BetterGrades Precalculus · Unit 3 · Lesson

Products and quotients of functions

Multiply and divide functions pointwise while analyzing zeros, restrictions, and combined graph behavior.

Opening

Start with the situation

Product functions multiply outputs; quotient functions divide outputs and exclude every denominator-function zero.

Transformation language lets you read a complicated graph as a modified parent rather than a collection of disconnected points. That makes prediction possible before any calculator window is opened.

Before you begin

Prerequisite check

Recognize the parent function family.
Read domain, range, and key points.
Use coordinate mappings.

Core explanation

Explanation

Intersect domains, factor, locate product zeros, add quotient restrictions, and use signs to analyze the graph.

A cancelled denominator factor still excludes its original input.

A symbolic answer is not complete by itself. In this lesson, the same claim must also be readable through pointwise multiplication, zero-set union, or another equivalent representation.

Conceptual reading

What the idea is really doing

Transformations become reliable when they are treated as coordinate mappings. Outside operations change outputs; inside operations change the inputs that produce those outputs, which is why horizontal changes often appear to work in the opposite direction.

This lesson narrows that lens to one goal: multiply and divide functions pointwise while analyzing zeros, restrictions, and combined graph behavior. The point is not to memorize an isolated trick; it is to know what evidence makes the conclusion valid and how a second representation can check it.

Reusable method

A reliable route through the problem

Intersect domains.
Factor.
Locate product zeros.
Add quotient restrictions.

Verification: Track at least one landmark point from the parent graph to the transformed graph, then verify the new domain, range, intercepts, or asymptotes from the formula.

Foundation walkthrough

Plan before calculating

Problem

$f=x^2-1,g=x-1$

Plan: Start by identifying the mathematical structure in the prompt. Then use the lesson method rather than guessing from appearance: Intersect domains, factor, locate product zeros, add quotient restrictions, and use signs to analyze the graph.
Conclusion: $\frac{f}{g}=x+1$ for $x\ne 1,$ with a hole at $(1,2)$ .
Why the check works: Cancellation preserves the excluded input.

Worked examples

See the idea in three forms

foundation example

$f=x^2-1,g=x-1$

Solution $\frac{f}{g}=x+1$ for $x\ne 1,$ with a hole at $(1,2)$ .

Cancellation preserves the excluded input.

representation example

Domain $\frac{x+1}{x}^2$ .

Solution $x\ne 0$

This example expresses products and quotients of functions in a second form.

transfer example

If $f(3)=2,g(3)=-4,$ product.

Solution $-8$

A cancelled denominator factor still excludes its original input.

Pointwise multiplication. Follow the foundation example from its given information to the conclusion. The labels identify the mathematical feature that makes the result valid: Cancellation preserves the excluded input. — **Anchor figure · Pointwise multiplication**
Follow the foundation example from its given information to the conclusion. The labels identify the mathematical feature that makes the result valid: Cancellation preserves the excluded input.

Zero-set union. Read the numbered reasoning path in order. Each stage preserves the quantities, restrictions, or structural conditions needed for products and quotients of functions. — **Mechanism figure · Zero-set union**
Read the numbered reasoning path in order. Each stage preserves the quantities, restrictions, or structural conditions needed for products and quotients of functions.

Hole-asymptote preview. Compare the valid path with the tempting shortcut. The figure shows why forgetting the extra quotient condition g(x)≠0 leads to a false conclusion. — **Comparison and error figure · Hole-asymptote preview**
Compare the valid path with the tempting shortcut. The figure shows why forgetting the extra quotient condition $g(x)\ne 0$ leads to a false conclusion.

Common mistake

Find the first invalid move

A frequent error is forgetting the extra quotient condition $g(x)\ne 0$ .

Check yourself

Zeros of $x(x-5)$ .

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Practice

Ten concrete questions

Practice 101

Zeros of $x(x-5)$ .

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Practice 202

Domain $\frac{x+1}{x}^2$ .

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Practice 303

If $f(3)=2,g(3)=-4,$ product.

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Practice 404

Quotient.

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Practice 505

Explain why this conclusion is valid: $\frac{f}{g}=x+1$ for $x\ne 1,$ with a hole at $(1,2)$ . Use the foundation problem as evidence: $f=x^2-1,g=x-1$ .

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Practice 606

Solve the representation example, then name the feature of products and quotients of functions that it illustrates: Domain $\frac{x+1}{x}^2$

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Practice 707

Correct this reasoning and identify the first unsafe assumption: A frequent error is forgetting the extra quotient condition $g(x)\ne 0$ .

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Practice 808

Connect two representations for this example: $f=x^2-1,g=x-1$ . Describe what a graph, table, mapping, or algebraic form would have to show.

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Practice 909

Create a nearby example by changing one number or condition in this prompt: If $f(3)=2,g(3)=-4,$ product. Predict the effect, solve your new example, and compare it with the original.

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Practice 1010

Write a short verification checklist for products and quotients of functions, then apply it to one worked example from this lesson.

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Lesson close

Connect forward

The next lesson, Reconstructing formulas from transformed graphs, uses this result as part of a larger structure.

Source record

Original BetterGrades manuscript, rights-separated references.

Lippman and Rasmussen, Precalculus Volume 1
Utah College Algebra
Stitz and Zeager, Precalculus

No long source passage is reproduced.

f=x2−1,g=x−1f=x^2-1,g=x-1f=x2−1,g=x−1

f=x2−1,g=x−1f=x^2-1,g=x-1f=x2−1,g=x−1

Domain x+1x2\frac{x+1}{x}^2xx+1​2.

If f(3)=2,g(3)=−4,f(3)=2,g(3)=-4,f(3)=2,g(3)=−4, product.

Original BetterGrades manuscript, rights-separated references.

$f=x^2-1,g=x-1$

$f=x^2-1,g=x-1$

Domain $\frac{x+1}{x}^2$ .

If $f(3)=2,g(3)=-4,$ product.