblique asymptote is $y = x + 2$. As $x \rightarrow \infty$, $g(x) \rightarrow +\infty$, and as $x \rightarrow -\infty$, $g(x) \rightarrow -\infty$.

Applications of Rational Functions

Applications of rational functions

Rational functions model situations where quantities are divided, such as rates, mixtures, and averages. Common types include:

• Rate/work problems: Combining work rates of machines or people
• Mixture problems: Concentrations changing as substances are added
• Average cost problems: Total cost divided by number of units

Steps to solve application problems:

1. Identify what you know and what you're solving for.
2. Set up a rational equation that models the situation.
3. Solve the equation for the unknown.
4. Check that your answer makes sense in context (and isn't an excluded value).

Example (combined work rate): Pipe A fills a tank in 4 hours; Pipe B fills the same tank in 6 hours. How long to fill the tank with both pipes running?

1. Pipe A's rate: $\frac{1}{4}$ tank per hour. Pipe B's rate: $\frac{1}{6}$ tank per hour.
2. Combined rate equation: $\frac{1}{4} + \frac{1}{6} = \frac{1}{t}$
3. Find a common denominator (12): $\frac{3}{12} + \frac{2}{12} = \frac{5}{12} = \frac{1}{t}$, so $t = \frac{12}{5} = 2.4$ hours.
4. This makes sense: the combined time (2.4 hours) is less than either pipe alone, which is what you'd expect.

Advanced Concepts in Rational Functions

Limits, Continuity, and Discontinuity

These ideas give you precise language for describing the behavior of rational functions at tricky points.

A limit describes the value a function approaches as $x$ gets closer to some number (or as $x \rightarrow \pm\infty$). The function doesn't have to actually equal that value at the point; limits are about the trend.

A function is continuous at a point if there are no breaks, jumps, or missing points in the graph there. Rational functions are continuous everywhere in their domain.

Discontinuities in rational functions come in two types:

• Removable discontinuity (hole): Happens when a common factor cancels from numerator and denominator. The limit exists at that point, but the function is undefined there. You could "fill in" the hole to make the function continuous.
• Non-removable discontinuity (vertical asymptote): Happens at values where the denominator is zero and the factor does not cancel. The function blows up toward $\pm\infty$, so no limit exists (as a finite number).

Example: For $f(x) = \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1}$, the $(x-1)$ cancels, leaving $f(x) = x + 1$ with a hole at $x = 1$. The limit as $x \rightarrow 1$ is $1 + 1 = 2$, but $f(1)$ is undefined. This is a removable discontinuity.