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Centennial College MHF4U – Unit 1 Unit 1 Assessment, Part 2

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MHF4U – Unit 1 Unit 1 Assessment, Part 2 Copyright © 2019 The Ontario Educational Communications Authority. All rights reserved. 1 Unit 1 Assessment, Part 2 1. Use the characteristics of the given function to sketch it, and its reciprocal, by hand on the same set of axes. Explain your thinking process. [K6] [C2] f (x) = - x 2 + 7x – 6 TVO ILC MHF4U – Unit 1 Unit 1 Assessment, Part 2 Copyright © 2019 The Ontario Educational Communications Authority. All rights reserved. 2 - firstly, draw the parent function and apply the transformations to sketch the given function, draw the same function but the reciprocal (since function has negative sign, remove negative sign and sketch the same way by applying transformations), lastly erase parent function. 2. a. Fill in the table and sketch the graph of the following equation. [A10], [C2] f (x) = - 2?-5 3?+18 Vertical asymptotes Horizontal asymptotes x - intercept y - intercept Domain =3x+18=0 3x=18 3 x=6 y=- 2 3 -2x-5=0 -2x=5 x-intercept =- 5 2 f (0)= - 2(0)-5 3(0)+18 y-intercept= 5 18 D{? ≠ - 5 2} 2 b. State positive and negative intervals for f (x) = - 2?-5 3?+18 [K2] Positive interval: no positive intervals Negative intervals: 2 Negative intervals: (- , -6) & (-6, ) 3. Find the real roots of the following rational equations. [K8] [C2] a. - 7? 9?+11 -12= 1 ? - 7?-12(9?+11) 9?+11 = 1 ? - 7?-12(9?+11) 9?+11 = 1 ? (9x+11) TVO ILC MHF4U – Unit 1 Unit 1 Assessment, Part 2 Copyright © 2019 The Ontario Educational Communications Authority. All rights reserved. 3 - 115-132 9?+11 (x)= 1 ? (x)(9x+11) (-115x2-132x)=(9x+11) -115x2-132x-9x-11=0 -115x2-141x-11=0 ? = -? ± √?2 - 4?? 2? ? = -(-141) ± √(-141)2 - 4(-115)(-11) 2(-115) ? = 141 ± √19881 - 5060 13225 ? = 141 ± √14821 13225 = 141+121.7415 13225 =0.0199 = 141-121.7415 13225 =1.45 x 10-3 b. ?-1 ?+2 = 3?+8 5?-1 (? - 1)(5? - 1) = (3? + 8)(? + 2) 5x2-x-5x+1=3x2+14x+16 5x2-6x+1=3x2+14x+16 2x2-20x-15=0 ? = -?±√?2-4?? 2? TVO ILC MHF4U – Unit 1 Unit 1 Assessment, Part 2 ? = -(-20)±√(-20)2-4(2)(-15) 2(2) ? = 20±√400-120 4 ? = 20±√280 4 ? = 20+16.733 4 = 9.183 ? = 20-16.733 4 = 0.817 4. Solve the following inequality algebraically. Explain your process. [A4] [C2] 8 x -3 ≤ 2x +1 ≤ 17x – 8 8x-3 ≤ 2x+1 2x+1 ≤ 17x-8 8x-2x-3≤1 1≤ 17x-2x-8 6x-3≤ 1 1≤ 15x-8 3 5 ≤ x ≤ 2 3 6x ≤ 1+3 1+8 ≤ 15x 6x ≤ 4 6 9 15 ≤ 15x x ≤ 2 3 3 5 ≤ x 5. Solve the following inequality algebraically. Include an interval chart in your solution. [A6] [C2] 5?+4 ?-1 < 5?-7 ?+13 5?+4 ?-1 - 5?-7 ?+13 < 0 = (5?+4)(?+13) (?-1)(?+13) - (5?-7)(?-1) (?+13)(?-1) < 0

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