Rational functions and graphing: VCE Specialist Mathematics Unit 3

What this dot point is asking

VCAA wants you to sketch rational functions by locating asymptotes (vertical, horizontal and oblique), intercepts and turning behaviour, and to understand how the reciprocal transformation $y \mapsto 1/y$ and the modulus transformation $y \mapsto |y|$ reshape a known graph. These skills feed into curve sketching with calculus and into the analysis of functions throughout Units 3 and 4.

Anatomy of a rational function

A rational function is a ratio of polynomials $f(x) = \dfrac{P(x)}{Q(x)}$ where $Q$ is not the zero polynomial. To sketch it, work through a fixed checklist.

Domain and vertical asymptotes. The function is undefined where $Q(x) = 0$. If $P$ and $Q$ have no common factor there, the line $x = a$ is a vertical asymptote and the graph shoots to $\pm\infty$ on each side. If $P$ and $Q$ share the factor $(x - a)$, there is instead a hole (point discontinuity) at $x = a$.

Intercepts. The $y$-intercept is $f(0)$ (when $0$ is in the domain). The $x$-intercepts are the zeros of the numerator that are not also zeros of the denominator, that is, the solutions of $P(x) = 0$ with $Q(x) \neq 0$.

End behaviour: horizontal and oblique asymptotes

Compare the degree of the numerator, $\deg P$, with the degree of the denominator, $\deg Q$.

• If $\deg P < \deg Q$, then $f(x) \to 0$ as $x \to \pm\infty$, so $y = 0$ is a horizontal asymptote.
• If $\deg P = \deg Q$, then $f(x) \to \dfrac{a_P}{a_Q}$ (ratio of leading coefficients), a horizontal asymptote.
• If $\deg P = \deg Q + 1$, there is an oblique (slant) asymptote. Divide $P$ by $Q$ to write $f(x) = (mx + c) + \dfrac{R(x)}{Q(x)}$ where $\deg R < \deg Q$; the line $y = mx + c$ is the asymptote since the remainder term tends to $0$.

A graph may cross a horizontal or oblique asymptote at finite $x$; the asymptote only describes behaviour as $x \to \pm\infty$.

The reciprocal transformation

Given the graph of $y = f(x)$, the graph of $y = \dfrac{1}{f(x)}$ behaves as follows:

• where $f(x) = 0$, the reciprocal has a vertical asymptote;
• where $f(x)$ has a vertical asymptote, the reciprocal has a zero (it crosses the $x$-axis);
• the points where $f(x) = 1$ or $f(x) = -1$ are fixed, since $1/1 = 1$ and $1/(-1) = -1$;
• a local maximum of $f$ (above the axis) becomes a local minimum of $1/f$, and vice versa, because reciprocation reverses the order of positive values;
• the sign of $1/f$ matches the sign of $f$ everywhere.

The modulus transformation

The graph of $y = |f(x)|$ keeps every part of $y = f(x)$ that is already on or above the $x$-axis and reflects every part below the $x$-axis up across the $x$-axis. The result is non-negative everywhere, with sharp corners where the original graph crossed the $x$-axis. By contrast $y = f(|x|)$ takes the part of $f$ for $x \geq 0$ and reflects it across the $y$-axis, producing an even function.

Examples in context

Example 1. Equal degrees. For $f(x) = \dfrac{2x^2 - 1}{x^2 + 3}$, the degrees match, so the horizontal asymptote is $y = \frac{2}{1} = 2$. The $y$-intercept is $\frac{-1}{3}$.

Example 2. Reciprocal of a parabola. For $f(x) = x^2 - 4$, the reciprocal $\frac{1}{x^2 - 4}$ has vertical asymptotes at $x = \pm 2$ (zeros of $f$) and a maximum at $x = 0$ where $f$ has a minimum of $-4$, giving $1/f = -\frac14$.

Try this

Q1. State all asymptotes of $f(x) = \dfrac{3x - 6}{x + 1}$. [2 marks]

• Cue. Vertical $x = -1$; equal degrees so horizontal $y = 3$.

Q2. Find the oblique asymptote of $f(x) = \dfrac{2x^2 + x}{x - 1}$. [2 marks]

• Cue. Division gives $2x + 3 + \frac{3}{x-1}$, so $y = 2x + 3$.

Q3. Describe how $y = |x^2 - 1|$ differs from $y = x^2 - 1$. [2 marks]

• Cue. The dip between $x = -1$ and $x = 1$ (where $y < 0$) is reflected up, giving two corners at $(\pm 1, 0)$.