Introduction to Rational Functions

Rational functions are fractions where both the numerator and denominator are polynomials. They're written as $f(x) = \frac{N(x)}{D(x)}$, where $D(x)$ can never equal zero. The simplest rational function is $f(x) = \frac{1}{x}$, which serves as the parent function for all others.

The domain of a rational function includes all real numbers except values that make the denominator zero. This makes sense - you can't divide by zero! Rational functions are continuous throughout their domain, meaning you can draw them without lifting your pencil (except at vertical asymptotes or holes).

When identifying rational functions, check if both numerator and denominator are polynomials with no radicals or other non-polynomial elements. For example, $f(x) = \frac{3x}{x^2 + 1}$ is rational with domain $\mathbb{R}$, while $h(x) = \frac{\sqrt{25x^4 + 4}}{7x^3 + 6x}$ is not rational because of the radical.

Remember This! Always find the domain by determining where the denominator equals zero, then exclude those values. This is crucial for understanding where the function "breaks" on the graph.

Finding Holes and Vertical Asymptotes

One tricky feature of rational functions is holes - points where the function should exist but doesn't. To spot a hole, look for common factors in both numerator and denominator. When you factor out $f(x) = \frac{2x^2-4x-6}{x^2-9x+18}$, you get $\frac{2(x-3)(x+1)}{(x-6)(x-3)}$ - the common factor $(x-3)$ indicates a hole at $x = 3$.

Vertical asymptotes occur when the denominator equals zero (but the numerator doesn't). These create vertical lines that the function approaches but never touches. For example, in $f(x) = \frac{3(x+1)(x+1)}{(x-6)(x+2)}$, the vertical asymptotes are at $x = 6$ and $x = -2$.

You can find vertical asymptotes by setting the denominator equal to zero and solving for x. Just remember that if the same value makes both numerator and denominator zero, you'll have a hole instead of an asymptote.

Quick Tip: When analyzing a rational function, always factor both numerator and denominator completely first. This makes finding holes and asymptotes much easier!

Intercepts and Horizontal Asymptotes

X-intercepts occur when the function equals zero, which happens when the numerator equals zero (assuming that value isn't also a hole). Simply set the numerator equal to zero and solve for x. For example, in $f(x) = \frac{3(x+1)(x+1)}{(x-6)(x+2)}$, the x-intercept is at $(-1,0)$.

Finding the y-intercept is straightforward - just substitute $x = 0$ into the function. In the same example, $f(0) = \frac{3}{-12} = -\frac{1}{4}$, so the y-intercept is $(0, -\frac{1}{4})$.

Horizontal asymptotes depend on the degrees of the numerator and denominator:

• If denominator degree > numerator degree: horizontal asymptote at $y = 0$
• If degrees are equal: horizontal asymptote at $y = \frac{a_n}{b_n}$ (ratio of leading coefficients)
• If numerator degree > denominator degree: no horizontal asymptote

For instance, in $f(x) = \frac{3x^2 + 6x + 3}{x^2 - 4x - 12}$, both polynomials have degree 2, so the horizontal asymptote is $y = \frac{3}{1} = 3$.

Study Strategy: Create a checklist for analyzing rational functions: domain, holes, asymptotes, and intercepts. Working through this checklist for each problem will ensure you don't miss any important features!

Putting It All Together

Sketching rational functions becomes easier when you identify all key features first. For example, with $f(x) = \frac{3(x+1)(x+1)}{(x-6)(x+2)}$, we've found:

• Domain: all reals except $x = 6$ and $x = -2$
• Vertical asymptotes: $x = 6$ and $x = -2$
• Horizontal asymptote: $y = 3$
• X-intercept: $(-1, 0)$
• Y-intercept: $(0, -\frac{1}{4})$

These features serve as a framework for your sketch. Start by drawing the asymptotes as dashed lines. Plot the intercepts, then connect the points while respecting the asymptotes. The function can't cross vertical asymptotes but will approach the horizontal asymptote as $x$ approaches infinity.

For $g(x) = \frac{x - 5}{x^2(x+4)}$, we have vertical asymptotes at $x = 0$ and $x = -4$, a horizontal asymptote at $y = 0$, and an x-intercept at $(5,0)$ with no y-intercept. This function will approach the x-axis $y = 0$ as x gets very large in either direction.

Visualization Tip: Think of vertical asymptotes as "walls" the function can't cross, and horizontal asymptotes as "ceilings" or "floors" the function approaches but never quite reaches as x gets very large.

Slant Asymptotes

When the degree of the numerator is exactly one more than the degree of the denominator, the rational function has a slant asymptote instead of a horizontal one. This creates a diagonal line that the function approaches as x approaches infinity.

To find a slant asymptote, use polynomial long division to divide the numerator by the denominator. The quotient gives you the equation of the slant asymptote. For example, with $f(x) = \frac{x^2 - x}{x + 1}$, dividing gives $x - 2$ with a remainder, so the slant asymptote is $y = x - 2$.

In practice, when graphing $f(x) = \frac{2x^2 - 3x - 5}{x - 2}$, we can factor to get $f(x) = \frac{(2x-5)(x+1)}{x-2}$. Using long division, we find the slant asymptote is $y = 2x + 1$. With the vertical asymptote at $x = 2$ and x-intercepts at $x = \frac{5}{2}$ and $x = -1$, we have enough information to sketch the graph.

Real-World Connection: Functions with slant asymptotes often model situations where there's a steady growth or decline with an initial adjustment period. For example, a company's profit might approach a linear trend after initial startup costs.

Practice with Complex Rational Functions

Let's analyze $f(x) = \frac{4x^2 - 4x - 8}{2x^2 + 6x - 20} = \frac{4(x-2)(x+1)}{2(x-2)(x+5)}$. After factoring, we can immediately identify:

• The common factor $(x-2)$ indicates a hole at $x = 2$
• The remaining factor in the denominator gives us a vertical asymptote at $x = -5$
• Since numerator and denominator have the same degree, the horizontal asymptote is $y = \frac{4}{2} = 2$
• The x-intercept is at $(-1,0)$ from the factored numerator
• The y-intercept is $(0, \frac{2}{5})$ by substituting $x = 0$

When sketching this function, start with the horizontal asymptote at $y = 2$ and vertical asymptote at $x = -5$. Plot the intercepts and remember there's a hole at $x = 2$ (mark it with an open circle). The function will approach but never cross the vertical asymptote and will get closer to the horizontal asymptote $y = 2$ as $x$ gets very large.

Master Tip: When you see a common factor in both numerator and denominator, cancel it immediately but remember to mark the hole in your graph. This simplifies the function while preserving its key features.