Rational Functions

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Rational Functions · Grade 12 · Worksheet 2

1. lim(x→∞) (5x⁴ - 3x³ + 2x - 7)/(2x⁴ + 4x² - x + 9) = ? Answer: 5/2 Solution: Numerator degree: 4 (from 5x⁴) Denominator degree: 4 (from 2x⁴) Since degrees are equal, the limit is the ratio of leading coefficients Leading coefficient of numerator: 5 Leading coefficient of denominator: 2 5/2 The answer is 5/2.
   Full step-by-step solution

   Step 1: Identify the degrees of numerator and denominator Numerator degree: 4 (from 5x⁴) Denominator degree: 4 (from 2x⁴) Step 2: Since degrees are equal, the limit is the ratio of leading coefficients Leading coefficient of numerator: 5 Leading coefficient of denominator: 2 Step 3: Calculate the ratio 5/2 The answer is 5/2.

2. lim(x→∞) (3x² - 2x + 1)/(x² + 5x - 3) = ? Answer: 3 Solution: lim(x→∞) (3x² - 2x + 1)/(x² + 5x - 3) Identify the highest power of x in the denominator The denominator is x² + 5x - 3. The highest power is x². This is the standard method for limits at infinity of rational functions.
   Full step-by-step solution

   Let's find the limit step by step. We want: lim(x→∞) (3x² - 2x + 1)/(x² + 5x - 3) --- **Step 1: Identify the highest power of x in the denominator** The denominator is x² + 5x - 3. The highest power is x². --- **Step 2: Divide numerator and denominator by x²** This is the standard method for limits at infinity of rational functions. Numerator: (3x² - 2x + 1) / x² = 3 - 2/x + 1/x² Denominator: (x² + 5x - 3) / x² = 1 + 5/x - 3/x² So the expression becomes: [3 - 2/x + 1/x²] / [1 + 5/x - 3/x²] --- **Step 3: Take the limit as x → ∞** As x → ∞: - 2/x → 0 - 1/x² → 0 - 5/x → 0 - 3/x² → 0 So the expression becomes: (3 - 0 + 0) / (1 + 0 - 0) = 3/1 = 3 --- **Step 4: Conclusion** The limit is 3. --- **Final Answer:** 3

3. Isabella is analyzing the efficiency of a chemical reaction in a new industrial process. The reaction rate, in moles per second, is modeled by the rational function R(x) = (3x² - 7x + 2) / (x² - 7x + 12), where x represents the temperature in hundreds of degrees Celsius. To ensure safe operation, Isabella must identify any temperatures where the reaction rate becomes undefined (vertical asymptotes) and determine the long-term reaction rate as temperature increases (horizontal asymptote). What are the vertical asymptotes and the horizontal asymptote of this reaction rate function? Additionally, describe the behavior of R(x) near the larger vertical asymptote from both sides. Answer: Vertical asymptotes at x = 3 and x = 4; horizontal asymptote at y = 3; near x = 4, as x approaches 4 from the left, R(x) approaches positive infinity; as x approaches 4 from the right, R(x) approaches negative infinity. Solution: Factor the denominator: x^2 - 7x + 12 = (x - 3)(x - 4). Set each factor equal to zero: x - 3 = 0 gives x = 3; x - 4 = 0 gives x = 4. Check that the numerator does not equal zero at these x-values.
   Full step-by-step solution

   Step 1: Factor the denominator: x^2 - 7x + 12 = (x - 3)(x - 4). Set each factor equal to zero: x - 3 = 0 gives x = 3; x - 4 = 0 gives x = 4. Step 2: Check that the numerator does not equal zero at these x-values. Factor the numerator: 3x^2 - 7x + 2 = (3x - 1)(x - 2). At x = 3, numerator = (9 - 1)(1) = 8, not zero. At x = 4, numerator = (12 - 1)(2) = 22, not zero. So vertical asymptotes at x = 3 and x = 4. Step 3: For horizontal asymptote, the degrees of numerator and denominator are both 2. The leading coefficient of numerator is 3, of denominator is 1. So horizontal asymptote is y = 3/1 = 3. Step 4: Behavior near x = 4. For x just less than 4, say x = 3.9: denominator = (3.9 - 3)(3.9 - 4) = (0.9)(-0.1) = -0.09 (negative). Numerator = (3*3.9 - 1)(3.9 - 2) = (11.7 - 1)(1.9) = (10.7)(1.9) = 20.33 (positive). Positive divided by negative gives negative, so R(x) approaches negative infinity from the left? Let's re-evaluate: As x approaches 4 from the left (x -> 4-), (x-4) is negative small, (x-3) is positive small. Product is negative small. Numerator is positive (since 3x-1 and x-2 are both positive for x near 4). Positive over negative small gives negative large, so R(x) -> -infinity. For x just greater than 4, say x = 4.1: denominator = (4.1-3)(4.1-4) = (1.1)(0.1) = 0.11 positive. Numerator = (12.3-1)(2.1) = (11.3)(2.1) = 23.73 positive. Positive over positive small gives positive large, so R(x) -> +infinity. Therefore, as x approaches 4 from the left, R(x) approaches negative infinity; as x approaches 4 from the right, R(x) approaches positive infinity. The answer: Vertical asymptotes at x = 3 and x = 4; horizontal asymptote at y = 3; near x = 4, as x approaches 4 from the left, R(x) approaches negative infinity; as x approaches 4 from the right, R(x) approaches positive infinity.

4. lim(x→∞) (5x³ - 2x² + 3x - 1)/(2x³ + 4x² - x + 7) = ? Answer: 5/2 Solution: Identify the degrees of numerator and denominator. Both are degree 3. For rational functions where degrees are equal, the limit as x approaches infinity equals the ratio of the leading coefficients.
   Full step-by-step solution

   Step 1: Identify the degrees of numerator and denominator. Both are degree 3. Step 2: For rational functions where degrees are equal, the limit as x approaches infinity equals the ratio of the leading coefficients. Step 3: Leading coefficient of numerator is 5, leading coefficient of denominator is 2. Step 4: The limit is 5/2. The answer is 5/2.

5. f(x) = (4x^2 - 16)/(2x^2 - 8x + 8). Identify all vertical and horizontal asymptotes, and describe the behavior of f(x) near each vertical asymptote. Answer: Vertical asymptote at x = 2 (from both sides, f(x) → -∞); horizontal asymptote at y = 2. Solution: Factor numerator and denominator. Numerator: 4x^2 - 16 = 4(x^2 - 4) = 4(x - 2)(x + 2). Denominator: 2x^2 - 8x + 8 = 2(x^2 - 4x + 4) = 2(x - 2)^2.
   Full step-by-step solution

   Step 1: Factor numerator and denominator. Numerator: 4x^2 - 16 = 4(x^2 - 4) = 4(x - 2)(x + 2). Denominator: 2x^2 - 8x + 8 = 2(x^2 - 4x + 4) = 2(x - 2)^2. Step 2: Simplify the rational function. f(x) = [4(x - 2)(x + 2)] / [2(x - 2)^2] = [2(x + 2)] / (x - 2), for x ≠ 2. Step 3: Find vertical asymptote(s). Set denominator of simplified form to zero: x - 2 = 0 → x = 2. Since the factor (x - 2) remains in the denominator after cancellation (with multiplicity 1), there is a vertical asymptote at x = 2. Step 4: Analyze behavior near x = 2. As x → 2⁺ (from the right), (x - 2) → 0⁺, so f(x) = 2(x + 2)/(x - 2) → 2(4)/(0⁺) = 8/0⁺ → +∞. As x → 2⁻ (from the left), (x - 2) → 0⁻, so f(x) = 2(x + 2)/(x - 2) → 2(4)/(0⁻) = 8/0⁻ → -∞. Step 5: Find horizontal asymptote. Degrees: numerator degree = 2, denominator degree = 2 (in original form). Leading coefficients: 4 and 2. Ratio = 4/2 = 2. So horizontal asymptote at y = 2. Final answer: Vertical asymptote at x = 2 (f(x) → +∞ from right, f(x) → -∞ from left); horizontal asymptote at y = 2.

6. Tane is a medical researcher studying how the concentration of a new antibiotic decays in the bloodstream. The concentration, C(t), in milligrams per liter, t hours after injection, is modeled by the rational function C(t) = (4t² - 20) / (t² - 25). The hospital needs to know the safe concentration level the drug approaches over a long period, and at what exact times the model becomes invalid (showing infinite concentration). Determine the horizontal asymptote and all vertical asymptotes of C(t). Answer: Horizontal asymptote: y = 4; Vertical asymptotes: t = 5 and t = -5 Solution: Identify the degrees of the numerator and denominator. The numerator is 4t² - 20 (degree 2). The denominator is t² - 25 (degree 2).
   Full step-by-step solution

   Step 1: Identify the degrees of the numerator and denominator. The numerator is 4t² - 20 (degree 2). The denominator is t² - 25 (degree 2). Since the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients: 4/1 = 4. So the horizontal asymptote is y = 4. This means over many hours, the drug concentration stabilizes at 4 mg/L. Step 2: Find vertical asymptotes by setting the denominator equal to zero: t² - 25 = 0. Solve: t² = 25, so t = 5 or t = -5. Step 3: Check if the numerator is also zero at these points. At t = 5: 4(5)² - 20 = 4(25) - 20 = 100 - 20 = 80 (not zero). At t = -5: 4(-5)² - 20 = 4(25) - 20 = 100 - 20 = 80 (not zero). Since the numerator is not zero at either value, both are vertical asymptotes. Step 4: Conclusion. The function has a horizontal asymptote at y = 4, and vertical asymptotes at t = 5 and t = -5. The vertical asymptote at t = -5 is mathematically valid but not physically relevant since time cannot be negative. The vertical asymptote at t = 5 hours means the model is invalid at that exact moment, suggesting a mathematical breakdown (infinite concentration) in the simple model. Answer: Horizontal asymptote: y = 4; Vertical asymptotes: t = 5 and t = -5.

7. lim(x→∞) (4x³ - 5x² + 2x - 1)/(2x³ + 3x² - x + 7) = ? Answer: 2 Solution: Identify the degrees of numerator and denominator. Both are degree 3. For rational functions where degrees are equal, the limit at infinity equals the ratio of leading coefficients.
   Full step-by-step solution

   Step 1: Identify the degrees of numerator and denominator. Both are degree 3. Step 2: For rational functions where degrees are equal, the limit at infinity equals the ratio of leading coefficients. Step 3: Leading coefficient of numerator is 4, leading coefficient of denominator is 2. Step 4: Calculate the ratio: 4/2 = 2. Step 5: Therefore, the limit is 2.