mhf4u_unit1_assessment_part2 (1)

Copyright © 20 ±² The Ontario Educational Communications Authority. All rights reserved. MHF4U /earning $ctivity ².²³ Unit 1 Assessment, Part 2 TVO ILC 1 f ( x ) =  − x 2 +  7x – 6 Unit 1 Assessment, Part 2 1. Use the characteristics of the given function to graph the function and it’s reciprocal on the same set of axes, either using online graphing technology or by hand. Find and state all aspects needed to sketch the function and its reciprocal, where possible . Explain your thinking process. [K6] [C2] x y 5 5 -6 -5 Given function is: f(x) = − x + 7x – 6 Function Reciprocal: 1 -x + 7x - 6 x-intercepts: x=1 and x=6 (this is for the regular function) (3.5,6.25) (3.5, 0.16) f(x) has a maximum at (3.5,6.25) for the regular funtion and for 2 2

Copyright © 20 ±² The Ontario Educational Communications Authority. All rights reserved. MHF4U /earning $ctivity ².²³ Unit 1 Assessment, Part 2 TVO ILC 2 f _ -2x - 5 ( x ) = 3x + 18 2. a. Fill in the table and sketch the graph of the following equation. [A10], [C2] Vertical asymptotes Horizontal asymptotes x- intercept y- intercept Domain Range Vertical asymptotes of reciprocal: x=1 and x=6 Horizontal asymptotes of reciprocal: y=0 Intervals where graphs is above x-axis x= -6 y= - 2 3 5 2 x = - x = - 5 18

Copyright © 20 ±² The Ontario Educational Communications Authority. All rights reserved. MHF4U /earning $ctivity ².²³ Unit 1 Assessment, Part 2 TVO ILC 4 5. Solve the following inequality algebraically. Include an interval chart in your solution. [A6] [C2] ____ 5x+4 < ____ 5x–7 x–11 x+13 6. The dimensions of a box are 3 cm × 5 cm × 7 cm. The width, length, and height of the box must be increased by the same amount x in order for the box to have a volume of 693 cm cubed. Determine the value of x that will produce a box with a volume of 693 cm 3 . [T4] [C2] 3 (7+x)(5+x)(3+x) = 693 (7+x)(x +8x+15) = 693 (x +8x 15x+7x +56x 105 = 693 (x +15x +71x+105) = 693 x + 15x + 71x + 105 - 693 = 0 x + 15x + 71x - 588 = 0 (4) + 15(4) +71(4) - 588 = 0 64 + 240 + 284 - 588 = 0 2 3 2 2 2 3 2 3 2 2 2 √ x + 15x + 71x - 588 - x - 4x (x-4) x +19x +147 3 2 3 2 2 19x +71x - 19x -76x 2 2 147x - 588 - 147x - 588 0 Therefore the (x - 4) is the divisor x = -19 + √(19) - 4(1)(147) 2(1) x = -19 + √227 2 x = -19 - √227 2 i i = -9.5 + 7.5332 i = -9.5 - 7.5332 i x minium value -9.5 y minium value 56.75 √-227 are imgainary part in this question: [ i = √-1 ] 3 = √-227 = √-1X227 = √-1 X √227 = i √227 x = 4 (5x+4) (x+13) < (5x-7) (x-11) 5x^2 +65x +4x + 52 < 5x^2 - 55x - 7x +77 5x^2 +69x + 52 < 5x^2 - 62x + 77 69x + 62x < 77 - 52 131x < 25 x < 25 131 Value of X ∈(-∞, -13) U (25/131 ,11)

Copyright © 20 ±² The Ontario Educational Communications Authority. All rights reserved. MHF4U /earning $ctivity ².²³ Unit 1 Assessment, Part 2 TVO ILC 5 7. 7he speci¿cations for a storage container state that the length is ² metre more than triple the width and the height is 5 metres less than double the width. Find the range of possible dimensions for a volume of at least 8436   m 3 using the algebraic method. [T6] [C2] width = x length = (3x + 1) height = (2x -5) To find Volume formula is v = w x l x h v = (x)(2x-5)(3x+1) Since the question states V > 8436m (x)(2x-5)(3x+1) > 8436 x(6x - 15x + 2x - 5) > 8436 x(6x -13x - 5) > 8436 6x - 13x - 5x - 8436 > 0 0 = 6(12) - 13(12) - 5(12) - 8436 > 0 Synthetic Division will be used (x-12)(6x +59x +703) > 0 Since x = width therefore, width > 12m Width is atleast 12m To find height, the width will be subsituted into the height formula 2(12) - 5 = 24 - 5 = 19 < Height: 2x - 5 > 19 Height is atleast 19m 3(12) + 1 = 36 + 1 = 37 < Length: 3x + 1 > 37 Length is atleast 37m 12m x 19m x 37m = 8436m 2 2 3 2 2 3 3 2 12 6 -13 72 -5 708 -8436 8436 6 59 703 0 3