Free Printable WorksheetMathAlgebra IIMath CurriculumALG2.139

ALG2.139 Interpreting Rational Inequality Intervals Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Algebra II Math worksheet provides structured practice on ALG2.139 Interpreting Rational Inequality Intervals. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

ALG2.139 Interpreting Rational Inequality Intervals - Practice questions for Interpreting Rational Inequality Intervals.

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Grade

Algebra II

Subject

Math

Domain

Math Curriculum

Formats

PDF & DOCX

Sample Practice Problems

Preview the types of problems and questions covered in this worksheet

Below are sample questions representative of the exercises included on this worksheet and related ALG2.139 Interpreting Rational Inequality Intervals skill assessments:

Problem 1Difficulty: Mastery

Mastery: use multiplicities (sign-change vs. no sign-change) to interpret the intervals: Critical numbers (in increasing order): x = -3 (undefined), x = 1 (zero), x = 5 (zero), x = 6 (undefined). For x > 6, R(x) is positive. At x = -3 (undefined value), the sign changes (odd multiplicity). At x = 1 (zero), the sign does not change (even multiplicity). At x = 5 (zero), the sign does not change (even multiplicity). At x = 6 (undefined value), the sign does not change (even multiplicity). Solve R(x) > 0. Which option correctly gives the solution set and correctly justifies inclusion/exclusion of every endpoint?

ASolution set: [-3, 1) ∪ (1, 5) ∪ (5, 6] ∪ [6, ∞) Reason: Include x=-3 because the inequality includes equality. (This ignores that R is undefined there.)

BSolution set: (-3, 1] ∪ [1, 5] ∪ [5, 6) ∪ (6, ∞) Reason: Include x=1 because it is a zero. (This ignores that the inequality is strict, so 0 is not allowed.)

CSolution set: (-∞, -3) Reason: Pick the intervals with the wrong sign (mixing up where R(x) is positive vs. negative).

DSolution set: (-3, 1) ∪ (1, 5) ∪ (5, 6) ∪ (6, ∞) Reason: Keep the intervals where R(x) is positive. Do not include zeros because the inequality is strict. Exclude all values where R(x) is undefined. Any undefined value like -3, 6 must stay excluded.

Solution & Explanation:Start with the sign information for R(x) on each interval between the critical numbers. Keep the intervals where R(x) is positive, because the inequality is R(x) > 0. Because the inequality is strict, any x-value where R(x)=0 is excluded. Exclude any x-value where R(x) is undefined (denominator zero), even if it would make the inequality true. Write the remaining x-values as a union of intervals: (-3, 1) ∪ (1, 5) ∪ (5, 6) ∪ (6, ∞).

Problem 2Difficulty: Mastery

Very carefully interpret the solution set from parity (sign-change) and excluded values: Critical numbers (in increasing order): x = -12 (undefined), x = -8 (zero), x = -1 (zero), x = 1 (undefined), x = 6 (zero). For x > 6, R(x) is negative. At x = -12 (undefined value), the sign does not change (even multiplicity). At x = -8 (zero), the sign does not change (even multiplicity). At x = -1 (zero), the sign changes (odd multiplicity). At x = 1 (undefined value), the sign changes (odd multiplicity). At x = 6 (zero), the sign changes (odd multiplicity). Solve R(x) > 0. Which option correctly gives the solution set and correctly justifies inclusion/exclusion of every endpoint?

ASolution set: (-∞, -12] ∪ [-12, -8) ∪ (-8, -1) ∪ [1, 6) Reason: Include x=-12 because the inequality includes equality. (This ignores that R is undefined there.)

BSolution set: (-∞, -12) ∪ (-12, -8] ∪ [-8, -1] ∪ (1, 6] Reason: Include x=-8 because it is a zero. (This ignores that the inequality is strict, so 0 is not allowed.)

CSolution set: (-∞, -12) ∪ (-12, -8) ∪ (-8, -1) ∪ (1, 6) Reason: Keep the intervals where R(x) is positive. Do not include zeros because the inequality is strict. Exclude all values where R(x) is undefined. Any undefined value like -12, 1 must stay excluded.

DSolution set: (-1, 1) ∪ (6, ∞) Reason: Pick the intervals with the wrong sign (mixing up where R(x) is positive vs. negative).

Problem 3Difficulty: Mastery

A rational expression R(x) has these critical numbers and multiplicity information: Critical numbers (in increasing order): x = -12 (undefined), x = 0 (undefined), x = 6 (undefined), x = 11 (zero). On (11, ∞), R(x) is negative. At x = -12 (undefined value), the sign does not change (even multiplicity). At x = 0 (undefined value), the sign does not change (even multiplicity). At x = 6 (undefined value), the sign changes (odd multiplicity). At x = 11 (zero), the sign changes (odd multiplicity). Solve R(x) > 0. Which option correctly gives the solution set and correctly justifies inclusion/exclusion of every endpoint?

ASolution set: [6, 11) Reason: Include x=-12 because the inequality includes equality. (This ignores that R is undefined there.)

BSolution set: (6, 11) Reason: Keep the intervals where R(x) is positive. Do not include zeros because the inequality is strict. Exclude all values where R(x) is undefined. Any undefined value like -12, 0 must stay excluded.

CSolution set: (6, 11] Reason: Include x=11 because it is a zero. (This ignores that the inequality is strict, so 0 is not allowed.)

DSolution set: (-∞, -12) ∪ (-12, 0) ∪ (0, 6) ∪ (11, ∞) Reason: Pick the intervals with the wrong sign (mixing up where R(x) is positive vs. negative).

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QWhat grade level and standards does this worksheet align with?

This worksheet is designed specifically for Algebra II Math students. It aligns with educational standard ALG2.139 (Practice questions for Interpreting Rational Inequality Intervals.).