Answers Chapter 1, Study notes of Pre-Calculus

Download Answers Chapter 1 and more Study notes Pre-Calculus in PDF only on Docsity! 408 MHR • Answer Key 978-0-07-073882-9 Answers Chapter 1 1.1 Arithmetic Sequences, pages 1–13 1. a) arithmetic; d = 5; 29, 34, 39 b) not arithmetic c) arithmetic; d = –4; –5, –9, –13 d) not arithmetic e) arithmetic; d = 3; 7, 10, 13 f) arithmetic; d = –13; –30, – 43, –56 2. a) 5, 11, 17, 23; tn = –1 + 6n b) 50, 41, 32, 23; tn = 59 – 9n c) 4.5, 3, 1.5, 0; tn = 6 – 1.5n d) 1 __ 5, 3 __ 5, 1, 7 __ 5; tn = – 1 __ 5 + 2 __ 5 d 3. a) t1 = 4 b) t8 = –11 c) t15 = 80.5 d) t20 = – 18 ___ 7 4. a) d = 12; t1 = –9; –9, 3, and 15 b) d = –8; t1 = 22; 22, 14, and –10 c) d = 3.2; t1 = 16.2; 16.2, and 22.6, 25.8 d) d = 1; t1 = – 3 __ 2 ; – 3 __ 2, and 1 __ 2, 3 __ 2 5. a) n = 12 b) n = 24 c) n = 16 d) n = 38 6. a) yes; n is a whole number: n = 17 b) no; n is not a whole number, so 89 is not a term in the sequence c) yes; n is a whole number: n = 6 d) yes; n is a whole number: n = 11 7. a) d = 4, t1 = 5; tn = 4n + 1 b) d = 6, t1 = – 4; tn = 6n – 10 c) d = –3, t1 = –8; tn = –3n – 5 d) d = 4.5x, t1 = 3 – 22x; tn = 4.5xn + 3 – 26.5x 8. 822 members 9. 7 m 10. x = 5; 5, 9.5, 14 11. after 16 years 12. 4, 11, 18, 25 13. a) an arithmetic sequence; because the common difference between consecutive terms is 1; tn = n b) an arithmetic sequence; because the common difference between consecutive terms is 7; tn = – 4 + 7n c) yes; they have a common difference, 8; tn = – 6 + 8n 14. a) Yes. The points (x, y) represent a sequence where the x-values represent n and the y-values represent the terms of the sequence. The sequence is arithmetic because the points form a straight line, which means that the difference between points is constant. The points in the sequence are 7, 5, 3, 1, –1, –3, –5, –7. b) by substituting into the formula for the general term, tn = t1 + (n – 1)d; tn = 9 – 2n c) t60 = 9 – 2(60) = –111; t300 = 9 – 2(300) = –591 d) The slope is –2, which is the coefficient of n in the formula tn = 9 – 2n. e) The y-intercept is 9, which is the constant term in the formula tn = 9 – 2n. 1. 2 Arithmetic Series, pages 14–21 1. a) S6 = 153 b) S7 = 385 c) S9 = 441 d) S10 = –110 2. a) S30 = 280.5 b) S30 = 1762.5 c) S30 = 445 ____ 3 3. a) S18 = –9 b) S23 = 1715.8 4. a) t1 = –101 b) t1 = 10 c) t1 = 15 d) t1 = 5 5. a) n = 20 b) n = 20 c) n = 16 d) n = 9 6. 204 cans 7. 2 + 6 + 10 + 14 + 18 8. 2 + 5 + 8 + 11 + 14 9. Sn = n __ 2 (3n – 1) 10. a) First determine t1, which is the first multiple of 5 greater than one: t1 = 5. Then determine tn, the last multiple of 5 less than 999: tn = 995. Finally, determine n, the total multiples of 5: n = 995 ____ 5 = 199. Substitute into the formula, Sn = n __ 2 (t1 + tn); S199 = 99 500. 11. Job A pays a total of $4350 and Job B pays a total of $4650. The student should select Job B because it pays $300 more. 1.3 Geometric Sequences, pages 22–31 1. a) not geometric b) geometric; r = 3; 81, 243, 729 c) geometric; r = 0.1; 0.0003, 0.000 03, 0.000 003 d) not geometric e) geometric; r = 1 __ 5; 1 ___ 25, 1 ____ 125, 1 ____ 625 f) geometric; r = –0.5; –0.5, 0.25, –0.125 2. a) tn = 3(4)n – 1 b) tn = 36 ( – 1 __ 3 ) n – 1 c) tn = 4.5(–1.5)n – 1 d) tn = 1 __ 5( – 2 __ 5 ) n – 1 3. a) t10 = –1024 b) t9 = 32 768 c) t11 = – 0.000 000 1 d) t200 = 1 4. a) E b) D c) A d) B e) F f) C 5. a) tn = –3(–2)n – 1 b) tn = (4)n – 1 or tn = –1(–4)n – 1 c) tn = 512(0.5)n – 1 PC11_WB_AK_408-434.indd Page Sec1:408 12/27/11 5:29:25 PM u-s035 /Volumes/101/MHR00089_R1/PRE_CALCULUS_11_STUDENT_WORKBOOK/SE/PRE_CALCULUS_11_... 978-0-07-073882-9 Pre-Calculus 11 Student Workbook • MHR 409 6. 8, 8 √ __ 3 , 16 7. a) n = 7 b) n = 10 c) n = 8 d) n = 7 8. a) two; There are two sequences. b) 2, 6, 18, 54, 162, … and 2, – 6, 18, –54, 162, … 9. 24 576, 12 288, 6144, 3072 10. a) ±8 and ±32 b) 12 and 72 c) ±6 and ±24 11. approximately 352 514 people 12. x = 10 13. a) $1000 b) $1562.50 14. a) The x-values of the points on the graph correspond to n, and the y-values are the terms of the sequence for each value of x. The first five terms of the sequence are the y-values that correspond to the x-values 1, 2, 3, 4, 5: 48, 24, 12, 6, 3. There is a common ratio of 0.5 between these y-values, so the points represent a geometric sequence. b) yes; tn = 48(0.5)n – 1 15. a) The sequence is 20 000, 2000, 200, 20, 2. This is a geometric sequence with common ratio r = 0.1. b) Use the general term of the sequence or write the terms until the seventh term is found (by multiplying the previous term by 0.1). The volume on the seventh day is 0.02 cm3. 1.4 Geometric Series, pages 32–40 1. a) yes; a common ratio of 3; S7 = 6558 b) yes; a common ratio of –3; S7 = 4376 c) no; no common ratio d) yes: a common ration of 1 __ 2; S7 = 127 ____ 192 2. a) t1 = 2, r = 3, n = 8; S8 = 6560 b) t1 = 2.1, r = –2, n = 9; S9 = 359 1 ___ 10 c) t1 = 30, r = – 1 __ 6, n = 7; S7 = 25 5555 _____ 7776 d) t1 = 24, r = – 3 __ 4, n = 6; S6 = 11 35 ____ 128 3. a) S12 = –2730 b) S8 = 29 999 999 700 4. a) Sn = 335 923 b) Sn = 3 333 333 _________ 2500 c) Sn = 78 642 d) Sn = 6315 _____ 6571 5. a) Sn = 3n – 1 b) Sn = 5(2n – 1) c) Sn = 4n – 1 d) Sn = 4(2n – 1) 6. a) n = 11; S11 = 6141 b) n = 6; S6 = –92.969 7. S7 = 5465 employees 8. 13 terms 9. 664.78 cm 10. 397 mg 11. S10 = 2046 12. approximately 89.5 turns per second 13. approximately 183 m 14. Example: The sum of the first three terms can be written as t1 + t1r + t1r2 = 93. From the given term, t2 = 15, or t1r = 15, write an expression for t1. Substitute this expression into the quadratic equation and solve: r = 5 or r = 1 __ 5; 3 + 15 + 75 + … or 75 + 15 + 3 + … 15. a) a geometric series; terms have common ratio 0.7 b) use Sn = t1(rn – 1) ________ r – 1 and substitute t1 = 50, r = 0.7; S7 ≈ 153 m 1.5 Infi nite Geometric Series, pages 41–49 1. a) convergent; S∞ = – 243 ____ 4 b) convergent; S∞ = 32 c) divergent; sum does not exist d) convergent; S∞ = 0.2 e) divergent; sum does not exist f) divergent; sum does not exist 2. a) S∞ = – 4 b) sum does not exist c) S∞ = –108 d) S∞ = 1728 _____ 19 e) S∞ ≈ 3.31 f) sum does not exist 3. a) 1 __ 3 b) 25 ___ 99 c) 447 ____ 99 d) 556 ____ 495 4. yes; S∞ = 3 + 9 ___ 10 ____ 9 ___ 10 = 4 5. a) S∞ = 5 b) S∞ = 21 ___ 11 6. a) –1 < x < 1 b) –2 < x < 2 7. 260 cm 8. 80 m 9. a) To solve for t1, substitute the known values S∞ = 105, and r = – 2 __ 3 into the formula S∞ = t1 ____ 1 – r: t1 = 175. b) 175 – 350 ____ 3 + 700 ____ 9 – ... 10. a) Solve for r by substituting the known values S∞ = – 45 and t1 = –18 into the formula S∞ = t1 ____ 1 – r: r = 3 __ 5. b) –18 – 54 ___ 5 – 162 ____ 25 – 486 ____ 125 – ... Chapter 1 Review, pages 50–54 1. a) not arithmetic b) arithmetic; d = 1 __ 2, tn = 1 1 __ 2 + 1 __ 2 n c) not arithmetic d) arithmetic; d = –3x2, tn = x2 – 3nx2 PC11_WB_AK_408-434.indd Page Sec1:409 12/27/11 5:29:29 PM u-s035 /Volumes/101/MHR00089_R1/PRE_CALCULUS_11_STUDENT_WORKBOOK/SE/PRE_CALCULUS_11_... 412 MHR • Answer Key 978-0-07-073882-9 10. a) x y 0 P(12, -3) θ b) 14° c) 346° 11. Trigonometric Ratio sin θ = y __ r cos θ = x __ r tan θ = y __ x 0° 0 1 0 Quadrant I + + + 90° 1 0 undefined Quadrant II + – – 180° 0 –1 0 Quadrant III – – + 270° –1 0 undefined Quadrant IV – + – 360° 0 1 0 2.3 The Sine Law, pages 86–93 1. a) 15.4 b) 24.6 c) 26° d) 11° 2. a) 15.3 m b) 11.7 km 3. a) 38° b) 15° c) 112° 4. a) 68.40 b) 9.64 m c) 2.20 d) 43.30 5. a) right angle, 1 triangle b) 0 triangles c) a = b, 0 triangles d) h < a < b, 2 triangles e) b ≤ a, 1 triangle 6. 1.7754 Å 7. ∠A = 35°, ∠B = 29°, ∠C = 116°, a = 120, b = 100, c = 188 8. acute △ABC: ∠A = 41°, ∠B = 56°, ∠C = 83°, a = 12.3 cm, b = 15.6 cm, c = 18.61 cm obtuse △ABC: ∠A = 41°, ∠B = 124°, ∠C = 15°, a = 12.3 cm, b = 15.6 cm, c = 4.85 cm 9. Answers may vary. 2.4 The Cosine Law, pages 98–102 1. a) a2 = b2 + c2 – 2bc cos A b) c2 = a2 + b2 – 2ab cos C c) l 2 = j 2 + k2 – 2jk cos L d) y2 = x2 + z2 – 2xz cos Y 2. 246 3. 44 mm or 4.4 cm 4. 145 mm or 14.5 cm 5. a) 39° b) 121° c) 36° 6. 8.1 cm 7. ∠A = 22°, ∠B = 142°, ∠C = 16°, a = 21, b = 35, c = 16 8. 28.7 ft 9. Given Information Solve For Formula m, ∠L, n (SAS) l l 2 = m2 + n2 – 2mn cos L l, ∠M, n (SAS) m m2 = l 2 + n2 – 2ln cos M l, ∠N, m (SAS) n n2 = l 2 + m2 – 2lm cos N l, m, n (SSS) ∠L cos L = l 2 – m2 – n2 __________ – 2mn l, m, n (SSS) ∠M cos M = m 2 – l 2 – n2 __________ – 2ln l, m, n (SSS) ∠N cos N = n 2 – l 2 – m2 __________ – 2lm Chapter 2 Review, pages 103–105 1. a) x y 0 35° θR I, θR = 35° b) x y 0 165° θR II, θR = 15° c) x y 0 216° θR III, θR = 36° PC11_WB_AK_408-434.indd Page Sec1:412 12/27/11 5:29:33 PM u-s035 /Volumes/101/MHR00089_R1/PRE_CALCULUS_11_STUDENT_WORKBOOK/SE/PRE_CALCULUS_11_... 978-0-07-073882-9 Pre-Calculus 11 Student Workbook • MHR 413 2. a) –1 b) 1 ___ √ __ 3 c) – 1 ___ √ __ 2 3. sin θ = 5 ____ √ ___ 41 , cos θ = – 4 ____ √ ___ 41 , tan θ = – 5 __ 4 4. cos θ = – 8 ___ 17, tan θ = – 15 ___ 8 5. a) 54°, 306° b) 300°, 240° 6. a) 19° b) 205 7. a) 0 triangles b) 2 triangles 8. a) 13.4 b) 27° Chapter 3 3.1 Investigating Quadratic Functions in Vertex Form, pages 110–119 1. a) (3, 5); no x-intercepts b) (0, 1); two x-intercepts c) (11, 0); one x-intercept d) ( – 1 __ 2, 7 __ 3 ) ; two x-intercepts 2. a) opens upward, x = 5; minimum value is –8 b) opens downward, x = –3; maximum is –5 3. a) a = 1, p = –4, q = –2; opens upward, minimum value –2; translated 4 units to the left and 2 units down; domain {x | x ∊ R}, range {y | y ≥ –2, y ∊ R} 0 2 4 x y -2 -2-4-6 2 y = (x + 4)2 - 2 b) a = –4, p = –7, q = 2; opens downward, maximum value 2; translated 7 units to the left and 2 units up; domain {x | x ∊ R}, range {y | y ≤ 2, y∊R} 0 2 x y -4 -2 -2-4-6-8 y = -4(x + 7)2 + 2 4. a) 0 2 4 x y -4 -2 -2-4-6-8-10-12 y = -2(x + 5)2 + 4 x = -5 (-5, 4) domain {x | x ∊ R}, range {y | y ≤ 4, y ∊ R} b) 0 2 4 x y -4 -2 -2 2 4 6 8 y = (x - 3)2 - 41 2(3, -4) x = 3 domain {x | x ∊ R}, range {y | y ≥ – 4, y ∊ R} 5. a) y = –(x + 5)2 + 2 b) y = 1 __ 2 (x + 6)2 6. a) y = 5x2 + 4 b) y = –2(x – 3)2 c) y = 4(x – 1)2 – 1 7. a) (–6, 0) b) (1, –1) c) (–2, 4.5) 8. a) y = 1 __ 9x2 b) y = 1 __ 9 (x – 3)2 – 1 c) Example: The vertical stretch and domain remain the same while the vertex, axis of symmetry, and range are different. 9. a) There are infinitely many parabolas. For example, any parabola with vertex on the line x = 4. b) Example: maximum at (4, 16) gives y = –(x – 4)2 + 16 c) Example: maximum at (4, 8) gives y = – 1 __ 2(x – 4)2 + 8 d) One or more of p and q change when the vertex of a parabola changes, depending on how the vertex is moved. The value of a may change when the location of the vertex changes. 10. a) Example: If θ = 30°, then d = –0.088v2. This graph has its vertex at the origin and is vertically stretched compared to the graph of y = x2, but opens downward. b) Answers may vary. c) Example: All the parabolas will have vertices at the origin. The vertical stretch factor will change depending on the angles. The domain and range will be the same for all graphs. 11. a) $3700 b) y x 3000 2000 1000 40 4000 5000 8 12 16 20 y = -10(x - 9)2 + 4510 c) Example: As prices initially rise more revenue will result, but with each price increase fewer people will buy a shirt. So, eventually prices can be so high that total revenue will decrease. PC11_WB_AK_408-434.indd Page Sec1:413 12/27/11 5:29:35 PM u-s035 /Volumes/101/MHR00089_R1/PRE_CALCULUS_11_STUDENT_WORKBOOK/SE/PRE_CALCULUS_11_... 414 MHR • Answer Key 978-0-07-073882-9 d) The vertex (9, 4510) is the maximum point on the graph. So, when x = 9 the price of each T-shirt is $21 and the maximum revenue is $4510. 3.2 Investigating Quadratic Functions in Standard From, pages 123–132 1. a) quadratic: f (x) = –5x2 – 20x – 12 b) not quadratic c) quadratic: f (x) = 8x2 – 29x – 55 d) not quadratic 2. a) vertex (1, 4), axis of symmetry x = 1, y-intercept 3, x-intercepts –1 and 3, maximum value 4, domain {x | x ∊ R}, range {y | y ≤ 4, y ∊ R} b) vertex (–3, 0), axis of symmetry x = –3, y-intercept 9, x-intercept –3, minimum value 0, domain {x | x ∊ R}, range {y | y ≥ 0, y ∊ R} c) vertex (–2, 2), axis of symmetry x = –2, y-intercept –6, x-intercepts –3 and –1, maximum value 2, domain {x | x ∊ R}, range {y | y ≤ 2, y ∊ R} 3. a) vertex (2.5, –7.3), axis of symmetry x = 2.5, minimum value –7.3, domain {x | x ∊ R}, range {y | y ≥ –7.25, y ∊ R}, x-intercepts –0.2 and 5.2, y-intercept –1 b) vertex (0.3, 3.1), axis of symmetry x = 0.3, maximum value 3.1, domain {x | x ∊ R}, range is {y | y ≤ 3.1, y ∊ R}, x-intercepts 1.5 and –1, y-intercept 3 c) vertex (11, –3), axis of symmetry x = 11, minimum value –3, domain {x | x ∊ R}, range {y | y ≥ –3, y ∊ R}, x-intercepts 7.5 and 14.5, y-intercept 27.3 4. a) (2, –16) b) (–1, –4) c) (4, 41) d) ( 3 __ 2, – 19 ___ 2 ) 5. a) two x-intercepts b) two x-intercepts c) one x-intercept d) two x-intercepts e) no x-intercepts f) no x-intercepts 6. a) 0 m b) 5 m after 1 s c) 2 s d) domain {t | 0 ≤ t ≤ 2, t ∊ R}, range {h | 0 ≤ h ≤ 5, h ∊ R} 7. a) a = v 2 ___ 25 b) vertex (0, 0), axis of symmetry x = 0, x- and y-intercept of 0 0 2 4 v a 6 8 4 8 12 a = v2 25 c) domain {v | v ≥ 0, v ∊ R}; since speed is positive d) range {a | a ≥ 0, a ∊ R}; since the graph opens up e) The curve does not fit the criteria. Example: When v = 14, a = 7.84 m/s2 which is not 6 m/s2. 8. a) Yes, it is possible to have more than one correct answer. Example: 0 2 4 x y -8 -10 -6 -4 -2 -2-4 2 4 6 8 b) 0 2 4 x y -6 -4 -2 -2-4 2 4 6 8 9. a) (12, 200) b) 0 50 100 p d 150 200 10 20 30 40 d = -p2 + 24p + 56 c) The vertex indicates that maximum demand is 200 at a price of $12. d) Demand initially increases as price increases, but when price exceeds $12 demand decreases. 10. a) A = w(12 – w); polynomial of degree 2 b) 0 10 20 w A 30 40 4 8 12 A = -w2 + 12w c) (6, 36); represents the maximum area of the dog enclosure d) 6 m by 6 m; maximum area of 36 m2 11. a) Knowing the location of the vertex and direction of opening allows you to visualize the parabola and its number of x-intercepts. PC11_WB_AK_408-434.indd Page Sec1:414 12/27/11 5:29:36 PM u-s035 /Volumes/101/MHR00089_R1/PRE_CALCULUS_11_STUDENT_WORKBOOK/SE/PRE_CALCULUS_11_... 978-0-07-073882-9 Pre-Calculus 11 Student Workbook • MHR 417 8. a) – 4.5t2 + 8.3t + 2.1 = 0 b) 2.1 s 0 2 4 x y 6 -2 -2-4 2 4 6 8 (-0.2, 0.0) (2.1, 0.0) f (x ) = -4.5x2 + 8.3x + 2.1 9. Examples: a) The value of a is the same in both cases. b) The form y = a(x – p)2 + q is more useful for finding the vertex. 10. Examples: a) y = (x – 7)2 b) The vertex is on the x-axis, so x = 7 is the only root. 4.2 Factoring Quadratic Equations, pages 160–164 1. a) (x – 6)(x – 3) b) 5(b – 3)(b + 2) 2. a) (n – 4)(3n + 1) b) (x + 2)(4x + 3) c) (t – 6)(2t – 5) d) (3x + 2)(4x – 3) 3. a) 1 __ 2(x – 6)(x + 2) b) 1 __ 4(x – 4)(x + 6) c) 0.1(a – 6)(a + 5) d) 0.1(z – 10)(5z – 4) 4. a) (0.9x + 0.5y)(0.9x – 0.5y) b) (1.1k – 0.1x)(1.1k + 0.1x) c) ( 1 __ 5d – 1 __ 7 f ) ( 1 __ 5d + 1 __ 7 f ) d) 2(2a – 3b)(2a + 3b) 5. a) (x + 6)(x – 2) b) 4(x + 4)(x + 3) c) (x + 6)(2x – 1) d) (10x – 7)(10x – 3) 6. a) –9, 5 b) – 9 __ 2, 4 c) – 3 __ 4, 11 ___ 2 d) 0, 14 ___ 3 7. a) 4, 5 b) –6 c) –6, 7 __ 3 d) 3 __ 4, 3 __ 2 e) 0, –6 f ) –9, –6 8. a) w(w + 3) = 154 b) 11 in. by 14 in. 9. a) x + 1 b) x(x + 1) = 156 c) 12 and 13, or –13 and –12 10. a) –0.ld 2 + 4.8d = 0 b) 48 m 11. a) 140 m2 b) (2x + 8)(2x + 12) = 140; 1 m 12. If c = 0, then one of the factors equals zero. Example: x2 – 3x = 0 factors to x(x – 3), and the factors are 0 and 3. 13. Example: Graph the corresponding function and determine the x-intercepts. 4.3 Solving Quadratic Equations by Completing the Square, pages 168–171 1. a) 36 b) 9 __ 4 c) 1 ___ 64 d) 0.16 2. a) –1, 9 b) –9, 3 c) –2 √ __ 5 – 5, 2 √ __ 5 – 5 d) –2 √ ___ 10 + 1, 2 √ ___ 10 + 1 e) – √ __ 6 – 3 __ 2, √ __ 6 – 3 __ 2 3. a) –1, 5 b) –12, 2 c) – 5 __ 3, 1 __ 3 d) 1 __ 5, 13 ___ 5 4. a) –5, 11 b) – √ ___ 10 + 2, √ ___ 10 + 2 c) –2 √ __ 2 + 5, 2 √ __ 2 + 5 d) –3 √ __ 2 – 2, 3 √ __ 2 – 2 5. a) A = π(x + 5)2 b) 1.9 m 6. a) The room is a rectangle with sides 8 + x and 6 + x. b) (6 + x)(8 + x) = 144 c) 11.0 ft by 13.0 ft 7. line 2 should be x2 – x + 1 __ 4 = 5 + 1 __ 4; – √ ___ 21 + 1 ________ 2 , √ ___ 21 + 1 _______ 2 8. 8, 15, 17 9. 100 yd 10. Example: x2 + 4x = –10; Any equation that requires the square root of a negative number cannot be solved. 11. Example: It depends on whether the zeros of the quadratic function or the vertex of the graph are required. 4.4 The Quadratic Formula, pages 177–180 1. a) 1 root b) 2 roots c) 0 roots d) 1 root 2. a) 3 – √ __ 3 ______ 3 , √ __ 3 + 3 ______ 3 b) – 2 __ 3, 1 __ 2 c) 2 + √ __ 2 ______ 3 , 2 – √ __ 2 ______ 3 d) 5 + √ ___ 57 ________ 8 , 5 – √ ___ 57 _______ 8 3 a) –0.6, 1.6 b) 0.2, 2.2 c) –1.5, 10 d) –1.5, 3.5 4. 0 or 6 5. 3, 12 6. 5 cm 7. a) 4.34 s b) 4.30 s c) In both cases, one root is negative, which does not make sense in this context. 8. 69.28 m 9. a) row 1: 125; 5 row 2: 100; 2 row 3: (100)(125); (125 + 5n)(100 – 2n) b) (125 + 5n) represents the cost of a calculator for any number of price increases of $5. (100 – 2n) represents the fact that for each price increase, the store sells two fewer calculators. The revenue is the price multiplied by the number sold, so r (n) = (125 + 5n)(100 – 2n). c) Set the revenue equation equal to 14 000 and solve for n. Since n = 10 and n = 15, there can be 10 to 15 price increases, resulting in a price range of $175 to $200. PC11_WB_AK_408-434.indd Page Sec1:417 12/27/11 5:59:06 PM u-s035 /Users/u-s035/Desktop/mdghazali_27-12-11/precalculus11/ak 418 MHR • Answer Key 978-0-07-073882-9 Chapter 4 Review, pages 181–186 1. a) –1, 5 b) 3 2. a) Example: The location of the vertex and the direction of opening determine the number of zeros for the quadratic function. In this case, the graph would intersect the x-axis in two places. b) Example: The location of the vertex is on the x-axis. c) Example: The minimum is above the x-axis, or the maximum is below the x-axis, meaning that the graph does not intersect the x-axis. 0 2 4 x y 6 8 10 -2 -4 -6 -8 -2-4-6-8 2 4 6 8 two roots one root no roots 3. a) 2.8, 7.2 0 2 4 x y 6 8 -2 -4 -6 -2 2 4 6 8 10 (2.8, 0) (7.2, 0) b) 8.6, 13.4 0 2 4 x y 6 8 10 -2 -2 4 8 12 16 20 (8.6, 0) (13.4, 0) c) 0.1, 3.1 0 2 4 x y 6 8 10 -2 -4 -2-4-6 2 4 6 (0.1, 0) (3.1, 0) 4. a) (a – 7b + 68)(a + 7b – 58) b) (x + 3)(x – 5) c) ( 3m ___ 4 + 10n ____ 9 ) ( 3m ___ 4 – 10n ____ 9 ) 5. a) –4, –2 b) 2 __ 3, 1 c) 3 __ 2, 9 __ 2 d) – 3 __ 2, 3 __ 2 6. 9 in. by 12 in. 7. a) ±13 b) –18, 4 c) –4 √ __ 5 + 12, 4 √ __ 5 + 12 d) –5, 3 8. a) – √ ___ 23 – 4, √ ___ 23 – 4; –8.8, 0.8 b) –3 √ __ 2 + 5, 3 √ __ 2 + 5; 0.8, 9.2 9. 30th day 10. a) 2 roots b) 2 roots c) 1 root d) 0 roots 11. a) 5 – √ ___ 15 , 5 + √ ___ 15 ; 1.1, 8.9 b) –1 + √ ___ 41 ________ 5 , –1 – √ ___ 41 ________ 5 ; –1.5, 1.1 12. a) –7, 3 b) – 2 __ 5, 3 c) –9 + √ ___ 57 ________ 4 , –9 – √ ___ 57 ________ 4 Chapter 5 5.1 Working With Radicals, pages 193–198 1. Mixed Radical 5 √ __ 5 3 √ __ 7 –2 3 √  7 Entire Radical √ ____ 252 – √ ____ 375 4 √  486 2. a) b 3 √  by b) √ ___ 4a , a ≥ 0 c) –2x 3 √  6y d) 2a 4 √  2a, a ≥ 0 e) √ _____ 98y2z , y ≥ 0, z ≥ 0 f) – √ _____ 45m5 , m ≥ 0 3. √ ___ 63 = 3 √ __ 7 , so √ ___ 63 and √ __ 7 , are like radicals; – √ ___ 27 = –3 √ __ 3 and 3 √ ___ 75 = 15 √ __ 3 , so – √ ___ 27 and 3 √ ___ 75 are like radicals; 3 √  250 = 5 3 √  2 and 4 3 √  16= 8 3 √  2, so 3 √  250 and 4 3 √  16 are like radicals PC11_WB_AK_408-434.indd Page Sec1:418 12/27/11 5:59:08 PM u-s035 /Users/u-s035/Desktop/mdghazali_27-12-11/precalculus11/ak 978-0-07-073882-9 Pre-Calculus 11 Student Workbook • MHR 419 4. a) Example: –5 √ __ 7 b) Example: 5 3 √  2 c) Example: 10 √ ___ 2m d) Example: 8a 4 √  2a e) Example: 5x √ __ 2 f) Example: –4y √ ____ 5xy 5. a) – √ ___ 60 , –8, –2 √ ___ 17 , –5 √ __ 3 b) 18, 3 ___ 10 √ _____ 3500 , 9 √ ___ 15 ___ 4 , √ ____ 300 c) 3 4 √  8, 5, 4 √  615, 4 4 √  9 __ 4 6. a) 8 √ __ 7 b) 15 √ __ 7 – 17 √ __ 2 c) 3 √  9 + 2 d) 0 7. a) 13 √ __ m , m ≥ 0 b) –x √ ___ 3x , x ≥ 0 c) –3ab √ ___ 2b , a ≥ 0, b ≥ 0 d) 2y √ __ y , y ≥ 0 8. 3 in. 9. 5 s 10. 8 √ __ 2 ft 11. 4 √ __ 2 ft 12. 20 √ ___ 42 ft 13. The error made was that the unlike radicals were added. You must simplify and then add. The answer should be 6 √ __ b . 14. 25 √ __ 5 – 3 √ __ 3 15. The two legs of the triangle are s ___ √ __ 2 . A = 1 __ 2ab = 1 __ 2( s ___ √ __ 2 ) ( s ___ √ __ 2 ) = s 2 __ 4 5.2 Multiplying and Dividing Radical Expressions, pages 203–209 1. a) 30 √ __ 6 b) 12a2 √ __ 6 c) –70x2 d) ( 3y2 ) 3 √  y 2. a) 2 √ ___ 30 + 4 √ __ 5 b) –5 √ ___ 30 + 3 √ __ 2 c) 6 √ ____ 35x – 2x √ __ 7 d) 5 √ ____ 21a – 7a √ __ 3 + 6 √ ___ 7a 3. a) –22 + 13 √ __ 3 b) 58 + 12 √ __ 6 c) 4 d) 3 √ ___ 21 – 2 √ __ 7 – 12 √ __ 3 – 1 4. a) –9k + 19 √ ___ 3k – 20, k ≥ 0 b) 2 – 6 √ ____ 10m + 45m, m ≥ 0 c) – 40x + 5x √ ___ 10 – 4 √ ____ 10x + 5 √ __ x , x ≥ 0 d) 16y + 40y 3 √  4y2 – y 3 √  2y – 5y2, any real number 5. a) 3 b) –20 c) 3 √ __ 7 ____ 4 d) –3 √ __ 6 _____ 16 e) √ __ 6 ___ 2y f) 3 √ ___ 2a – 3a √ __ 6 ___________ 16a 6. row 1: 4 – √ __ 5 , 11; row 2: –5 + 3 √ __ 3 , –2; row 3: 7 √ __ 5 – 4 √ __ 2 , 213; row 4: 2 √ __ z + √ __ 3 , 4z – 3 7. a) – ( 4 √ __ 2 + 28 ) ___________ 47 b) – ( 3 √ ___ 30 – 12 √ __ 5 ) ______________ 10 c) – 4 √ __ 5 + 8 √ __ 2 + 10 √ ___ 10 + 25 ______________________ 109 d) 36 + 9 √ __ x ________ 16 – x 8. 60 √ ___ 10 9. 32π √ ___ 13 10. a) 10 √ __ 2 cm2 b) 242 cm2 c) 12 √ __ 6 cm2 11. Tineka did not apply the distributive rule properly when expanding the squared binomial. The proper solution is 23 – 6 √ ___ 10 . 12. ( 2 – √ __ 7 ) 2, ( √ __ 7 + 2 ) ( √ __ 7 – 2 ) , √ __ 7 ( √ __ 7 + 2 ) , ( √ __ 7 + 2 ) 2 13. 2.26 m 14. Anthony made an error when multiplying the numerator. The product should be 4 √ __ m + m – 12. Also, the denominator should be 36 – m. The correct solution is 4 √ __ m + m – 12 ____________ 36 – m , m ≥ 0, m ≠ 36. 15. i) All possible factors have been removed from each radical. ii) The radical contains no fractions. iii) The denominator is a rational number. 16. False. The radicand of a radical with an even index must be positive. Therefore, √ ___ –8 and √ ___ –2 are not real numbers. 5.3 Radical Equations, pages 216–222 1. a) 16x2 – 40x + 25 b) 7y c) 2x – 3 d) 324m e) 9n – 48 √ __ n + 64 2. x = 2 is correct, x = 6 is extraneous 3. a) x = 83, x ≥ 2 b) m = 10 000, m ≥ 0 c) a = 6, a ≥ 1 d) p = 1, p ≤ 2; p = –2 is an extraneous root e) n = 5; n ≥ –19 ____ 6 ; n = –3 is an extraneous root f) c = 10 and c = 9, c ≥ 54 ___ 7 4. a) x = 0 b) x = –3 + 3 √ __ 3 ; x ≤ –3 √ __ 2 _____ 2 or x ≥ 3 √ __ 2 ____ 2 c) x = 16, x ≥ 7 5. a) n = 1 b) x = 20 c) y = 3, –1 d) c = 2 6. The term (m – 4) was squared incorrectly. The correct squaring is m2 – 8m + 16. You can factor this equation, resulting in roots of m = 3 and m = 2, both of which are extraneous. 7. a) 19.44 m/s b) 70 km/h 8. 39.06 ft 9. 1.528 m 10. 8 cm 11. a) (a – b)2 = (a – b)(a – b) = a2 – 2ab + b2 Therefore, (a – b)2 ≠ a2 – b2. b) Example: Let a = 10 and b = 6 Left side = 16; Right side = 64 The left side is not equal to the right side. c) If a = b, then (a – b)2 = a2 – b2. PC11_WB_AK_408-434.indd Page Sec1:419 12/27/11 5:59:09 PM u-s035 /Users/u-s035/Desktop/mdghazali_27-12-11/precalculus11/ak 422 MHR • Answer Key 978-0-07-073882-9 Chapter 7 7.1 Absolute Value, pages 280–284 1. a) 5 __ 6 b) 1 1 __ 4 c) 0.8 d) 12 e) 1.2 f ) 0 2. 2 3 __ 5, |–2.5|, ∣ –15 ____ 7 ∣ , |–2.09|, –2, –2 6 __ 9 3. a) A B B -2-3-4 -1-6-5 0 1 2 3 4 5 6 7 8 or b) A B B -4-5-6 -3-2 -1-8-7 0 1 2 3 4 5 6 or c) B B -2-3-4 -1-6-5 0 1 2 3 4 5 6 7 8 or A 4. a) 0.3 b) –8 c) 8 d) –1 e) 0.09 f ) 4 5. a) |156 – (–458)| or |–458 – 156| b) 614 m 6. a) |–39 – 357| or |357 – (–39)| b) 396 7. a) |–23 – (–27)| or |–27 – (–23)| b) |15 – 35| or |35 – 15| 8. a) |25.98 – 26.83| or |26.83 – 25.98| b) rise of $0.85 9. Differences: use of absolute value symbols; Similarities: both expressions will produce the same result, that is: (x1 – x2)2 = |x1 – x2| 2 10. Answers will vary. 7.2 Absolute Value Functions, pages 290–296 1. a) 0, 6, 12, 14 b) 7, 4, 1, 2, 5 2. (–1, 10), (0, 6), (1, 2), (2, 2) 3. a) 0 2 4 6 x y -4 -2 -2-4 2 4 6 y = f(x) y = |f(x)| b) 0 2 4 6 8 x y -4 -2 -2 2 4 6 8 10 y = f(x) y = |f(x)| 4. a) 0 2 4 x y -4 -2 -2-4 2 4 y = f(x) y = |f(x)| b) 0 2 4 6 -4 -2 -2-4 2 4 x y y = f(x) y = |f(x)| 5. a), b) 0 2 4 6 8 x y 10 12 -2 -2-4-6 2 4 6 8 10 12 y = |x + 2| y = |-(x - 1)2 + 5| For y = |x + 2|: x-intercept –2, y-intercept 2, domain {x | x ∈ R}, range {y | y ≥ 0, y ∈ R} For y = | – (x – 1)2 + 5|: x-intercepts 1 ± √ __ 5 , y-intercept 4, domain {x | x ∈ R}, range {y | y ≥ 0, y ∈ R} 6. a) y = x – 5, if x ≥ 5 –(x – 5), if x < 5 b) y = 1 __ 3x + 2, if x ≥ –6 – ( 1 __ 3x + 2 ) , if x < –6 7. a) y = –x2 + 2, if – √ __ 2 ≤ x ≤ √ __ 2 –(–x2 + 2), if x < – √ __ 2 or x > √ __ 2 b) Example: y = 2(x – 1)2 – 2, if x ≤ 0 or x ≥ 2 –[2(x – 1)2 – 2], if 0 < x < 2 8. a) x y = ∣ 1 __ 2x − 6 ∣ 0 6 12 0 2 5 4 4 18 or 6 3 –2 7 –6 9 PC11_WB_AK_408-434.indd Page Sec1:422 12/27/11 5:59:12 PM u-s035 /Users/u-s035/Desktop/mdghazali_27-12-11/precalculus11/ak 978-0-07-073882-9 Pre-Calculus 11 Student Workbook • MHR 423 b) 0 2 4 6 x y 2 4 6 8 10 12 c) Domain: {x | x ∈ R}; Range: {y | y ≥ 0, y ∈ R} d) y = 1 __ 2x – 6, if x ≥ 12 – ( 1 __ 2x – 6 ) , if x < 12 9. a) y = |(x + 2)2 – 7| b) (–2, 7), (–2 – √  7, 0), (–2 + √  7, 0), (0, 3) c) 0 2 4 6 8 x y 10 -2-4-6-8 2 d) Domain: {x | x ∈ R}; Range: {y | y ≥ 0, y ∈ R} e) y = x2 + 4x –3, if x ≤ –2 – √ __ 7 or x ≥ –2 + √ __ 7 –(x2 + 4x –3), if –2 – √ __ 7 < x < –2 + √ __ 7 10. a) none; none; none b) The graphs are the same. c) 0 2 4 6 8 x y 10 -2 -2 2 4 6 8 11. a) y = |2x| b) y = |3x + 6| 12. a) Yes, by definition, 2 – x = 2 – x and 2 – x = –2 + x, both equations have x = 2 as the solution. b) Yes, ∣ x2 ∣ = x2, by definition, x2 = x2 for all real values of x, and x2 = – ( x2 ) , only if x = 0. 13. a) Cell Phone Sales and Profit 0 x y 400 800 1200 1600 2000 2400 2800 3200 20 40 60 80 100 120 (75, 3025) -20 -400 Pr ofi t ($ ) Number of Phones b) y = |–x2 + 150x – 2600| c) $3025 d) No, sales can go to infinity according to the function from part b); so, profit can go to infinity. 7.3 Absolute Value Equations, pages 303–308 1. a) 11, –11 b) 6, –6 c) 2, –2 d) 9.5, –9.5 2. a) 7, –3 b) 6, –12 c) 13, –3 3. a) 10, –76 ____ 7 b) No solution c) No solution 4. a) 6, –14 b) ± √ __ 5 , –1 5. a) Example: 0.5 = |x – 32| b) 31.5 oz and 32.5 oz 6. a) Example: 0:15 = |x – 08:00| b) 8:15 to 7:45 7. a) Example: 0.01 = |x – 2.5| b) 2.49 mm to 2.51 mm 8. a) Example: 0.002 = |i – 0.036| b) 0.038 A to 0.034 A 9. Example: 0.35 = |x – 9.6|; 9.25 lb to 9.95 lb 10. 24 screens, $64 000 11. No; 4 incorrectly grouped with –9 in step 1. 4 – 9|–6 – x| = –15 –9| –6 – x| = –19 |–6 – x| = 19 ___ 9 –6 – x = 19 ___ 9 x = –73 ____ 9 –6 – x = –19 ____ 9 x = –35 ____ 9 12. a) No; d = |a| + |b| + |c|. b) Example: let a = 7, b = –5, c = –1 d = |7| + |–5| + |–1| d = 7 + 5 + 1 d = 13 d = |7 + (–5) + (–1)| d = |1| d = 1 |a + b + c| ≠ |a| + |b| + |c| PC11_WB_AK_408-434.indd Page Sec1:423 12/27/11 9:59:31 PM u-s035 /Volumes/101/MHR00089_R1/PRE_CALCULUS_11_STUDENT_WORKBOOK/SE/PRE_CALCULUS_11_... 424 MHR • Answer Key 978-0-07-073882-9 13. a) Example: d = |m1 – m2| b)– c) Answers may vary. 14. a) |4x + 7| + 8 = 2 Isolate the absolute value |4x + 7| = –6; This is not possible according to the definition of absolute value. b) Example: |2x – 8| = –10; This is not possible according to the definition of absolute value. 7.4 Reciprocal Functions, pages 318–323 1. y = f (x) y = 1 ____ f(x) y = –x y = 1 ___ –x y = 3x – 1 y = 1 ______ 3x – 1 y = x2 – 4x + 4 y = 1 __________ x2 – 4x + 4 y = x2 y = 1 __ x2 y = x __ 2 y = 2 __ x 2. a) x = 0 b) x = –5 c) x = 2 __ 7 d) x = 4 and x = –4 e) x = –3 and x = 2 3. y = f (x) y = 1 ____ f (x) Invariant Point(s) y = x – 8 y = 1 _____ x – 8 (9, 1), (7, –1) y = 3x – 4 y = 1 ______ 3x – 4 ( 5 __ 3, 1 ) , (1, –1) y = x2 – 25 y = 1 ______ x2 – 25 ( √ ___ 26 , 1 ) , (– √ ___ 26 , 1), ( √ ___ 24 , –1 ) , (– √ ___ 24 , –1) y = x2 – x – 29 y = 1 __________ x2 – x – 29 (6, 1), (–5, 1), ( 1 – √ ____ 113 ________ 2 , –1 ) , ( 1 + √ ____ 113 ________ 2 , –1 ) 4. a) {x | x ≠ 0, x ∈ R}; {y | y ≠ 0, y ∈ R}; x = 0 b) {x | x ≠ 1, x ∈ R}; {y | y > 0, y ∈ R}; x = 1 5. a) 0 2 4 6 x y -4 -6 -2 -2-4-6 2 4 y = f(x) y = 1 f(x) b) 0 2 4 x y -4 -6 -2 -2-4 2 4 y = f(x) y = 1 f(x) 6. a) 0 2 4 6 x y -4 -6 -2 -2-4-6 2 4 6 y = f(x) y = 1 f(x) x = –3, y = 0; (–2, 1), (– 4, –1); f(x): (–3, 0), (0, 3); 1 ____ f(x) : ( 0, 1 __ 3 ) b) 0 2 x y -6 -4 -2 -2-4 2 4 y = f(x) y = 1 f(x) x = 2, x = –3, y = 0; ( –1 ± √ ___ 29 ________ 2 , 1 ) , ( –1 ± √ ___ 21 ________ 2 , –1 ) ; f(x): (2, 0), (–3, 0), (0, –6); 1 ____ f(x): ( 0, –1 ___ 6 ) 7. a) 0 2 4 x y -4 -2 -2-4 2 4 6 y = f(x) y = 1 f(x) Example: Use the vertical asymptote to fi nd the zero of the function. Then, use the given point to fi nd the equation. y = 2x – 4 b) 0 2 4 6 x y -2 2 4 6 y = f(x) y = 1 f(x) Example: Use the vertical asymptote to fi nd the zero of the function. Then, use an invariant point to fi nd the equation. y = (x – 2)2 PC11_WB_AK_408-434.indd Page Sec1:424 12/27/11 5:59:14 PM u-s035 /Users/u-s035/Desktop/mdghazali_27-12-11/precalculus11/ak 978-0-07-073882-9 Pre-Calculus 11 Student Workbook • MHR 427 x-intercepts: for f(x): 2, –2; for 1 ____ f(x) : none y-intercept: for f(x): –4, for 1 ____ f(x): –1 ___ 4 14. a) 0 2 4 6 y x -4 -6 -2 -2-4 2 4 y = f(x) y = 1 f(x) f (x) 1 ____ f (x) Asymptotes N/A x = – 4 ___ 3 , y = 0 x-intercept – 4 ___ 3 None y-intercept 4 1 __ 4 Invariant points (–1, 1), ( –5 ___ 3 , –1 ) Domain {x | x ∈ R} x | x ≠ – 4 ___ 3 , x ≠ 0, x ∈ R Range {y | y ∈ R} {y | y ≠ 0, y ∈ R} b) 0 2 4 x y -4 -6 -8 -10 -12 -2 -2-4 2 4 y = f(x) y = 1 f(x) f (x) 1 ____ f (x) Asymptotes None x = 4, x = –3, y = 0 x-intercepts 4, –3 None y-intercept –12 –1 ___ 12 Invariant points ( 1 ± √ ___ 53 _______ 2 , 1 ) , ( 1 ± 3 √ __ 5 _______ 2 , –1 ) Domain {x | x ∈ R} {x | x ≠ 0, x ≠ 4, x ≠ –3, x ∈ R} Range {y | y ≥ –12.25, y ∈ R} {y | y ≠ 0, y ∈ R} 15. a) R = P __ I 2 b) 125 ohms c) 5 amperes d) 150 watts Chapter 8 8.1 Solving Systems of Equations Graphically, pages 338–344 1. a) (0, –3) b) (–1, 5) c) (–2, 11) 2. a) quadratic-quadratic; (–2, –2) and (2, –2) b) linear-quadratic; (0, –4) c) quadratic-quadratic; no solution 3. a) (–4, –3) and (2, 3) y 2 20-2-4-6-8 4 6 8 x -2 -4 4 2y - x2 - 4x + 6 = 0 y - x - 1 = 0 (-4, -3) (2, 3) b) ( 1 __ 2, 7 __ 2 ) and (1, 3) y 2 20-2-4-6-8 4 6 8 x -2 -4 4 6 8 y - 2x2 + 4x - 5 = 0 y + x - 4 = 0 1 2 7 2 ⎛ ⎝ ⎛ ⎝ , (1, 3) c) no solution y 20-2-4-6-8 4 6 8 x -2 -4 -6 -8 -10 -12 2 4 y - 4x2 - 7x + 5 = 0 10y - 32x + 90 = 0 PC11_WB_AK_408-434.indd Page Sec1:427 12/27/11 5:59:17 PM u-s035 /Users/u-s035/Desktop/mdghazali_27-12-11/precalculus11/ak 428 MHR • Answer Key 978-0-07-073882-9 4. a) (–1, 7) and (2, –2) 2 20-2-4-6-8 4 6 8 x -2 -4 4 6 8 10 12 14 y y + 4x2 - x = 12 (-1, 7) (2, -2) y - 3x2 + 6x = -2 b) no solution y 2 20-2-4-6-8 4 6 8 x -2 -4 -6 -8 4 6 8 y - 3x2 + 24 = 30 y + 2x2 + 8x = -9 c) (3, 4) y 2 20-2-4 4 6 8 9 x -2 -4 4 6 8 y - x2 + 6x = 13 y + 2x2 - 12x = -14 (3, 4) 5. Example: a) y x y = -2(x + 3)2 + 5 y = -4x - 5 (-2, 3) (0, -5) 0 2 4 6 -2-4-6-8-10 2 -2 -4 -6 4 b) (–2, 3) and (0, –5); y = – 4x – 5 c) (–2, 3) 6. Example: a) 2 20-2-4-6-8 4 6 8 x -2 -4 -6 -8 4 6 y y = 3x - 4 y = x2 - 4 (0, -4) (3, 5) b) y = x2 – 4 c) (0, −4) and (3, 5) 7. a) (–2, 15) and (2, 19) 2 20-2-4-6-8 4 6 8 x -2 8 10 12 14 16 18 20 y y = x2 + x + 13 y = x + 17 (-2, 15) (2, 19) b) (–2, 15) cannot be used because the question states that both numbers are positive. 8. They intersect at (5, 75). After 5 s, they are both 75 m above the ground. H ei gh t (m ) Time (s) y 20 20 4 6 8 10 12 14 16 18 x 40 60 80 100 120 140 160 180 200 y = -5x2 + 50x - 50 y = -5x2 + 100x - 300 (5, 75) PC11_WB_AK_408-434.indd Page Sec1:428 12/27/11 10:00:33 PM u-s035 /Volumes/101/MHR00089_R1/PRE_CALCULUS_11_STUDENT_WORKBOOK/SE/PRE_CALCULUS_11_... 978-0-07-073882-9 Pre-Calculus 11 Student Workbook • MHR 429 9. a) No solution. The parabola opens upward. The y-intercept of the line is below the vertex. b) An infi nite number of solutions. When the fi rst equation is expanded, it is exactly the same as the second equation. c) One solution. The parabolas share the same vertex at (−2, 4). One parabola opens upward, the other downward. d) Two solutions. The vertex of one parabola is directly above the other. One parabola has a smaller vertical stretch factor. 8.2 Solving Systems of Equations Algebraically, pages 350–355 1. a) (2, 2) b) (0, 1) and (1, 2) c) (–2, –1) and (3, 14) d) ( – 1 __ 2, 2 ) and (1, 5) 2. a) (–1, 3) and (3, –37) b) ( – 1 __ 4, 1 __ 4 ) c) (–1, 7) and ( – 10 ___ 3 , – 7 __ 9 ) d) no solution 3. m = 4, n = 16 4. a) x + y = 11 b) x2 – 5x + 6 = 5y c) x = 7, y = 4 d) base = 8 cm, height = 5 cm, and width = 6 cm 5. 11 and 8 6. (1.45, 3.80) and (3.22, –0.91) Chapter 8 Review, pages 356–360 1. a) no solution 1 solution y 0 x y 0 x 2 solutions y 0 x b) no solution 1 solution y 0 x y 0 x 2 solutions y 0 x 2. a) (0, 4) y x 2y + 6x2 + x - 8 = 0 2y + x - 8 = 0 (0, 4) 0 2 4 6 -2-4-6-8 2 -2 -4 -6 4 6 b) (4, 4) and (1, –5) y 2 20-2-4 4 6 8 9 x -2 -4 -6 -8 4 6 8 (4, 4) (1, -5) y - x2 + 2x + 4 = 0 y - 3x + 8 = 0 PC11_WB_AK_408-434.indd Page Sec1:429 12/27/11 10:00:51 PM u-s035 /Volumes/101/MHR00089_R1/PRE_CALCULUS_11_STUDENT_WORKBOOK/SE/PRE_CALCULUS_11_... 432 MHR • Answer Key 978-0-07-073882-9 2. a) 0 2 4 x y -4 -6 -2 -2-4-6 2 4 6 f(x) = x2 + 2x - 3 {x  x ≤ −3 or x ≥ 1, x ∈ R} b) 0 2 4 x y 6 8 10 -2 -2-4-6 2 4 6 8 f(x) = -x2 + 4x + 5 {x  x < −1 or x > 5, x ∈ R} c) 0 2 4 x y 6 8 -2 -2-4-6-8-10 2 f(x) = 2x2 + 10x + 12 {x −3 ≤ x ≤ −2, x ∈ R} d) 0 2 4 x y 6 8 -2 -2 2 4 6 8 f(x) = x2 - 6x + 9 no solution 3. a) {x −2 < x < −1, x ∈ R} b) {x  3 __ 4 < x < 1, x ∈ R} c) {x  x ≤ − 1 __ 3 or x ≥ 1 __ 2, x ∈ R} d) {x  x ≤ − 1 __ 2 or x ≥ 4, x ∈ R} 4. x must be approximately greater than or equal to 7 cm. 5. approximately {x | −3.7 ≤ x ≤ 13.7, x ∈ R} (Note: The number of possible increases is from 0 to 13 since you cannot have a “negative” increase.) The price per ticket could be set at $10 up to $23. 6. approximately {x | 36.8 ≤ x ≤ 163.2, x ∈ R}; possible measures for x are from 36.8 m up to 163.2 m. 7. {x | 10 ≤ x ≤ 20, x ∈ R} 8. a) Example: 2x2 – 7x – 4 < 0 or (2x + 1)(x – 4) < 0 b) Example: 8x2 + 10x – 3 ≥ 0 or (2x + 3)(4x – 1) ≥ 0 c) Example: 3x2 – 4x ≤ 0 or (x)(3x – 4) ≤ 0 9. Answers may vary. 9.3 Quadratic Inequalities in Two Variables, pages 394–400 1. a) (0, 0) and (2, 1) b) (−4, −5) and (−2, 10) c) (−2, −10), (2, −7), and (0, 36) d) (1, 9) and (−1, 17) 2. a) Example: (−5, −5) is a solution; (0, 5) is not a solution b) Example: (5, −5) is a solution; (1, −15) is not a solution c) Example: (4, −7) is a solution; (−1, 20) is not a solution d) Example: (−10, −3) is a solution; (0, 0) is not a solution 3. a) 0 2 4 6 x y -4 -6 -2 -2-4-6-8 2 4 6 8 y > x2 + x - 6 b) 0 2 4 6 x y -4 -2 -2-4-6-8 2 4 6 8 y ≤ x2 - x - 2 c) 0 2 4 6 8 x y -2-4-6-8 2 4 y ≥ 2x2 + 9x + 10 PC11_WB_AK_408-434.indd Page Sec1:432 12/27/11 5:32:17 PM u-s035 /Volumes/101/MHR00089_R1/PRE_CALCULUS_11_STUDENT_WORKBOOK/SE/PRE_CALCULUS_11_... 978-0-07-073882-9 Pre-Calculus 11 Student Workbook • MHR 433 d) 0 2 4 6 8 x y -2 -2-4-6-8 2 4 6 8 y < 6x2 - x - 1 4. a) 0 2 4 6 8 x y -2 -2-4-6-8 2 4 6 8 y > (x - 1)2 + 1 b) 0 2 4 x y -4 -6 -8 -2 -2-4-6-8 2 4 6 8 y ≤ (x + 2)2 -9 c) 0 2 4 6 8 x y 10 12 -2 -2-4-6 2 4 6 8 y ≥ 2(x - 3)2 + 8 d) 0 2 4 6 8 x y 10 12 -2 -2-4-6-8 2 4 6 8 y < (x - 1)2 + 2 5. a) y > 2x2 – 4x – 1 b) y ≤ −3x2 – 4x – 1 c) y > −4x2 + 5x + 3 d) y ≤ 2x2 – 3x + 2 6. a) y ≥ −2(x – 4)2 + 8 b) y < 1.5(x + 3)2 – 6 c) y > 0.25(x + 3)2 – 4 d) y ≤ −0.5(x + 4)2 + 2 7. a) 0 100 200 300 400 x y 500 600 700 800 5 10 15 20 25 30 To ta l R ev en ue ($ ) Number of Additional Campers (17, 512) y ≤ (50 - 2x)(15 + x) b) from 0 to 17 additional campers; 15 to 32 campers in total 8. Example: To solve a system of quadratic equations graphically, graph each equation and determine the point(s) of intersection. This system would have two solutions, as there are two points where the graphs intersect. 0 2 4 6 8 x y -2 -2-4-6 2 4 6 y = 2x2 - x + 2 y = -2x2 - x + 6 Solving a system of quadratic inequalities means determining the intersection; the solution will be an area of intersection, not a point (or points) of intersection. Using the same functions, but changing them to inequalities, the solution might look like this: 0 2 4 6 8 x y -2 -2-4-6 2 4 6 PC11_WB_AK_408-434.indd Page Sec1:433 12/27/11 5:32:21 PM u-s035 /Volumes/101/MHR00089_R1/PRE_CALCULUS_11_STUDENT_WORKBOOK/SE/PRE_CALCULUS_11_... 434 MHR • Answer Key 978-0-07-073882-9 9. 0 2 4 6 8 x y -4 -6 -8 -2 -2-4-6-8 2 4 6 8 y ≤ -x2 - 4x + 4 y ≥ x2 - 4x - 4 10. Answers may vary. Chapter 9 Review, pages 401–406 1. a) (0, 7), (5, −3) b) (7, 0) (−3, −4) (5, −1) 2. a) 0 2 x y -4 -6 -2 -2-4-6-8 2 4 6 8 2x - 3y ≥ 12 b) 0 2 4 x y -4 -2 -2-4-6-8 2 4 6 8 - 5x - y < 0 3. a) y > − 3 __ 4x − 2 b) y ≤ 4 __ 5x + 3 4. 0 20 40 60 x y 80 100 20 40 Hours of Part-time Work H ou rs B ab ys it ti ng 60 80 (50, 40) 15x + 10y ≥ 1000 Possible solution: (50, 40); Amber could work 50 h at her part-time job and babysit for a total of 40 h. This would help her save $1150. 5. 0 2 4 6 8 x y -2 -2-4-6 2 4 6 8 y = -2x2 + 3x + 7 approximately {x | −1.3 < x < 2.8, x ∈ R} 6. a) {x | 0 < x < 1, x ∈ R} b) {x | x < 0 or x > 1, x ∈ R} c) {x | x ≤ 0 or x ≥ 1, x ∈ R} d) {x | 0 ≤ x ≤ 1, x ∈ R} 7. {x | −3 ≤ x ≤ 4, x ∈ R} 8. a) (1, 6), (6, 12) b) (−1, −2), (3, 4) 9. a) 0 2 x y -4 -6 -8 -2 -2-4-6-8 2 4 6 8 y > -3x2 -3x + 1 b) 0 2 x y -4 -6 -8 -10 -12 -2 -2-4-6-8-10 2 4 y ≤ 0.5x2 + 4x - 3 10. a) y < (x + 1)2 – 4 b) y ≤ – (x – 1)2 + 9 PC11_WB_AK_408-434.indd Page Sec1:434 12/27/11 5:32:24 PM u-s035 /Volumes/101/MHR00089_R1/PRE_CALCULUS_11_STUDENT_WORKBOOK/SE/PRE_CALCULUS_11_...