Unit 1 · Topic 1.10 · about 25 minutes · 7 parts · 11 scored checks, 1 of them multiple choice

Rational Functions and Holes

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Graph of r(x) = (x + 4)(x - 3) / ((x + 2)(x - 3)). An open circle marks a hole at (3, 1.4), where the shared factor (x - 3) cancels. A dashed vertical asymptote at x = -2 comes from the factor (x + 2), which is only in the denominator.

Part 1 Think first

A line with a gap

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

Let $r (x) = \frac{x^{2} - 9}{x - 3}$ . For every input except 3, $r (x)$ equals $x + 3$ . What does the graph of $r$ look like at $x = 3$ ?

In short. A hole is a single missing point on the graph of a rational function, where a zero of the denominator is cancelled completely by the numerator.

If a real zero's multiplicity in the numerator is greater than or equal to its multiplicity in the denominator, the graph has a hole there instead of a vertical asymptote.
The hole at $x = c$ sits at $(c, L)$ , where $L$ is the value the outputs approach as the inputs approach $c$ .
In limit notation, $\lim_{x \to c} r (x) = L$ , and the limits from the left and from the right are both $L$ .

Part 2 Learn it

Three short ideas, each checked as you go

2a. Hole or asymptote?

When $c$ is a zero of both the numerator and the denominator, compare multiplicities, as in 1.9.

If the multiplicity of $c$ in the numerator is greater than or equal to its multiplicity in the denominator, every factor $(x - c)$ on the bottom cancels. The graph has a hole at $x = c$ .
If the denominator has more, a factor is left on the bottom and the graph has a vertical asymptote at $x = c$ .

For $\frac{(x - 2) (x + 1)}{x - 2}$ the multiplicities are equal, 1 and 1, so there is a hole at $x = 2$ . For $\frac{(x - 2)^{3}}{(x - 2)^{2}}$ the numerator has more, so there is a hole there too. The function is still not defined at $c$ , because the original denominator is zero.

Quick check · Sort it

No calculator. Tap a function, then what its graph has at $x = 2$ .

A hole at x = 2

A vertical asymptote at x = 2

2b. Where exactly is the hole?

A hole has coordinates $(c, L)$ . The $x$ -coordinate is the cancelled zero, $c$ . The $y$ -coordinate is the value the outputs approach as the inputs get close to $c$ .

To find $L$ , cancel the common factors and evaluate what is left at $x = c$ . For $r (x) = \frac{x^{2} - 9}{x - 3} = \frac{(x - 3) (x + 3)}{x - 3}$ , what is left is $x + 3$ , and at $x = 3$ that is 6. The hole is at $(3, 6)$ .

The cancelled form is not the same function as $r$ : it has 3 in its domain and $r$ does not (1.1, equal functions need equal domains). It is only a tool for finding $L$ .

Quick check · Calculate

No calculator. The graph of $r (x) = \frac{x^{2} - x - 6}{x - 3}$ has a hole at $x = 3$ . What is the $y$ -coordinate of the hole?

Your answer

2c. Holes in limit notation

If input values sufficiently close to $c$ give output values arbitrarily close to $L$ , the hole is at $(c, L)$ , and we write

\lim_{x \to c} r (x) = L

Unlike at a vertical asymptote, the two sides agree: $\lim_{x \to c^{-}} r (x) = \lim_{x \to c^{+}} r (x) = L$ .

You can see this in a table. For $r (x) = \frac{x^{2} - 9}{x - 3}$ : $r (2.9) = 5.9$ , $r (2.99) = 5.99$ , $r (3.01) = 6.01$ , $r (3.1) = 6.1$ . From both sides the outputs close in on 6, even though $r (3)$ does not exist.

Quick check · Fill the statement

No calculator. A rational function $r$ has a hole at $(- 1, 4)$ . Tap a blank, then tap an answer.

As $x$ approaches from either side, the outputs of $r$ approach , and $r (- 1)$ is .

Part 3 Watch it done

A hole and an asymptote in one function

Watch each step, then do the last one yourself.

Find every hole and vertical asymptote of $r (x) = \frac{x^{2} + x - 12}{x^{2} - x - 6}$ , and locate the hole.

Step 1 · Factor both
$r (x) = \frac{(x + 4) (x - 3)}{(x + 2) (x - 3)}$ . The denominator is zero at $x = 3$ and $x = - 2$ .
Step 2 · Classify each zero
At $x = 3$ the factor $(x - 3)$ appears once on top and once on the bottom, so it cancels completely: a hole. At $x = - 2$ the numerator is not zero: a vertical asymptote.
Why this step?
Why does the hole get a $y$ -coordinate, while the asymptote does not?
Near $x = 3$ , after cancelling, the outputs approach a single number, so the missing point has a location. Near $x = - 2$ the denominator approaches zero with nothing to cancel it, so the outputs grow without bound and approach no number at all.
Your step · Calculate

No calculator. Finish it. What is the $y$ -coordinate of the hole? Give a decimal.

Your answer

Part 4 Try it

Practice finding holes

Holes turn up where you might not expect them. In 1.2 you estimated a falling object's speed with average rates of change over shorter and shorter intervals. Written as a function of the interval's end, that average has a hole exactly at the moment you care about.

Try it · Calculate

No calculator. A ball rolls down a ramp and travels $d (t) = 5 t^{2}$ meters in $t$ seconds. Its average speed from 2 seconds to $t$ seconds is $s (t) = \frac{5 t^{2} - 20}{t - 2}$ , which is not defined at $t = 2$ . The graph of $s$ has a hole at $t = 2$ . What is the hole's $y$ -coordinate, in meters per second?

Your answer m/s

Then practice on generated rational functions, each with a factor shared by the top and the bottom. Before you type, decide whether each zero of the denominator is a hole or an asymptote.

Practice

Practice until it is automatic

Holes and vertical asymptotes. For a hole, cancel the shared factor and evaluate what is left.

Asymptotes, zeros and holes practice page

Part 5 Explain it

Write it the way the exam asks

Nothing you type leaves this page.

Free response is 37.5% of the exam score. This one is shaped like Question 1, Function Concepts.

Free-response practice · Question 1 style

Graphing calculator required. Let $r (x) = \frac{2 x^{2} - 8}{x^{2} + x - 6}$ . Explain why the graph of $r$ has a hole at $x = 2$ , and find its coordinates. Use limit notation.

Your answer stays on this page. It is never sent or saved.

Model answer

Factored, $r (x) = \frac{2 (x - 2) (x + 2)}{(x - 2) (x + 3)}$ . The zero 2 has multiplicity 1 in the numerator and 1 in the denominator; since the numerator's multiplicity is greater than or equal to the denominator's, the factor cancels completely and there is a hole, not an asymptote. For $x \neq 2$ , $r (x) = \frac{2 (x + 2)}{x + 3}$ , which approaches $\frac{8}{5}$ as $x$ approaches 2. So $\lim_{x \to 2} r (x) = \frac{8}{5}$ , and the hole is at $(2, \frac{8}{5})$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Factors both polynomials and compares the multiplicity of 2 on top and bottom.
Explains that equal multiplicity means the factor cancels completely, so a hole and not an asymptote.
Finds $\lim_{x \to 2} r (x) = \frac{8}{5}$ from the cancelled form.
Gives the hole as $(2, \frac{8}{5})$ .

Part 6 Check yourself

Finding a hole from a table

The two case questions use the same data. Answer them from the table.

The function $r (x) = \frac{x^{2} + 3 x - 10}{x - 2}$ is not defined at $x = 2$ . The table gives its values close to 2.

Values of $r$ near $x = 2$
$x$	$r (x)$
1.9	6.9
1.99	6.99
2	not defined
2.01	7.01
2.1	7.1

Case question 1 · Calculate

No calculator. The graph of $r$ has a hole at $x = 2$ . What is its $y$ -coordinate?

Your answer

Case question 2 · Multiple choice

No calculator. Which statement describes $r$ near $x = 2$ ?

The rest stand on their own and do not use the case.

Check yourself · Decide and justify

No calculator. Let $r (x) = \frac{(x - 4)^{2}}{(x - 4)^{2} (x + 1)}$ . What does the graph of $r$ have at $x = 4$ ?

Which reason supports your answer?

Check yourself · Spot the error

No calculator. A student locates the hole in the graph of $r (x) = \frac{x^{2} - 1}{x + 1}$ . Tap the line with the mistake.

Part 7 Keep it

Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 1.9 Rational Functions and Vertical Asymptotes · Decide and justify

No calculator. Let $r (x) = \frac{(x - 1)^{2} (x + 3)}{(x - 1)^{3}}$ . Does the graph of $r$ have a vertical asymptote at $x = 1$ ?

Which reason supports your answer?

From 1.7 Rational Functions and End Behavior · Match

No calculator. Tap a function, then what its graph does at the ends.

In three lines

A shared zero with multiplicity on top at least as great as on the bottom gives a hole, not an asymptote.
Cancel the common factor and evaluate what is left at $x = c$ to find the hole's height $L$ .
$\lim_{x \to c} r (x) = L$ , from both sides, even though $r (c)$ does not exist.

Vocabulary

Hole: A single missing point on a graph, at an input where a shared factor cancels completely from the denominator.
Limit at a point: $\lim_{x \to c} r (x) = L$ means outputs get arbitrarily close to $L$ for inputs sufficiently close to $c$ , from both sides.
Cancelled form: The rational expression left after removing common factors; it matches $r$ everywhere except at the cancelled zero.

Course alignment, for teachers

AP Precalculus topic 1.10, Unit 1: Polynomial and Rational Functions.

Skill 3.C: Support conclusions or choices with a logical rationale or appropriate data.

This lesson maps all 2 essential knowledge statements of the topic to the part that teaches each one: 2 are checked by a scored question. The map is in the lesson file.