Cogito
Algebra 2 · Chapter 4 · Lesson 1
Rational Functions and Extraneous Solutions
Where the graph breaks, and where answers go wrong.
10 problems · about 21 minutes · A-APR.D.6, A-REI.A.2, F-IF.C.7
What this lesson teaches
The student identifies asymptotes and solves rational equations, checking for extraneous solutions.
- A rational function breaks where its denominator is zero.
- Clear the denominators to solve.
- Always check: some answers are extraneous.
Warm-Up Review
From earlier lessons. Loosen up before the new work.
2 problemsReview — Solving Polynomial Equations: x³ − 4x = 0. How many real roots?
Answer 3
Why 3.
Review — Dividing Polynomials: f(x) = x² + 2, divided by (x − 3). Remainder?
Answer 11
Why 11.
Warm Up
Straightforward practice. Get the method working first.
4 problemsy = 1/(x − 7). At what x is it undefined?
Answer 7
Why x = 7.
Why can a rational equation produce an answer that does not work?
Answer Multiplying by an expression that could be zero adds solutions.
Why Clearing denominators can add solutions.
y = 1/(x + 3). At what x is it undefined?
Answer -3
Why Denominator zero.
Solve 10/x = 5.
Answer 2
Why 10 = 5x.
Build It Up
The same ideas with more to keep track of.
2 problemsSolve 12/x = 4.
Answer 3
Why 12 = 4x.
Must you check answers to a rational equation?
Answer Yes. Some are extraneous.
Why Clearing denominators can invent answers.
Stretch Yourself
Mixed problems. Work out what kind of question it is before you start.
2 problemsThe Asymptote: y = 1/(x − 5). At what x is it undefined?
Answer 5
Why x = 5.
The Solve: Solve 6/x = 3.
Answer 2
Why x = 2.