Find the hole in $f(x) = \frac{x^2 - 4}{x - 2}$.

Hole at $(2, 4)$; the graph is the line $y = x + 2$ with that point missing.

Find the hole and the vertical asymptote of $g(x) = \frac{x^2 + 3x}{x^2 - 9}$.

Hole at $(-3, \frac{1}{2})$, vertical asymptote $x = 3$.

For $h(x) = \frac{x^2 - 5x + 6}{x^2 - x - 2}$, find the hole, the asymptotes and the intercepts.

Hole $(2, -\frac{1}{3})$, vertical asymptote $x = -1$, horizontal asymptote $y = 1$, intercepts $(3, 0)$ and $(0, -3)$.

Does $k(x) = \frac{2x^2 - 2x}{x^2 - 2x + 1}$ have a hole?

No hole. Vertical asymptote $x = 1$, and the graph is that of $\frac{2x}{x - 1}$.

Write a rational function with a hole at $(3, 1)$, a vertical asymptote at $x = -1$ and a horizontal asymptote at $y = 2$.

$f(x) = \frac{(2x - 2)(x - 3)}{(x + 1)(x - 3)} = \frac{2x^2 - 8x + 6}{x^2 - 2x - 3}$

Grade 12 · MHF4U · Advanced Functions

MHF4U: Holes in Rational Functions

A hole is a single missing point on the graph of a rational function. It appears where a factor cancels from the numerator and the denominator: the function is still undefined there, but the graph does not blow up, it just skips one point.

How It Works

Factor the numerator and the denominator fully. Any factor $(x - a)$ that appears in both cancels, and $x = a$ is a hole. The domain still excludes $a$ , because the original denominator is zero there, so write the restriction down before you cancel.

To find the hole's $y$ -coordinate, substitute $x = a$ into the simplified function. For $f(x) = \frac{x^2 - 4}{x - 2}$ the simplified form is $x + 2$ , so the hole is at $(2, 4)$ and the graph is the line $y = x + 2$ with one open circle.

What does not cancel in the denominator gives the vertical asymptotes. So a rational function can have holes and asymptotes at once, and the test is the same: factor, cancel, and look at what is left on the bottom.

Hole or asymptote? Count the copies. If $(x - 1)$ appears twice on the bottom and once on top, cancelling one copy leaves another, and $x = 1$ is still a vertical asymptote. A hole needs the factor to disappear from the denominator completely.

Every other feature is read from the simplified function: intercepts, horizontal asymptote, the sketch. Just check that no intercept lands on the hole. If the cancelled factor was also the only zero of the numerator, the graph has an open circle on the axis instead of an $x$ -intercept.

Going backwards is a favourite test question: build a function with a given hole and asymptotes. Write the simplified function first, from the asymptotes and the hole's $y$ -value, then multiply top and bottom by $(x - a)$ to put the hole in.

5 Worked Examples

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

Easy
Find the hole in $f(x) = \frac{x^2 - 4}{x - 2}$ .
Show solution
1. Factor: $f(x) = \frac{(x - 2)(x + 2)}{x - 2}$ . The domain excludes $x = 2$ .
2. Cancel: $f(x) = x + 2$ for $x \ne 2$ . At $x = 2$ that gives $y = 4$ .
Answer: Hole at $(2, 4)$ ; the graph is the line $y = x + 2$ with that point missing.
Easy
Find the hole and the vertical asymptote of $g(x) = \frac{x^2 + 3x}{x^2 - 9}$ .
Show solution
1. Factor: $g(x) = \frac{x(x + 3)}{(x - 3)(x + 3)}$ . Domain: $x \ne \pm 3$ .
2. Cancel $(x + 3)$ : $g(x) = \frac{x}{x - 3}$ . The hole is at $x = -3$ , where $y = \frac{-3}{-6} = \frac{1}{2}$ .
3. $(x - 3)$ did not cancel, so $x = 3$ is a vertical asymptote.
Answer: Hole at $(-3, \frac{1}{2})$ , vertical asymptote $x = 3$ .
Medium
For $h(x) = \frac{x^2 - 5x + 6}{x^2 - x - 2}$ , find the hole, the asymptotes and the intercepts.
Show solution
1. Factor: $h(x) = \frac{(x - 2)(x - 3)}{(x - 2)(x + 1)}$ . Domain: $x \ne 2, -1$ .
2. Cancel $(x - 2)$ : $h(x) = \frac{x - 3}{x + 1}$ . Hole at $x = 2$ : $y = \frac{-1}{3}$ .
3. Vertical asymptote $x = -1$ . Degrees are equal, so the horizontal asymptote is $y = \frac{1}{1} = 1$ .
4. $x$ -intercept where the simplified numerator is zero: $x = 3$ . $y$ -intercept: $h(0) = \frac{-3}{1} = -3$ .
Answer: Hole $(2, -\frac{1}{3})$ , vertical asymptote $x = -1$ , horizontal asymptote $y = 1$ , intercepts $(3, 0)$ and $(0, -3)$ .
Medium
Does $k(x) = \frac{2x^2 - 2x}{x^2 - 2x + 1}$ have a hole?
Show solution
1. Factor: $k(x) = \frac{2x(x - 1)}{(x - 1)^2}$ .
2. Cancel one $(x - 1)$ : $k(x) = \frac{2x}{x - 1}$ . A copy of $(x - 1)$ is still in the denominator.
3. So the function still blows up at $x = 1$ : it is a vertical asymptote, not a hole.
Answer: No hole. Vertical asymptote $x = 1$ , and the graph is that of $\frac{2x}{x - 1}$ .
Test level
Write a rational function with a hole at $(3, 1)$ , a vertical asymptote at $x = -1$ and a horizontal asymptote at $y = 2$ .
Show solution
1. Build the simplified function first. A vertical asymptote at $x = -1$ needs $(x + 1)$ on the bottom; a horizontal asymptote $y = 2$ needs equal degrees with leading coefficients in ratio $2$ : try $\frac{2x + c}{x + 1}$ .
2. The hole's $y$ -value is the simplified function at $x = 3$ : $\frac{6 + c}{4} = 1$ , so $c = -2$ . The simplified function is $\frac{2x - 2}{x + 1}$ .
3. Put the hole in by multiplying top and bottom by $(x - 3)$ : $\frac{(2x - 2)(x - 3)}{(x + 1)(x - 3)}$ .
4. Expand if asked: $\frac{2x^2 - 8x + 6}{x^2 - 2x - 3}$ . Check: at $x = 3$ the simplified form gives $\frac{4}{4} = 1$ , the bottom still has $x + 1$ , and the leading coefficients give $y = 2$ .
Answer: $f(x) = \frac{(2x - 2)(x - 3)}{(x + 1)(x - 3)} = \frac{2x^2 - 8x + 6}{x^2 - 2x - 3}$

Learn It on Martian Lab

Martian Lab teaches this in the Intercepts & Asymptotes of Rational Functions 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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Free Rational Functions practice test, with solutions

More Rational Functions topics: Vertical and Horizontal Asymptotes · Graphing Rational Functions · Rational Equations and Inequalities

More MHF4U topics: Long and Synthetic Division · Proving Trig Identities · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations · End Behaviour and Degree · Finding the Equation of a Polynomial · Even and Odd Functions · Remainder and Factor Theorem · Solving Polynomial Equations · Polynomial Inequalities · Sum and Difference of Cubes · Radian Measure · Unit Circle and Special Angles · Compound Angle Formulas · Double Angle Formulas · Graphing Sine and Cosine with Transformations · Graphs of Tan, Sec, Csc and Cot · Sinusoidal Word Problems · Solving Exponential Equations · Graphing Logarithmic Functions · Exponential and Log Word Problems · Sums, Products and Quotients of Functions · Composition of Functions · Average vs Instantaneous Rate of Change

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