Graphs of Rational Functions

A rational function is a ratio of two polynomials, written as $f(x) = \frac{P(x)}{Q(x)}$. Graphing these functions requires analyzing their behavior at undefined points and extreme values. To sketch a rational function accurately, you need to find its asymptotes, holes, and intercepts.

Key Features of Rational Graphs

Vertical Asymptotes (VA)

Vertical asymptotes occur where the denominator is zero, meaning the function is undefined. To find them, set the simplified denominator $Q(x)$ to zero and solve for $x$.

Holes (Removable Discontinuities)

If a factor can be canceled from both the numerator and the denominator, there is a "hole" in the graph at that $x$-value, rather than a vertical asymptote. For example, in $f(x) = \frac{(x-2)(x+1)}{x-2}$, there is a hole at $x = 2$.

Horizontal Asymptotes (HA)

Horizontal asymptotes describe the end behavior of the graph as $x$ approaches positive or negative infinity. Compare the degree of the numerator ($n$) to the degree of the denominator ($m$):

• If $n < m$: The horizontal asymptote is the x-axis, $y = 0$.
• If $n = m$: The horizontal asymptote is $y = \frac{a}{b}$, where $a$ and $b$ are the leading coefficients of the numerator and denominator.
• If $n > m$: There is no horizontal asymptote. (If $n$ is exactly one more than $m$, there is an oblique/slant asymptote found by polynomial division).

Intercepts

• y-intercept: Evaluate $f(0)$.
• x-intercepts: Set the simplified numerator $P(x)$ to zero and solve for $x$.

Example 1: Finding Asymptotes

Problem: Find all asymptotes of $f(x) = \frac{2x^2 + 1}{x^2 - 4}$.

Solution:

1. Vertical Asymptotes: Set the denominator to zero. $x^2 - 4 = 0 \implies (x - 2)(x + 2) = 0$ The vertical asymptotes are $x = 2$ and $x = -2$.
2. Horizontal Asymptote: The degree of the numerator (2) equals the degree of the denominator (2). Divide the leading coefficients: $y = \frac{2}{1} = 2$ The horizontal asymptote is $y = 2$.

Example 2: Analyzing Intercepts and Asymptotes

Problem: Analyze $f(x) = \frac{x - 1}{(x + 2)(x - 3)}$ to find all asymptotes and intercepts.

Solution:

1. Vertical Asymptotes: The denominator is zero at $x = -2$ and $x = 3$.
2. Horizontal Asymptote: The numerator has a degree of 1, and the denominator has a degree of 2 (since $x \cdot x = x^2$). Because $n < m$, the horizontal asymptote is $y = 0$.
3. x-intercept: Set the numerator to zero: $x - 1 = 0 \implies x = 1$. The x-intercept is $(1, 0)$.
4. y-intercept: Evaluate $f(0)$: $f(0) = \frac{0 - 1}{(0 + 2)(0 - 3)} = \frac{-1}{2 \cdot (-3)} = \frac{1}{6}$ The y-intercept is $(0, \frac{1}{6})$.