Polynomial and Rational Functions review games for Pre-Calculus.
This unit covers end behavior, zeros and multiplicity, asymptotes and partial fractions — essential concepts for Pre-Calculus. Use our interactive study games to test your understanding, or review questions in traditional format below.
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This unit covers end behavior, zeros and multiplicity, asymptotes and partial fractions — essential concepts for Pre-Calculus. Use our interactive study games to test your understanding, or review questions in traditional format below.
Key Concepts Breakdown
1 End Behavior
End behavior describes what happens to a polynomial function as x approaches positive or negative infinity. It is determined entirely by the leading term (highest degree and its coefficient). Students must be able to describe end behavior using arrow notation and match it to a graph.
Key Points
- Even degree, positive leading coefficient: both ends go UP (↑ ↑)
- Even degree, negative leading coefficient: both ends go DOWN (↓ ↓)
- Odd degree, positive leading coefficient: falls left, rises right (↓ ↑)
- Odd degree, negative leading coefficient: rises left, falls right (↑ ↓)
Describe the end behavior of f(x) = -2x³ + 5x - 1.
The leading term is -2x³, which has odd degree (3) and a negative leading coefficient (-2). For odd degree with negative leading coefficient, the left end rises and the right end falls. In arrow notation: as x → -∞, f(x) → +∞ and as x → +∞, f(x) → -∞.
2 Zeros And Multiplicity
A zero of a polynomial is an x-value where f(x) = 0, found by setting each factor equal to zero. Multiplicity refers to how many times a factor is repeated, and it determines whether the graph crosses or touches (bounces off) the x-axis at that zero. Students must connect zeros, multiplicity, and graph behavior.
Key Points
- Odd multiplicity (1, 3, 5…): graph CROSSES the x-axis at that zero
- Even multiplicity (2, 4, 6…): graph TOUCHES and BOUNCES at that zero
- Higher multiplicity = flatter curve near the zero
- The sum of all multiplicities equals the degree of the polynomial
Given f(x) = (x + 3)²(x - 1)(x - 4)³, identify all zeros, their multiplicities, and whether the graph crosses or bounces at each.
Setting each factor to zero gives zeros at x = -3 (multiplicity 2, even → bounces), x = 1 (multiplicity 1, odd → crosses), and x = 4 (multiplicity 3, odd → crosses). The total degree is 2 + 1 + 3 = 6, which is even, so both ends point in the same direction.
3 Asymptotes
Asymptotes are lines that a rational function approaches but never reaches (or crosses only in special cases). Students must find vertical asymptotes (VA), horizontal asymptotes (HA), and oblique (slant) asymptotes using degree comparison and polynomial division. These appear frequently on exams in both equation and graphing form.
Key Points
- Vertical asymptotes: set the denominator equal to zero (after fully canceling common factors); cancelled factors produce holes, not VAs
- HA rule: if degree of numerator < denominator → y = 0; if degrees equal → y = ratio of leading coefficients; if numerator degree > denominator → no HA
- Oblique asymptote: exists when numerator degree is exactly one more than denominator degree; find it by polynomial long division
- A function can cross a horizontal or oblique asymptote but NEVER a vertical asymptote
Find all asymptotes of f(x) = (3x² + 1) / (x² - 4).
For vertical asymptotes, set x² - 4 = 0, giving x = 2 and x = -2 (no common factors to cancel, so both are VAs). For the horizontal asymptote, the numerator and denominator have equal degrees (both 2), so HA is y = 3/1 = 3. There is no oblique asymptote because the degrees are equal, not differing by one.
4 Partial Fractions
Partial fraction decomposition rewrites a rational expression as a sum of simpler fractions. Students must factor the denominator, set up the correct form of the decomposition based on factor type, and solve for unknown constants. This topic is tested algebraically; students will not need to integrate.
Key Points
- Linear factor (ax + b): contributes a term of the form A / (ax + b)
- Repeated linear factor (ax + b)²: contributes A / (ax + b) + B / (ax + b)²
- Irreducible quadratic factor (ax² + bx + c): contributes (Ax + B) / (ax² + bx + c)
- Solve for constants by multiplying both sides by the common denominator, then substituting strategic x-values or matching coefficients
Decompose (5x + 3) / [(x - 1)(x + 2)] into partial fractions.
Set up: (5x + 3) / [(x - 1)(x + 2)] = A / (x - 1) + B / (x + 2). Multiply both sides by (x - 1)(x + 2) to get 5x + 3 = A(x + 2) + B(x - 1). Substituting x = 1 gives 8 = 3A so A = 8/3; substituting x = -2 gives -7 = -3B so B = 7/3. The decomposition is (8/3)/(x - 1) + (7/3)/(x + 2).
Questions, answered.
What is Polynomial and Rational Functions?
Polynomial and Rational Functions is Unit 2 of Pre-Calculus, covering end behavior, zeros and multiplicity, asymptotes and partial fractions.
How to study for Pre-Calculus Unit 2?
Start with the Quick Summary above, review the Key Concepts, then test yourself with our interactive study games. Aim for 80%+ accuracy before moving on.
How many questions are in this unit?
This unit has 28+ review questions across 5 different game modes.