Asymptotes

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Answer Key & Explanations

Asymptotes · Grade 12 · Worksheet 2

1. lim(x→∞) (4x³ - 2x² + 5x - 1)/(2x³ + 3x² - x + 7) = ? Answer: 2 Solution: Identify the degrees of numerator and denominator. Both are degree 3. Identify the leading coefficients.
   Full step-by-step solution

   Step 1: Identify the degrees of numerator and denominator. Both are degree 3. Step 2: Identify the leading coefficients. Numerator leading coefficient is 4, denominator leading coefficient is 2. Step 3: When degrees are equal, the limit equals the ratio of leading coefficients: 4/2 = 2. Step 4: Therefore, lim(x→∞) (4x³ - 2x² + 5x - 1)/(2x³ + 3x² - x + 7) = 2.

2. lim(x→∞) (3x² + 2x - 5)/(x² - 4x + 1) = ? Answer: 3 Solution: limit as x → ∞ of (3x² + 2x - 5) / (x² - 4x + 1) Identify the highest power of x in the denominator. Here, the highest power is x² in both numerator and denominator.
   Full step-by-step solution

   Let's find the limit step by step. We are given: limit as x → ∞ of (3x² + 2x - 5) / (x² - 4x + 1) Step 1: Identify the highest power of x in the denominator. Here, the highest power is x² in both numerator and denominator. Step 2: Divide every term in the numerator and denominator by x². Numerator: 3x²/x² + 2x/x² - 5/x² = 3 + 2/x - 5/x² Denominator: x²/x² - 4x/x² + 1/x² = 1 - 4/x + 1/x² So the expression becomes: (3 + 2/x - 5/x²) / (1 - 4/x + 1/x²) Step 3: Take the limit as x → ∞. As x → ∞, terms with 1/x or 1/x² go to 0. So: 2/x → 0 5/x² → 0 4/x → 0 1/x² → 0 Step 4: Substitute these limits into the expression: (3 + 0 - 0) / (1 - 0 + 0) = 3/1 = 3 Therefore, the limit is 3. Final answer: 3

3. A biomedical engineer 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 - 2)/(t² - 4), where t represents hours after administration. To determine the long-term behavior of the medication concentration, the engineer needs to find all vertical and horizontal asymptotes of this function. What are the equations of these asymptotes? Answer: Vertical asymptotes: x = 2 and x = -2; Horizontal asymptote: y = 3 Solution: Horizontal asymptotes represent the long-term behavior or equilibrium state of a system, while vertical asymptotes indicate values where the model breaks down or becomes undefined.
   Full step-by-step solution

   Rational functions often model real-world scenarios where quantities approach limiting values. Horizontal asymptotes represent the long-term behavior or equilibrium state of a system, while vertical asymptotes indicate values where the model breaks down or becomes undefined. To find horizontal asymptotes, compare the degrees of the numerator and denominator polynomials. For vertical asymptotes, identify values that make the denominator zero but not the numerator, unless there are common factors that cancel.

4. Noah is an aerospace engineer analyzing the flight path of a new drone. The drone's altitude in meters over time t (in seconds) is modeled by the rational function A(t) = (7t³ + 9t - 11) / (t² - 6t + 9). To ensure the drone avoids sudden drops or climbs at critical moments and to predict its long-term flight behavior, Noah must determine all vertical, horizontal, and slant asymptotes of this function. What are the equations of all asymptotes? Answer: Vertical asymptote: t = 3; Slant asymptote: A = 7t + 21; No horizontal asymptote Solution: Find vertical asymptotes by setting the denominator equal to zero. Denominator: t² - 6t + 9 = 0 Factor: (t - 3)(t - 3) = 0 Solve: t = 3 (multiplicity 2) Check numerator at t = 3: 7(27) + 9(3) - 11 = 189 + 27 - 11 = 205 ≠ 0 So t = 3 is a vertical asymptote.
   Full step-by-step solution

   Step 1: Find vertical asymptotes by setting the denominator equal to zero. Denominator: t² - 6t + 9 = 0 Factor: (t - 3)(t - 3) = 0 Solve: t = 3 (multiplicity 2) Check numerator at t = 3: 7(27) + 9(3) - 11 = 189 + 27 - 11 = 205 ≠ 0 So t = 3 is a vertical asymptote. Step 2: Determine horizontal or slant asymptote by comparing degrees. Numerator degree: 3 (highest power t³) Denominator degree: 2 (highest power t²) Since numerator degree > denominator degree, there is no horizontal asymptote. Since numerator degree = denominator degree + 1, there is a slant asymptote. Step 3: Find slant asymptote by polynomial long division. Divide 7t³ + 0t² + 9t - 11 by t² - 6t + 9. First term: 7t³ / t² = 7t Multiply: 7t(t² - 6t + 9) = 7t³ - 42t² + 63t Subtract: (7t³ + 0t² + 9t - 11) - (7t³ - 42t² + 63t) = 42t² - 54t - 11 Second term: 42t² / t² = 42 Multiply: 42(t² - 6t + 9) = 42t² - 252t + 378 Subtract: (42t² - 54t - 11) - (42t² - 252t + 378) = 198t - 389 Quotient: 7t + 42, Remainder: 198t - 389 The slant asymptote is the quotient: A = 7t + 42. Final answer: Vertical asymptote: t = 3; Slant asymptote: A = 7t + 42; No horizontal asymptote.

5. Mason is an aerospace engineer analyzing the flight path of a drone. The drone's altitude in meters over time (in seconds) is modeled by the function A(t) = (4t² + 9t - 13)/(t² - 11t + 28). To ensure the drone avoids no-fly zones and predict its long-term behavior, Mason must find all vertical and horizontal asymptotes of this altitude function. What are the equations of these asymptotes? Answer: Vertical asymptotes: t = 4, t = 7; Horizontal asymptote: A = 4 Solution: Find vertical asymptotes by setting the denominator equal to zero. At t = 4: numerator = 4(16) + 9(4) - 13 = 64 + 36 - 13 = 87 (not zero) At t = 7: numerator = 4(49) + 9(7) - 13 = 196 + 63 - 13 = 246 (not zero) So both t = 4 and t = 7 are vertical asymptotes.
   Full step-by-step solution

   Step 1: Find vertical asymptotes by setting the denominator equal to zero. Denominator: t² - 11t + 28 = 0 Factor: (t - 4)(t - 7) = 0 Solve: t = 4 or t = 7 Step 2: Verify that the numerator is not zero at these points. At t = 4: numerator = 4(16) + 9(4) - 13 = 64 + 36 - 13 = 87 (not zero) At t = 7: numerator = 4(49) + 9(7) - 13 = 196 + 63 - 13 = 246 (not zero) So both t = 4 and t = 7 are vertical asymptotes. Step 3: Find horizontal asymptote by comparing degrees. Numerator degree: 2 (highest power t²) Denominator degree: 2 (highest power t²) When degrees are equal, horizontal asymptote is y = (leading coefficient of numerator)/(leading coefficient of denominator). Leading coefficient of numerator: 4 Leading coefficient of denominator: 1 Horizontal asymptote: A = 4/1 = 4 Final answer: Vertical asymptotes at t = 4 and t = 7; Horizontal asymptote at A = 4.

6. A rational function is graphed on a coordinate plane. The function is f(x) = (2x² - 5x - 3) / (x² - 9). Describe all vertical and horizontal asymptotes of this function using proper mathematical notation. Answer: Vertical asymptotes: x = 3, x = -3; Horizontal asymptote: y = 2 Solution: When analyzing rational functions, vertical asymptotes are found by setting the denominator equal to zero and solving for x, excluding any values that also make the numerator zero (which would indicate removable discontinuities). If the numerator's degree is less than the denominator's, the…
   Full step-by-step solution

   When analyzing rational functions, vertical asymptotes are found by setting the denominator equal to zero and solving for x, excluding any values that also make the numerator zero (which would indicate removable discontinuities). Horizontal asymptotes are determined by comparing the degrees of the numerator and denominator polynomials. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the numerator's degree is less than the denominator's, the horizontal asymptote is y=0. If the numerator's degree is greater, there is no horizontal asymptote (though there may be an oblique asymptote).

7. Aroha is an environmental engineer studying the rate at which a wetland filters pollutants. She models the rate of filtration (in liters per hour) using the rational function R(t) = (5t³ - 3t + 7) / (t² - 1), where t represents time in days since monitoring began. To predict long-term behavior and identify times when the model breaks down, Aroha needs to find all vertical and slant (oblique) asymptotes of this function. What are the equations of all asymptotes? Answer: Vertical asymptotes: t = 1 and t = -1; Slant asymptote: R = 5t Solution: Find vertical asymptotes by setting the denominator equal to zero. Denominator: t² - 1 = 0 Factor: (t - 1)(t + 1) = 0 Solutions: t = 1 and t = -1 Check that the numerator is not zero at these points.
   Full step-by-step solution

   Step 1: Find vertical asymptotes by setting the denominator equal to zero. Denominator: t² - 1 = 0 Factor: (t - 1)(t + 1) = 0 Solutions: t = 1 and t = -1 Step 2: Check that the numerator is not zero at these points. At t = 1: numerator = 5(1)³ - 3(1) + 7 = 5 - 3 + 7 = 9 ≠ 0 At t = -1: numerator = 5(-1)³ - 3(-1) + 7 = -5 + 3 + 7 = 5 ≠ 0 So vertical asymptotes exist at t = 1 and t = -1. Step 3: Determine if there is a slant asymptote. Degree of numerator: 3 Degree of denominator: 2 Since 3 = 2 + 1, there is a slant asymptote. Step 4: Perform polynomial long division of (5t³ - 3t + 7) by (t² - 1). Divide the leading term: 5t³ ÷ t² = 5t Multiply: 5t * (t² - 1) = 5t³ - 5t Subtract: (5t³ - 3t + 7) - (5t³ - 5t) = 2t + 7 Since the remainder (2t + 7) has degree 1, which is less than degree 2 of the divisor, the division stops. Step 5: The quotient is 5t, so the slant asymptote is R = 5t. Final answer: Vertical asymptotes at t = 1 and t = -1; Slant asymptote at R = 5t.