Rational Functions

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Rational Functions · Grade 12 · Worksheet 3

1. Hana is a marine biologist studying the feeding rate of a filter-feeding organism in a tidal estuary. She models the rate at which the organism filters water using the rational function F(t) = (4t² + 8t - 12)/(t² - 16), where F(t) represents the filtration rate in liters per hour and t represents the time in hours after low tide. Hana needs to understand the long-term filtration rate as time increases and identify any times when the model becomes undefined due to tidal changes. Determine all vertical asymptotes and the horizontal asymptote of this filtration rate function, and describe the behavior of F(t) near each vertical asymptote. Answer: Vertical asymptotes at t = -4 and t = 4; horizontal asymptote at y = 4. As t approaches -4 from the left, F(t) approaches +infinity; from the right, -infinity. As t approaches 4 from the left, F(t) approaches -infinity; from the right, +infinity. Solution: Factor the numerator and denominator. Numerator: 4t² + 8t - 12 = 4(t² + 2t - 3) = 4(t + 3)(t - 1). Denominator: t² - 16 = (t - 4)(t + 4).
   Full step-by-step solution

   Step 1: Factor the numerator and denominator. Numerator: 4t² + 8t - 12 = 4(t² + 2t - 3) = 4(t + 3)(t - 1). Denominator: t² - 16 = (t - 4)(t + 4). So F(t) = 4(t + 3)(t - 1)/((t - 4)(t + 4)). Step 2: Find vertical asymptotes. Set denominator equal to zero: (t - 4)(t + 4) = 0, so t = 4 and t = -4. Check that numerator is not zero at these points. At t = 4: 4(4 + 3)(4 - 1) = 4(7)(3) = 84 ≠ 0. At t = -4: 4(-4 + 3)(-4 - 1) = 4(-1)(-5) = 20 ≠ 0. So both are vertical asymptotes. Step 3: Determine behavior near t = 4. As t approaches 4 from the left (e.g., t = 3.9), denominator is (3.9 - 4)(3.9 + 4) = (-0.1)(7.9) which is negative, numerator is positive, so F(t) approaches -infinity. As t approaches 4 from the right (e.g., t = 4.1), denominator is (0.1)(8.1) positive, numerator positive, so F(t) approaches +infinity. Step 4: Determine behavior near t = -4. As t approaches -4 from the left (e.g., t = -4.1), denominator is (-4.1 - 4)(-4.1 + 4) = (-8.1)(-0.1) = positive, numerator at t = -4.1 is 4(-1.1)(-5.1) = positive, so F(t) approaches +infinity. As t approaches -4 from the right (e.g., t = -3.9), denominator is (-3.9 - 4)(-3.9 + 4) = (-7.9)(0.1) = negative, numerator positive, so F(t) approaches -infinity. Step 5: Find horizontal asymptote. Both numerator and denominator are degree 2. The leading coefficient ratio is 4/1 = 4. So horizontal asymptote is y = 4. Final answer: Vertical asymptotes at t = -4 and t = 4; horizontal asymptote at y = 4. Near t = -4: as t → -4⁻, F(t) → +∞; as t → -4⁺, F(t) → -∞. Near t = 4: as t → 4⁻, F(t) → -∞; as t → 4⁺, F(t) → +∞.

2. An environmental engineer is modeling the rate of pollutant absorption in a wetland system using the function R(x) = (2x² - 5x + 3)/(x² - 9), where x represents the concentration of pollutants in parts per million and R(x) is the absorption rate in grams per day. The engineering team needs to understand the system's maximum absorption capacity and identify any pollutant concentrations where the absorption model becomes undefined. Determine all vertical and horizontal asymptotes of this absorption rate function. Answer: y = 2, x = 3, x = -3 Solution: Find horizontal asymptote by comparing degrees of numerator and denominator Both numerator and denominator have degree 2, so horizontal asymptote is y = (leading coefficient of numerator)/(leading coefficient of denominator) = 2/1 = 2 Find vertical asymptotes by setting denominator equal to zero…
   Full step-by-step solution

   Step 1: Find horizontal asymptote by comparing degrees of numerator and denominator Both numerator and denominator have degree 2, so horizontal asymptote is y = (leading coefficient of numerator)/(leading coefficient of denominator) = 2/1 = 2 Step 2: Find vertical asymptotes by setting denominator equal to zero x² - 9 = 0 x² = 9 x = 3 or x = -3 Step 3: Verify these are vertical asymptotes (not holes) Check if factors cancel: numerator factors as (2x - 3)(x - 1), denominator factors as (x - 3)(x + 3) No common factors, so both x = 3 and x = -3 are vertical asymptotes Final answer: Horizontal asymptote at y = 2, vertical asymptotes at x = 3 and x = -3

3. Consider the graph of the rational function f(x) = (3x^2 - 15) / (x^2 - 25). On a coordinate plane, describe the locations of the vertical asymptotes, the horizontal asymptote, and explain the behavior of the function near each vertical asymptote (including whether the function approaches positive or negative infinity from each side). Answer: Vertical asymptotes at x = -5 and x = 5; horizontal asymptote at y = 3. Near x = -5: as x approaches -5 from the left, f(x) approaches positive infinity; as x approaches -5 from the right, f(x) approaches negative infinity. Near x = 5: as x approaches 5 from the left, f(x) approaches negative infinity; as x approaches 5 from the right, f(x) approaches positive infinity. Solution: Factor the function. Numerator: 3x^2 - 15 = 3(x^2 - 5). Denominator: x^2 - 25 = (x - 5)(x + 5).
   Full step-by-step solution

   Step 1: Factor the function. Numerator: 3x^2 - 15 = 3(x^2 - 5). Denominator: x^2 - 25 = (x - 5)(x + 5). So f(x) = 3(x^2 - 5) / ((x - 5)(x + 5)). Step 2: Find vertical asymptotes. Set denominator equal to zero: (x - 5)(x + 5) = 0 gives x = 5 and x = -5. Check numerator at these points: at x = 5, numerator = 3(25 - 5) = 60 (not zero); at x = -5, numerator = 3(25 - 5) = 60 (not zero). So both are vertical asymptotes. Step 3: Find horizontal asymptote. Degrees of numerator and denominator are both 2. Leading coefficient of numerator is 3, leading coefficient of denominator is 1. So horizontal asymptote is y = 3/1 = 3. Step 4: Behavior near x = -5. For x just left of -5 (e.g., x = -5.1): denominator = (-10.1)(-0.1) = 1.01 (positive), numerator = 3(26.01 - 5) = 63.03 (positive). So f(x) is positive, approaches +infinity. For x just right of -5 (e.g., x = -4.9): denominator = (-9.9)(0.1) = -0.99 (negative), numerator positive. So f(x) is negative, approaches -infinity. Step 5: Behavior near x = 5. For x just left of 5 (e.g., x = 4.9): denominator = (-0.1)(9.9) = -0.99 (negative), numerator = 3(24.01 - 5) = 57.03 (positive). So f(x) is negative, approaches -infinity. For x just right of 5 (e.g., x = 5.1): denominator = (0.1)(10.1) = 1.01 (positive), numerator positive. So f(x) is positive, approaches +infinity. The answer: Vertical asymptotes at x = -5 and x = 5; horizontal asymptote at y = 3. Near x = -5: left side positive infinity, right side negative infinity. Near x = 5: left side negative infinity, right side positive infinity.

4. A pharmaceutical company is modeling the concentration of a new medication in a patient's bloodstream over time using the rational function C(t) = (3t² + 12t) / (t² + 4t + 3), where C(t) represents the concentration in milligrams per liter and t represents time in hours after administration. The researchers need to determine the long-term concentration level that the medication approaches and identify any times when the concentration becomes undefined due to vertical asymptotes. What is the horizontal asymptote representing the long-term concentration, and at what time values does the concentration become undefined? Answer: Horizontal asymptote: y = 3; Vertical asymptotes: t = -3 and t = -1 Solution: C(t) = (3t² + 12t) / (t² + 4t + 3) Find the horizontal asymptote (long-term concentration) For a rational function (polynomial / polynomial), - If degree of numerator = degree of denominator, horizontal asymptote = (leading coefficient of numerator) / (leading coefficient of denominator).
   Full step-by-step solution

   Let's go step-by-step. We are given: C(t) = (3t² + 12t) / (t² + 4t + 3) --- **Step 1: Find the horizontal asymptote (long-term concentration)** For a rational function (polynomial / polynomial), - If degree of numerator = degree of denominator, horizontal asymptote = (leading coefficient of numerator) / (leading coefficient of denominator). Here: Numerator: 3t² + 12t → degree 2, leading coefficient 3 Denominator: t² + 4t + 3 → degree 2, leading coefficient 1 So horizontal asymptote: y = 3/1 = 3. Thus, long-term concentration approaches 3 mg/L. --- **Step 2: Find when C(t) is undefined (vertical asymptotes)** C(t) is undefined where denominator = 0 and numerator ≠ 0 (unless factor cancels). Denominator: t² + 4t + 3 = 0 Factor: (t + 3)(t + 1) = 0 So t = -3 or t = -1. --- **Step 3: Check if these make numerator zero (possible hole instead of asymptote)** Numerator: 3t² + 12t = 3t(t + 4) At t = -3: 3(-3)(-3 + 4) = (-9)(1) = -9 ≠ 0 At t = -1: 3(-1)(-1 + 4) = (-3)(3) = -9 ≠ 0 So no cancellation; both are vertical asymptotes. --- **Step 4: Interpret time values** t is time in hours after administration. t = -3 and t = -1 are before administration (t=0), so in a real-world sense, the model may only be intended for t ≥ 0. But mathematically, vertical asymptotes are at t = -3 and t = -1. --- **Final Answer:** Horizontal asymptote: y = 3 Vertical asymptotes: t = -3 and t = -1

5. A pharmaceutical company is modeling the concentration of a new medication in a patient's bloodstream over time. The concentration function is given by C(t) = (3t² + 5t) / (t² - 4), where t represents hours after administration. The researchers need to determine the long-term concentration level and identify any times when the concentration becomes undefined. What is the horizontal asymptote of this function, and at what time values does the vertical asymptote occur? Answer: Horizontal asymptote: y = 3; Vertical asymptotes: t = 2 and t = -2 Solution: C(t) = (3t² + 5t) / (t² - 4) Vertical asymptotes occur where the denominator is zero (and numerator is not zero at the same time, otherwise it might be a hole).
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Identify the function** We are given: C(t) = (3t² + 5t) / (t² - 4) --- **Step 2: Find vertical asymptotes** Vertical asymptotes occur where the denominator is zero (and numerator is not zero at the same time, otherwise it might be a hole). Denominator: t² - 4 = 0 t² = 4 t = 2 or t = -2 Check numerator at t = 2: 3(2)² + 5(2) = 12 + 10 = 22 ≠ 0 Check numerator at t = -2: 3(4) + 5(-2) = 12 - 10 = 2 ≠ 0 So both t = 2 and t = -2 make the denominator zero but not the numerator, so they are vertical asymptotes. --- **Step 3: Find horizontal asymptote** For a rational function (polynomial / polynomial): - If degree of numerator = degree of denominator, horizontal asymptote is y = (leading coefficient of numerator) / (leading coefficient of denominator). Here: numerator degree = 2, denominator degree = 2. Leading coefficient of numerator = 3 Leading coefficient of denominator = 1 So horizontal asymptote: y = 3/1 = 3. --- **Step 4: Interpret in context** t = time in hours, so t = -2 is not physically meaningful in the real-world situation (time before administration), but mathematically it's still a vertical asymptote of the function. The horizontal asymptote y = 3 means that in the long term, the concentration approaches 3 (in whatever units). --- **Final Answer:** Horizontal asymptote: y = 3 Vertical asymptotes: t = 2 and t = -2