Grade 12 · MHF4U · Advanced Functions

MHF4U: Vertical and Horizontal Asymptotes

An asymptote is a line the graph gets closer and closer to without settling on it. For a rational function, you can read every asymptote off the equation once it is factored.

How It Works

Vertical asymptotes live where the denominator is zero and the numerator is not. So factor first, top and bottom. If a factor cancels, its zero is not an asymptote: it is a hole, a single missing point. Find the hole's $y$ by plugging its $x$ into the simplified function.

Horizontal asymptotes describe the ends of the graph, and the degrees decide them:

Numerator degree lower than the denominator's: $y = 0$ .

Degrees equal: $y$ is the ratio of the leading coefficients. For $\frac{6x^2 - 5}{2x^2 - 8}$ that is $y = \frac{6}{2} = 3$ .

Numerator exactly one degree higher: no horizontal asymptote, but there is an oblique one. Long-divide, and the quotient is the line.

Numerator two or more degrees higher: no horizontal or oblique asymptote.

The reason is that far from the origin only the leading terms matter. In $\frac{2x + 1}{x - 3}$ , a huge $x$ makes it about $\frac{2x}{x} = 2$ .

Two traps. A cancelled factor is a hole, not an asymptote, and it stops being an $x$ -intercept too. And the graph can cross a horizontal asymptote in the middle; the asymptote only promises what happens at the ends. A vertical asymptote is never crossed, since the function is undefined there.

Tests also run it backwards: give you the asymptotes and intercepts, and ask for the equation. Each vertical asymptote $x = a$ is a factor $(x - a)$ on the bottom, each $x$ -intercept is a factor on top, and the horizontal asymptote fixes the leading coefficient.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Find the asymptotes of $f(x) = \frac{3}{x - 4}$ .
Show solution
1. The denominator is zero at $x = 4$ , and the numerator $3$ is never zero, so $x = 4$ is a vertical asymptote.
2. The numerator has degree 0 and the denominator degree 1. Lower on top means $y = 0$ .
Answer: Vertical asymptote $x = 4$ , horizontal asymptote $y = 0$ .
Easy
Find the asymptotes of $f(x) = \frac{2x + 1}{x - 3}$ .
Show solution
1. The denominator is zero at $x = 3$ , where the numerator is $7$ , so $x = 3$ is a vertical asymptote.
2. Both degrees are 1, so take the ratio of leading coefficients: $y = \frac{2}{1} = 2$ .
Answer: Vertical asymptote $x = 3$ , horizontal asymptote $y = 2$ .
Medium
Find the asymptotes and any hole of $f(x) = \frac{x^2 - 4}{x^2 - x - 2}$ .
Show solution
1. Factor: $f(x) = \frac{(x - 2)(x + 2)}{(x - 2)(x + 1)}$ .
2. $(x - 2)$ cancels, leaving $\frac{x + 2}{x + 1}$ with $x \ne 2$ . So $x = 2$ is a hole, not an asymptote.
3. The hole's $y$ : $\frac{2 + 2}{2 + 1} = \frac{4}{3}$ . The hole is at $(2, \frac{4}{3})$ .
4. The remaining denominator is zero at $x = -1$ : vertical asymptote.
5. Degrees are equal (2 and 2) with leading coefficients 1 and 1, so $y = 1$ .
Answer: Vertical asymptote $x = -1$ , horizontal asymptote $y = 1$ , hole at $(2, \frac{4}{3})$ .
Medium
Find the asymptotes of $f(x) = \frac{2x^2 + 3x - 1}{x + 2}$ .
Show solution
1. At $x = -2$ the denominator is zero and the numerator is $8 - 6 - 1 = 1$ , so $x = -2$ is a vertical asymptote.
2. The numerator is one degree higher, so there is no horizontal asymptote. Look for an oblique one.
3. Long divide: $\frac{2x^2}{x} = 2x$ , and $2x(x + 2) = 2x^2 + 4x$ . Subtract to get $-x - 1$ .
4. $\frac{-x}{x} = -1$ , and $-1(x + 2) = -x - 2$ . Subtract to get $1$ .
5. So $f(x) = 2x - 1 + \frac{1}{x + 2}$ . The fraction shrinks to 0 at the ends, so the graph approaches $y = 2x - 1$ .
Answer: Vertical asymptote $x = -2$ , oblique asymptote $y = 2x - 1$ , no horizontal asymptote.
Test level
Write a rational function with vertical asymptotes $x = 2$ and $x = -3$ , horizontal asymptote $y = -2$ , and $x$ -intercepts $1$ and $5$ .
Show solution
1. Vertical asymptotes give the denominator: $(x - 2)(x + 3)$ .
2. The $x$ -intercepts give the numerator: $(x - 1)(x - 5)$ .
3. Both are degree 2, so the horizontal asymptote is the ratio of leading coefficients. Put $a$ in front of the numerator: $y = \frac{a}{1} = a$ , so $a = -2$ .
4. $f(x) = \frac{-2(x - 1)(x - 5)}{(x - 2)(x + 3)}$ . Check: no factor cancels, so both asymptotes and both intercepts survive.
Answer: $f(x) = \frac{-2(x - 1)(x - 5)}{(x - 2)(x + 3)} = \frac{-2x^2 + 12x - 10}{x^2 + x - 6}$

Learn It on Martian Lab

Martian Lab teaches this in the Intercepts & Asymptotes lesson of its MHF4U Rational Functions unit. Watch it, then take the Rational Functions practice tests to see if it stuck.

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More MHF4U topics: Long and Synthetic Division · Proving Trig Identities · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations

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