Knowledge Notes | Carey AP Precalculus Practice Lab

THINK BEFORE YOU CALCULATE

If a factor cancels, does the missing input become allowed?

Compare $f(x)=\frac{x^2-1}{x-1}$ with $g(x)=x+1$ . They produce the same output at every input allowed for $f$ , but their domains differ.

Make a prediction. Keep it in mind as you test the representation.

Cancellation changes the rule's appearance, not the original domain.

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FROM OBSERVATION TO EVIDENCE

Build an answer that explains the mathematics

1
Start with the original domain
Set the original denominator equal to zero. This input is excluded before any cancellation takes place.
Write the evidence
$x-1=0\Rightarrow x=1$ , so $\operatorname{Dom}(f)=(-\infty,1)\cup(1,\infty)$ .
2
Simplify on the shared domain
Use the difference of squares, then cancel the common factor only for allowed inputs.
Write the evidence
$f(x)=\frac{(x-1)(x+1)}{x-1}=x+1\quad\text{for }x\ne1.$
3
Locate and explain the hole
The excluded input is the horizontal coordinate. Evaluate the simplified rule there to find the missing height. This value describes the hole; it is not an original function value.
Write the evidence
$1+1=2$ , so the graph of $f$ is the line $y=x+1$ with a hole at $(1,2)$ .

Read one complete model response

The original function is undefined at $x=1$ . For every $x\ne1$ , it simplifies to $x+1$ . The simplified rule gives $2$ at the excluded input, so $f$ has a removable discontinuity, shown by a hole at $(1,2)$ .

Carey teaching response. Use the evidence requested by your actual question; these steps are not additional official scoring points.

A COMMON WRONG TURN

Spot the exact reasoning error

Incomplete or incorrect response

“After cancellation, $f(1)=2$ , so there is no hole.”

What needs to change

Cancellation preserves outputs only on the original domain. The simplified expression cannot restore an input at which the original denominator was zero.

TRY THE IDEA IN A NEW SETTING

Can you transfer the method?

For $r(x)=\frac{(x-2)(x+3)}{(x-2)(x-4)}$ , give the original restrictions, the hole, and the vertical asymptote.

Compare your reasoning

Exclude $x=2$ and $x=4$ . The simplified rule is $\frac{x+3}{x-4}$ on that domain. At $x=2$ , it gives $-5/2$ , so the hole is $(2,-5/2)$ . The remaining denominator zero gives the vertical asymptote $x=4$ .

CONNECT THE WHOLE TOPIC

Complete topic reference

Use the guided example above, then connect it to the remaining ideas and representations in this topic.

What You Need to Know

Key idea

A hole is a single missing point created when a common factor cancels from a rational function.

Fast decision rule

Find the cancelled input, then evaluate the reduced function at that input to locate the missing point.

Worked example

If a factor cancels at $x=1$ and the reduced rule gives output $2$ there, the hole is at $\left(1,\,2\right)$ .

Common mistake

Putting the cancelled input back into the domain.

Mini check

What two coordinates are needed to name a hole?

Learning Objectives

Key idea

1.10.A determines holes in graphs of rational functions.

Fast decision rule

Factor, cancel, preserve the restriction, and evaluate the reduced rule.

Worked steps

$1$ . Factor numerator and denominator. $2$ . Identify the common factor. $3$ . Record its zero as excluded. $4$ . Simplify. $5$ . Evaluate the simplified rule at the excluded input.

Common mistake

Using the original $\frac{0}{0}$ expression as the hole’s y-value.

Mini check

Which rule should you use to find the hole’s output?

Key Concepts

Key idea

A hole is removable because the surrounding curve approaches a finite value; a vertical asymptote is not removable.

Fast decision rule

Fully reduce first. At an originally excluded input, a nonzero reduced denominator gives a hole; a remaining denominator zero gives a vertical asymptote.

Worked example

A rational graph can contain both a hole and an asymptote, because different denominator factors can have different factor status.

Common mistake

Treating every undefined input as an asymptote.

Mini check

What makes a discontinuity removable?

Relevant Representations

Key idea

Factored form shows the cancelled factor; the reduced rule gives the missing point; the graph shows the open circle.

Fast decision rule

Use the algebra to locate the exact hole and the graph to confirm that the curve is otherwise smooth there.

Identify the graph feature
Algebraic evidence	Graph feature	Report
common factor cancels	open circle on a smooth curve	hole $\left(c,\,k\right)$
denominator factor remains	branch diverges	vertical asymptote $x=c$
numerator factor remains	curve meets x-axis	zero $x=c$
no restriction	continuous point	no hole

Common mistake

Reading the open circle’s x-value but not its y-value.

Mini check

Which representation supplies the missing y-coordinate most reliably?

Essential Relationships or Formulas

Key idea

The hole’s x-value comes from the cancelled factor and its y-value comes from the reduced function.

r(x)=\frac{(x-c)u(x)}{(x-c)v(x)}=\frac{u(x)}{v(x)},\qquad x\ne c

v(c)\ne0\quad\Longrightarrow\quad\text{hole}=\left(c,\frac{u(c)}{v(c)}\right)

Fast decision rule

Evaluate after simplifying, but keep the original restriction in the final statement.

Worked example

For a reduced rule $x+1$ with $x\ne 1$ , the missing point is found by evaluating the reduced rule at $x=1$ .

Common mistake

Writing the simplified rule without “ $x\ne c$ .”

Mini check

What part of the final answer records that the point is missing?

Worked Examples

Key idea

A complete hole solution reports the restriction and the point on the reduced curve.

Fast decision rule

Do not stop after cancellation; compute the point.

Worked steps

$1$ . Factor. $2$ . Circle the common factor. $3$ . Record its zero. $4$ . Simplify. $5$ . Substitute the excluded input into the reduced rule. $6$ . Write the open-circle coordinate.

Worked example

For a function that reduces to $x+3$ with $x\ne 2$ , the hole lies at $\left(2,\,5\right)$ .

Common mistake

Using the original expression at the cancelled input, which produces $\frac{0}{0}$ .

Mini check

Why is the reduced rule allowed for the y-coordinate but not for the domain?

Common Mistakes

Key idea

The main errors are losing the restriction, confusing holes with asymptotes, and reporting only an x-value.

Fast decision rule

Always write the three-part record: cancelled input, reduced rule, missing point.

Worked example

An open circle on a finite smooth curve indicates a hole; a branch shooting upward or downward indicates an asymptote.

Common mistake

Calling a hole “undefined everywhere near $x=c$ .”

Mini check

What remains defined near a removable hole?

AP Exam Tips

Key idea

Hole questions are short when you separate cancellation from evaluation.

Fast decision rule

Factor $\to$ record excluded input $\to$ reduce $\to$ evaluate reduced rule $\to$ report the open circle.

Worked steps

$1$ . Find common factors. $2$ . Preserve the original domain restriction. $3$ . Simplify. $4$ . Evaluate the simplified rule. $5$ . Name the hole and any remaining asymptotes.

Common mistake

Reporting the reduced equation as the complete function.

Mini check

Does your answer include both the equation and the missing point?

Practice

Use the Questionbank items below to locate holes and distinguish them from asymptotes.

Take the next step

Use an independent practice question to test what you can now explain.

Practise rational-function holes Review vertical asymptotes

1.10 · Rational Functions and Holes

Topic 1.10 · Rational Functions and Holes

If a factor cancels, does the missing input become allowed?

Build an answer that explains the mathematics

Start with the original domain

Simplify on the shared domain

Locate and explain the hole

Spot the exact reasoning error

Can you transfer the method?

Complete topic reference