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Chapter 9 Rational FunctionsSection 9.1 Exploring Rational Functions Using TransformationsSection 9.1Page 442Question 1a k .x ha) Since the graph of the rational function has a vertical asymptote at x –1 and ahorizontal asymptote at y 0, h –1 and k 0. Then, the function is of the form2ay , which is B(x) .x 1x 1Compare each graph to the function form y b) Since the graph of the rational function has a vertical asymptote at x 0 and ahorizontal asymptote at y –1, h 0 and k –1. Then, the function is of the forma2y 1 , which is A(x) 1 .xxc) Since the graph of the rational function has a vertical asymptote at x 0 and ahorizontal asymptote at y 1, h 0 and k 1. Then, the function is of the forma2y 1 , which is D(x) 1 .xxd) Since the graph of the rational function has a vertical asymptote at x 1 and ahorizontal asymptote at y 0, h 1 and k 0. Then, the function is of the forma2y , which is C(x) .x 1x 1Section 9.1Page 442a) The base function for y Question 211is y . Compare thex 2xa k : a 1, h –2, and k 0.x hThe graph of the base function must be translated 2 units to theleft. So, the vertical asymptote is at x –2 and the horizontalasymptote is still at y 0.function to the form y b) The base function for y 11is y . Compare thex 3xa k : a 1, h 3, and k 0.x hThe graph of the base function must be translated 3 units tothe right. So, the vertical asymptote is at x 3 and thehorizontal asymptote is still at y 0.function to the form y MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 1 of 60

c) The base function for y 11is y 2 . Compare the2( x 1)xa k : a 1, h –1, and k 0.( x h) 2The graph of the base function must be translated 1 unit to theleft. So, the vertical asymptote is at x –1 and the horizontalasymptote is still at y 0.function to the form y 11is y 2 . Compare2( x 4)xa k : a 1, h 4, andthe function to the form y ( x h) 2k 0. The graph of the base function must be translated4 units to the right. So, the vertical asymptote is at x 4 andthe horizontal asymptote is still at y 0.d) The base function for y Section 9.1Page 442Question 36, a 6, h –1, and k 0. The graph of thex 11base function y must be stretched vertically by a factor ofx6 and translated 1 unit to the left. The domain is{x x –1, x R} and the range is {y y 0, y R}. There isno x-intercept. The y-intercept is 6. The vertical asymptote isat x –1 and the horizontal asymptote is at y 0.a) For y 4 1 , a 4, h 0, and k 1. The graph of thex1base function y must be stretched vertically by a factor ofx4 and translated 1 unit up. The domain is {x x 0, x R}and the range is {y y 1, y R}. The x-intercept is –4. Thereis no y-intercept. The vertical asymptote is at x 0 and thehorizontal asymptote is at y 1.b) For y MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 2 of 60

2 5 , a 2, h 4, and k –5. The graph of thex 41base function y must be stretched vertically by a factor ofx2 and translated 4 units to the right and 5 units down. Thedomain is {x x 4, x R} and the range is22, or 4.4. The y{y y –5, y R}. The x-intercept is511intercept is – , or –5.5. The vertical asymptote is at x 4 and the horizontal asymptote2is at y –5.c) For y 8 3 , a –8, h 2, and k 3. The graph ofx 21the base function y must be stretched vertically by axfactor of 8, reflected in the x-axis, and translated 2 units to theright and 3 units up. The domain is {x x 2, x R} and the14. Therange is {y y 3, y R}. The x-intercept is3y-intercept is 7. The vertical asymptote is at x 2 and thehorizontal asymptote is at y 3.d) For y Section 9.1Page 442Question 42x 1, the vertical asymptote is at x 4, thex 4horizontal asymptote is at y 2, the x-intercept is –0.5,and the y-intercept is –0.25.a) For y 3x 2, the vertical asymptote is at x –1,x 1the horizontal asymptote is at y 3, the x-intercept isabout 0.67, and the y-intercept is –2.b) For y MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 3 of 60

4 x 3, the vertical asymptote is at x –2,x 2the horizontal asymptote is at y –4, the x-intercept is0.75, and the y-intercept is 1.5.c) For y 2 6x, the vertical asymptote is at x 5, thex 5horizontal asymptote is at y –6, the x-intercept is about0.33, and the y-intercept is –0.4.d) For y Section 9.1Page 442Question 511x 12x12 11 x12 11x12For y 11 , a 12, h 0, and k 11. The vertical asymptote is at x 0, and thexhorizontal asymptote is at y 11.Substitute y 0.12y 11x120 11x12–11 x12x 1112The x-intercept is , or about –1.09, and there is no11y-intercept.a) y MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 4 of 60

xx 8x 8 8 x 8x 88 x 8 x 88 1 x 88 1x 88For y 1 , a –8, h –8, and k 1. The vertical asymptote is at x –8, and thex 8horizontal asymptote is at y 1.Substitute y 0.Substitute x 0.88y y 1 1x 8x 8880 1 10 8x 8 –1 18 1 0x 8x 0The x-intercept is 0, and the y-intercept is 0.b) y x 2x 6 x 6 6 2 x 6c) y ( x 6) 4x 64 ( x 6) x 6x 64 1 x 6 MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 5 of 60

4 1 , a 4, h –6, and k –1. The vertical asymptote is at x –6, and thex 6horizontal asymptote is at y –1.Substitute y 0.Substitute x 0.44y y 1 1x 6x 6440 1 10 6x 6421 –13x 6x 6 41 x –231The x-intercept is –2, and the y-intercept is , or about –0.33.3For y Section 9.1Page 442CharacteristicNon-permissiblevalueBehaviour near nonpermissible valueEnd behaviourQuestion 6f ( x) 1x2x 0As x approaches 0, y becomes verylarge.As x becomesvery large, yapproaches 0.g( x ) 8( x 6)2x –6As x approaches–6, y becomesvery large.As x becomesvery large, yapproaches 0.MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9h( x ) 4 3x 4x 42x 2As x approaches 2, y becomes very large.As x becomes verylarge, y approaches–3.Page 6 of 60

Domain{x x 2, x R}{x x 0, x R} {x x –6, x R}Range{y y 0, y R}{y y –3, y R}{y y 0, y R}Equation of verticalx 0x –6x 2asymptoteEquation ofy 0y 0y –3horizontal asymptoteEach function has a single non-permissible value, a vertical asymptote, and a horizontalasymptote. The domain of each function consists of all real numbers except for a singlevalue. The range of each function consists of a restricted set of the real numbers. y becomes very large for each function when the values of x approach the non-permissiblevalue for the function.Section 9.1Page 443Question 7a) From the graph, the vertical asymptote is at x 0 andthe horizontal asymptote is at y 0. So, h 0 and k 0.aThen, the equation of the function is of the form y .xUse one of the given points, say (2, –2), to determine thevalue of a.a–2 2a –4aThe equation of the function in the form y k is yx h4 .xb) From the graph, the vertical asymptote is at x –3 andthe horizontal asymptote is at y 0. So, h –3 and k 0.Then, the equation of the function is of the formay .x 3Use one of the given points, say (–2, 1), to determine thevalue of a.a1 2 3a 1a1The equation of the function in the form y . k is y x hx 3MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 7 of 60

c) From the graph, the vertical asymptote is at x 2 andthe horizontal asymptote is at y 4. So, h 2 and k 4.Then, the equation of the function is of the formay 4.x 2Use one of the given points, say (4, 8), to determine thevalue of a.a8 44 2a4 2a 8a8The equation of the function in the form y k is y 4.x hx 2d) From the graph, the vertical asymptote is at x 1 andthe horizontal asymptote is at y –6. So, h 1 andk –6. Then, the equation of the function is of the formay 6.x 1Use one of the given points, say (0, –2), to determine thevalue of a.a–2 60 14 –aa –4a4The equation of the function in the form y k is y 6.x hx 1Section 9.1Page 443Question 8a k passing through points (10, 1) and (2, 9)x 7For (10, 1),For (2, 9),aay ky kx 7x 7aa1 k9 k10 72 7aa1 k9 k3 53 a 3k –45 a – 5k Solve the system of equations.a) Given: y MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 8 of 60

3 a 3k–45 a – 5k48 8k – k 6Substitute k 6 into .3 a 3k3 a 3(6)a –15The equation of the function is y 15 6.x 7b)Section 9.1Page 443Question 9Examples:a) For asymptotes at x 2 and y –3, h 2 and k –3. Choose a 1, then the equationa1of the function in the form y k is y 3.x hx 2p ( x)Then, rewrite in the form y :q( x)1y 3x 213( x 2) x 2x 21 3x 2 x 23 3x x 2b)The domain is {x x 2, x R} and the range is{y y –3, y R}.MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 9 of 60

c) There are many possible functions that meet the given criteria, since any value of a(other than 0) will result in the same equations for the asymptotes.Section 9.1Page 443Question 10a) In the fourth line, Mira incorrectly factored –3 from –3x – 21. She should havegrouped –3x 21. The corrected solution is2 3xy x 7 3x 2y x 7 3x 21 21 2y x 7 3( x 7) 19y x 7 3( x 7) 19y x 7x 719y 3 x 719y 3x 7b) Example: Without technology, Mira could have discovered her error by substitutingthe same value of x into each form of the function. With technology, Mira could havegraphed the two functions to see if they were the same.Section 9.1Page 443Question 11a)x 22x 4x 2y 2( x 2)x 2 4y 2( x 2)x 24y 2( x 2) 2( x 2)12y 2 x 221y x 2 2y MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 10 of 60

211 , a –2, h –2, and k . Thex 2 221graph of the base function y must be stretchedxvertically by a factor of 2, reflected in the x-axis, and1translated 2 units to the left and unit up.2b) For y Section 9.1Page 443Question 12Determine the intercepts.Substitute x 0.Substitute y 0.3x 53x 5y y 2x 32x 33x 53(0) 50 2(0) 32x 30 3x – 55 53x 355The x-intercept is , and the y-intercept is .33Use technology to graph the function.The asymptotes are located at x –Section 9.1Page 44333and y .22Question 13500 000, as the value of p increases, the value of N decreases.pThis means that as the average price of a home increases, the number of buyers lookingto buy a home decreases.For the function N(p) MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 11 of 60

Section 9.1Page 444Question 14a) For a rectangle with constant area of 24 cm2,A lw24 lw24l wb) As the width increases, the length decreases tomaintain an area of 24 cm2.Section 9.1Page 444Question 15a) Let x represent the number of students who contribute. Let y represent the averageamount required per student to meet the goal. Then, a function to model this situation is4000xy 4000, or y .xb)c) As the number of students that contribute increases, the average amount required bystudent decreases.d) If the student council also received a 1000 donation from a local business, the4000function becomes y 1000. This represents a vertical translation of 1000 unitsxup of the original function graph.Section 9.1Page 444Question 16a) Let C represent the average cost per year. Let t represent the time, in years. Then, a500 100tfunction to model freezer one is C and a function to model freezer two ist800 60tC .tMHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 12 of 60

b)c) Both graphs have a vertical asymptote at x 0, but different horizontal asymptotes.The average cost for freezer one approaches 100/year, while the average cost for freezertwo approaches 60/year. The graph shows that the more years you run the freezer, theless the average cost per year is. Freezer one is cheaper to run for a short amount of time,while freezer two is cheaper if you run it for a longer period of time.d) The point of intersection of the two graphs can help Hanna decide which model tochoose. If Hanna wants to run the freezer for more than 7.5 years, she should choose thesecond model. Otherwise, she is better off with the first one.Section 9.1Page 444Question 17a) An equation for the current in the given circuit is I 12.15 xb) Since the variable resistor can be set anywhere from 0 Ω to 100 Ω, an appropriatedomain is {x 0 x 100, x R}. The graph does not have a vertical asymptote for thisdomain.c) Substitute I 0.2.12I 15 x120.2 15 x0.2(15 x) 1215 x 60x 45A setting of 45 Ω is required.12. The verticalxasymptote is at x 0, so the domain must change to {x 0 x 100, x R}.d) An equation for the current in the circuit without the bulb is I MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 13 of 60

Section 9.1Page 445Question 18a) Let C represent the average cost per hour. Let t represent the rental time, in hours.20 4tThen, a function to model renting from store one is C and a function to modelt10 5trenting from store two is C .tb) Both graphs have a vertical asymptote at x 0, but different horizontal asymptotes.The average cost for renting from store one approaches 4/h, while the average cost forrenting from store two approaches 5/h. The graph shows that the longer you rent thebike, the less the average cost per hour is. Store two is cheaper to rent for a short amountof time, while store one is cheaper if you rent for a longer period of time.c) No. The point of intersection of the two graphs indicates that if you rent a bike for lessthan 10 h, then you should choose store two. Otherwise, choose store one.Section 9.1Page 445Question 19a) Let v represent the average speed, in kilometres per hour, over the entire tripand t represent the time, in hours, since leaving the construction zone. Then, an80 100tequation for v as a function of t for this situation is v .2 tb) An appropriate domain for this situation is {t t 0, t R}.c) The equation of the vertical asymptote is t –2 and the equation of the horizontalasymptote is v 100. The vertical asymptote does not mean anything in this context,since time cannot be negative. The horizontal asymptote means that the average speedgets closer and closer to 100 km/h but never reaches it.MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 14 of 60

d) Substitute v 80.80 100tv 2 t80 100t80 2 t80(2 t) 80 100t160 80t 80 100t–20t –80t 4The truck will have to drive 4 h after leaving the construction zone before its averagespeed is 80 km/h.e) Example: By including an application on a GPS unit that calculates the average speedover the entire trip, a driver could adjust his/her speed accordingly to maintain a targetspeed that provides the best fuel economy for his/her vehicle.Section 9.1Page 445Question 20Given: vertical asymptote at x 6, horizontal asymptote at y –4, and x-interceptof –1aFor a function of the form y k , h 6 and k –4. Then, the equation becomesx hay 4 . Use the point (–1, 0) to determine the value of a.x 6a 4 1 6a4 7a –280 Rewrite the function y 28ax b 4 in the form y .cx dx 628 4x 628 4( x 6) x 6x 6 28 4 x 24 x 6 4 x 4 x 6y MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 15 of 60

Section 9.1Page 445Question 21b)a)x 3x 1x 3y x 1y 3x y 1x( y 1) y 3xy x y 3xy y x 32x 4x 52x 4y x 52y 4x y 5( x 4)( y 5) 2 yxy 4 y 5 x 20 2 yxy 6 y 5 x 20y ( x 6) 5 x 205 x 20y x 65 x 20f 1 ( x) x 6f ( x) f ( x) y ( x 1) x 3 x 3y x 1 x 3f 1 ( x) x 1Section 9.1Page 445Question 22Use graphing technology to graph y xx 4. x 2 x 2The graph has three branches separated by vertical asymptotes at x –2 and x 2. It alsoappears to have a horizontal asymptote at y 2 for x –2 and x 2.Section 9.1Page 445Question C1a k , thenx husing transformations is no different than with any other function. However, if theequation of the function is not in this form, it is more difficult to manipulate the equationto determine the transformations that have been applied.Example: If the equation of the rational function is given in the form y MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 16 of 60

Section 9.1Page 445Question C2200 000 p, an appropriate domain is100 p{p 0 p 100, p R}. The function is not defined at p 100, meaning that 100% ofthe emissions can never be eliminated.a) For the function C(p) b) The shape of the graph indicates that as the percentof emissions eliminated increases, so does the cost.c) For p 80,For p 40,200 000 p200 000 pC ( p) C ( p) 100 p100 p200 000(80)200 000(40)C (80) C (40) 100 80100 40C (80) 800 000C (40) 133 333.33It costs 6 times as much to eliminate 80% as it does to eliminate 40%.d) Is it not possible to completely eliminate all of the emissions according to thismodel because the graph of the function has a vertical asymptote at p 100.Section 9.1Page 445Question C32 4 , a 2, h 3, and k 4. For y 2 x 3 4 , a 2, h 3, and k 4.x 3Example: Both functions are vertically stretched by a factor of 2 and translated 3 unitsright and 4 units up. In the case of the rational function, the values of the parameters hand k represent the locations of asymptotes. For the square root function, the point (h, k)gives the location of the endpoint of the graph.For y MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 17 of 60

Section 9.2 Analysing Rational FunctionsSection 9.2Page 451Question 1a)x 4x 4. There is a vertical asymptote atis y x 6x 8( x 4)( x 2)x 2 because x – 2 is factor of the denominator only. There is a point of discontinuity at(4, 0.5) because x – 4 is a factor of both the numerator and the denominator.b) In factored form, y Section 9.2Page 4512Question 2Examples:x 2 3xx( x 3)a) In factored form, y . The non-permissible value is x 0.is y xxSince the function does not increase or decrease drastically as xapproaches the non-permissible value, it must be a point ofdiscontinuity.x 2 3x 10( x 5)( x 2)is y . The non-permissible value isb) In factored form, y x 2x 2x 2.Since the function changes sign at the non-permissible value and y increases, it must be a vertical asymptote.MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 18 of 60

c) In factored form, y 3x 2 4 x 4(3 x 2)( x 2)is y . The non-permissible value isx 4x 4x –4.Since the function changes sign at the non-permissible value and y increases, it must be a vertical asymptote.d) In factored form, y 5x2 4 x 1(5 x 1)( x 1)is y . The non-permissible value is5x 15x 1x 0.2.Since the function does not increase or decrease drastically as xapproaches the non-permissible value, it must be a point ofdiscontinuity.Section 9.2Page 451Question 3a)Both of the functions have a non-permissible value of –3. However, the graph of f(x) hasa vertical asymptote, while the graph of g(x) has a point of discontinuity.( x 3)( x 1)x2 2 x 3b) In factored form, f(x) is f(x) . In factored formx 3x 3( x 3)( x 1)x2 2x 3is g(x) .g(x) x 3x 3The graph of f(x) has a vertical asymptote at x –3, since x 3 is a factor of only thedenominator.Use the simplified form, g(x) x – 1, x –3, to determine the coordinates of the point ofdiscontinuity. Substituting x –3, the graph of g(x) has a point of discontinuity at(–3, –4). This is because x 3 is a factor of both the numerator and the denominator.MHR 978-0-07-0738850 Pre-Calculus 12 Solutions Chapter 9Page 19 of 60

Section 9.2Page 452Question 4x2 4 xa) y 2x 9 x 20x( x 4) ( x 4)( x 5)x, x 4, 5 x 5The graph will have a vertical asymptote at x –5. Substituting x –4, the graph willhave a point of discontinuity at (–4, –4). By substituting y 0 and x 0, the graph willhave an x-intercept of 0 and a y-intercept o