Asymptotes

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Asymptotes · Grade 12 · Worksheet 1

1. Tane is a fluid dynamics engineer analyzing the flow rate of water through a filtration system. The flow rate over time is modeled by the function R(t) = (3t³ + 5t² - 7t + 9)/(t² - 5t + 6), where R(t) represents the flow rate in liters per minute and t represents time in hours. To understand the system's behavior at critical times and its long-term trend, Tane needs to determine all vertical and slant asymptotes of this function. What are the equations of these asymptotes? Answer: Vertical asymptotes: t = 2 and t = 3; Slant asymptote: R = 3t + 20 Solution: Factor the denominator to find vertical asymptotes. Denominator: t² - 5t + 6 = 0 Factor: (t - 2)(t - 3) = 0 So t = 2 and t = 3 are potential vertical asymptotes. Check that the numerator is not zero at these points.
   Full step-by-step solution

   Step 1: Factor the denominator to find vertical asymptotes. Denominator: t² - 5t + 6 = 0 Factor: (t - 2)(t - 3) = 0 So t = 2 and t = 3 are potential vertical asymptotes. Step 2: Check that the numerator is not zero at these points. Numerator at t = 2: 3(8) + 5(4) - 7(2) + 9 = 24 + 20 - 14 + 9 = 39 ≠ 0 Numerator at t = 3: 3(27) + 5(9) - 7(3) + 9 = 81 + 45 - 21 + 9 = 114 ≠ 0 Both are non-zero, so vertical asymptotes exist at t = 2 and t = 3. Step 3: Determine the type of horizontal/slant asymptote by comparing degrees. Numerator degree: 3 Denominator degree: 2 Since numerator degree (3) is exactly one more than denominator degree (2), there is a slant asymptote (no horizontal asymptote). Step 4: Perform polynomial long division to find the slant asymptote. Divide 3t³ + 5t² - 7t + 9 by t² - 5t + 6. First term: 3t³ ÷ t² = 3t Multiply: 3t(t² - 5t + 6) = 3t³ - 15t² + 18t Subtract: (3t³ + 5t² - 7t + 9) - (3t³ - 15t² + 18t) = 20t² - 25t + 9 Next term: 20t² ÷ t² = 20 Multiply: 20(t² - 5t + 6) = 20t² - 100t + 120 Subtract: (20t² - 25t + 9) - (20t² - 100t + 120) = 75t - 111 Step 5: The quotient is 3t + 20, which is the equation of the slant asymptote. The remainder is 75t - 111, which becomes negligible as t approaches infinity. Final answer: Vertical asymptotes: t = 2 and t = 3; Slant asymptote: R = 3t + 20

2. A rational function is graphed on a coordinate plane with the equation f(x) = (3x² - 5x - 2)/(x² - 4). Describe all vertical and horizontal asymptotes of this function, providing their equations. Answer: Vertical asymptotes: x = 2, x = -2; Horizontal asymptote: y = 3 Solution: Rational functions have vertical asymptotes where the denominator is zero and the numerator is not zero at those same points.
   Full step-by-step solution

   Rational functions have vertical asymptotes where the denominator is zero and the numerator is not zero at those same points. Horizontal asymptotes depend on the relationship between the degrees of the numerator and denominator polynomials. When degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.

3. Find the vertical and horizontal asymptotes of f(x) = (4x² - 7x + 2)/(x² - 9) = ? Answer: x = 3, x = -3, y = 4 Solution: Find vertical asymptotes by setting denominator equal to zero x² - 9 = 0 x² = 9 x = 3 or x = -3 Numerator degree = 2, denominator degree = 2 When degrees are equal, horizontal asymptote is ratio of leading coefficients Leading coefficient of numerator = 4 Leading coefficient of denominator = 1…
   Full step-by-step solution

   Step 1: Find vertical asymptotes by setting denominator equal to zero x² - 9 = 0 x² = 9 x = 3 or x = -3 Step 2: Find horizontal asymptote by comparing degrees Numerator degree = 2, denominator degree = 2 When degrees are equal, horizontal asymptote is ratio of leading coefficients Leading coefficient of numerator = 4 Leading coefficient of denominator = 1 Horizontal asymptote: y = 4/1 = 4 Step 3: Final answer The vertical asymptotes are x = 3 and x = -3 The horizontal asymptote is y = 4

4. 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. Answer: Vertical asymptotes: x = 3 and x = -3; Horizontal asymptote: y = 2 Solution: Rational functions have vertical asymptotes where the denominator is zero and the numerator is not zero at those same points.
   Full step-by-step solution

   Rational functions have vertical asymptotes where the denominator is zero and the numerator is not zero at those same points. Horizontal asymptotes depend on the relationship between the degrees of the numerator and denominator polynomials. When degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.

5. f(x) = (9x² + 12x - 5) / (x² - 16). Find all asymptotes. Answer: Vertical asymptotes at x = 4 and x = -4; horizontal asymptote at y = 9 Solution: Find vertical asymptotes by setting denominator to zero: x² - 16 = 0 → x² = 16 → x = 4 or x = -4.
   Full step-by-step solution

   Step 1: Find vertical asymptotes by setting denominator to zero: x² - 16 = 0 → x² = 16 → x = 4 or x = -4. Step 2: Check if numerator is also zero at these points: at x = 4, 9(16) + 12(4) - 5 = 144 + 48 - 5 = 187 ≠ 0; at x = -4, 9(16) + 12(-4) - 5 = 144 - 48 - 5 = 91 ≠ 0. So both are vertical asymptotes. Step 3: Find horizontal asymptote: numerator degree = 2, denominator degree = 2. Since degrees are equal, horizontal asymptote is ratio of leading coefficients: 9/1 = 9. The answer is vertical asymptotes at x = 4 and x = -4; horizontal asymptote at y = 9.

6. f(x) = (4x^2 + 8x - 12) / (2x^2 - 8). Find all asymptotes. Answer: Vertical asymptotes at x = 2 and x = -2; horizontal asymptote at y = 2 Solution: Factor the numerator: 4x^2 + 8x - 12 = 4(x^2 + 2x - 3) = 4(x + 3)(x - 1). Factor the denominator: 2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2). There are no common factors to cancel, so the function is already simplified.
   Full step-by-step solution

   Step 1: Factor the numerator: 4x^2 + 8x - 12 = 4(x^2 + 2x - 3) = 4(x + 3)(x - 1). Step 2: Factor the denominator: 2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2). Step 3: There are no common factors to cancel, so the function is already simplified. Step 4: Find vertical asymptotes by setting the denominator equal to zero: 2(x - 2)(x + 2) = 0. This gives x = 2 and x = -2. Step 5: Find horizontal asymptote by comparing degrees. The numerator is degree 2, denominator is degree 2. The leading coefficients are 4 and 2. The horizontal asymptote is y = 4/2 = 2. Step 6: Since the degrees are equal, there is no slant asymptote. The answer is: Vertical asymptotes at x = 2 and x = -2; horizontal asymptote at y = 2.

7. 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

8. A biomedical engineer is modeling the concentration of a new drug 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. Determine all vertical and horizontal asymptotes of this function to understand the drug's long-term behavior and critical time points. Answer: Vertical asymptotes: x = 2, x = -2; Horizontal asymptote: y = 3 Solution: Asymptotes describe the behavior of rational functions as inputs approach certain critical values or infinity.
   Full step-by-step solution

   Asymptotes describe the behavior of rational functions as inputs approach certain critical values or infinity. Vertical asymptotes occur where the function approaches positive or negative infinity, typically where the denominator equals zero but the numerator doesn't. Horizontal asymptotes describe the function's end behavior as the input grows very large in either direction, determined by comparing the highest-degree terms in the numerator and denominator.