Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level by Tim Garry, Ibrahim Wazir, Kevin Frederick, Bryan Landmann, Jim Nakamoto, John Whalley (2019)

Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level by Tim Garry, Ibrahim Wazir, Kevin Frederick, Bryan Landmann, Jim Nakamoto, John Whalley (2019) Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level by Tim Garry, Ibrahim Wazir, Kevin Frederick, Bryan Landmann, Jim Nakamoto, John Whalley (2019) Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level by Tim Garry, Ibrahim Wazir, Kevin Frederick, Bryan Landmann, Jim Nakamoto, John Whalley (2019) Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level by Tim Garry, Ibrahim Wazir, Kevin Frederick, Bryan Landmann, Jim Nakamoto, John Whalley (2019) Preview Mathematics Analysis and Approaches for the IB Diploma Higher Level by Tim Garry, Ibrahim Wazir, Kevin Frederick, Bryan Landmann, Jim Nakamoto, John Whalley (2019)

L AN \ \ Y ¥‘ ? \\\ \ ¥ \ A Y \ ) LR - W\ M N ¥ eBook BN Mathematic Analysis and Approaches for the IB Diploma Pearson L o IBRAHIM WAZIR Mathematics Analysis and Approaches for the IB Diploma IBRAHIM WAZIR R Ner Published by Pearson Education Limited, 80 Strand, London, WC2R ORL wwwpearsonglobalschools.com Text © Pearson Education Limited 2019 Development edited by Jim Newall Copy edited by Linnet Bruce Proofread by Eric Pradel and Martin Payne Indexed by Georgie Bowden Designed by © Pearson Education Limited 2019 Typeset by © Tech-Set Ltd, Gateshead, UK Original illustrations © Pearson Education d 2019 Tlustrated by © Tech-Set Ltd, Gateshead, UK Cover design by © Pearson Education Limited 2019 ‘over images: Front: © Getty Images: Busi Photography Inside front cover: Shutterstock.com: Dmitry Lobanov “The rights of Ibrahim Wazir and Tim Garry to be identified as the authors of this work have been asscrtcd by them in accordance with the Copyright, Designs and Patents Act 1988 First published 2019 242322212019 IMP10987654321 British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library ISBN 978043519342 Copyright notice Al rights reserved No part o this publication may be reproduced in any form or by any means (including photocopying or storing it in any medium by clectronic means and whether or no transiently or incidentally to some other use of this publication) without the written permission of the copyright owner, exceptin accordance with the provisions of the Copyright, Designs and Patents Act 1988 or under the terms ofa licence issued by the Copyright Licensing Agency, Barnard's Inn, 86 Fetter Lane, London, EC4A 1EN (www.cla.co.uk) Applications for the copyright owner's written permission should be addresse to the publisher Printed in Slovakia by Neografia Acknowledgements The authors and publisher would like to thank the following individuals and organisations for their kind permission to reproduce copyright material Photographs < b-bottom; e-centre; left; r-right; t-top) Getty Images: JPL[Moment/Getty Images 1, baxsyl[Moment/Getty Images 57, d3sign/Moment/Getty Images 107, Paul Biris/Moment/Getty Images 161, Image Source/DittofGetty Images 203, Westend61 [Getty Images 235, Alberto Manuel Urosa Toledano|Moment/Getty Images 301, Jamil Caram/EyeEmGetty Images 345, Chris eberhardt/Getty Images 379, ] Broughton Photography/Moment/Getty Images 449, Howard Pugh (Marais) Moment|Getty Images 519, aaaaimages| Moment/Getty Images 573, Ryan Trussler/Getty Images 627, Johner Images|Getty Images 697, Gabricl Perez|Moment/Getty Images 765, Witthaya Prasongsin] Moment/Getty Images 835 Piet Mondrian: Figure Composition with Red, Blue and Yellow (1926) Piet Mondrian Al other images © Pearson Education ‘We are grateful to the following for permission to reproduce copyright material: Text pages 912-913, Edge Foundation Inc.: What Kind of Thing Is a Number? ATalk with Reuben Hersh, Wed, Oct 24, 2018, Used with permission of Edge Foundation Inc Text extracts elating to the IB syllabus and assessment have been reproduced from 1BO documents Our thanks go to the International Baccalaureate for permission to reproduice ts copyright “This work has been developed independently from and is not endorsed by the International Baccalaurcate (IB) International Baccalaurcate® is a registered trademark of the International Bacealaureate Organization “This work is produced by Pearson Education and is not endorsed by any trademark owner referenced in this publication Dedications dedicate this work the memory of my parenis and my brother, Saced, who passed avay during the arly stages of work on thisedition My specalthanks go t my wife, Lody, for standing besde me thrughout witing his book She has ben my inspiraton and motivation for continuing to improve my knowledge and move my career forward She i my rock, and | dedicate this book 10 hr My thanks go 1o al the students and eachers who used the carlir editons and sent s ther comments Ibrahim Wazir Inloving memory of my parens Iyvish express my deepest thanks and ove to my wif, Val, for her unflappable good nature and support and for swiling and laughing with me each day I an infniely thankfulfor ourwonderfuland kind-hearted children Bethany, Neil and Rhona My lov foryou ll s immeasurable Tim Garry Contents Introduction BEEEEEEEOOEOOOEDNDE Algebra and function basics Functions Sequences and series 57 107 Exponential and logarithmic functions 161 Proofs 203 Trigonometric functions and equations 235 Geometry and trigonometry 301 Complex numbers 345 Vectors, lines, and planes 379 Statistics 449 Probability 519 Differential calculus 573 Differential calculus 627 Integral calculus 697 Probability distributions 765 Integral calculus 835 Internal assessment 895 Theory of knowledge 902 Answers 926 Index 996 Introduction 1IB Mathematics: Analysis and Approaches Higher Level syllabus topics Number and Algebra Functions Geometry and This textbook comprehensively covers all of the material in the syllabus for the twoyear Mathematics: Analysis and Approaches Higher Level course of the International Baccalaureate (IB) Diploma Programme (DP) First teaching of this course starts in the autumn of 2019 with first exams occurring in May 2021 We, the authors, have strived to thoroughly explain and demonstrate the mathematical concepts and methods listed in the course syllabus ‘Trigonometry Statistics and Probability Calculus As you will see when you look at the table of contents, the five syllabus topics (see margin note) are fully covered, though some are split over different chapters in order to group the information as logically as possible This textbook has been designed so that the chapters proceed in a manner that supports effective learning of the course content Thus — although not essential — it is recommended that you read and study the chapters in numerical order It is particularly important that you thoroughly review and understand all of the content in the first chapter, Algebra and function basics, before studying any of the other chapters Other than the final two chapters (Theory of knowledge and Internal assessment), each chapter has a set of exercises at the end of every section Also, at the end of each chapter there is a set of practice questions, which are designed to expose you to questions that are more ‘exam-like’ Many of the end-of-chapter practice questions are taken from past IB exam papers Near the end of the book, you will find answers to all of the exercises and practice questions There are also numerous worked examples throughout the book, showing you how to apply the concepts and skills you are studying The Internal assessment chapter provides thorough information and advice on the required mathematical exploration component Your teacher will advise you on the timeline for completing your exploration and will provide critical support during the process of choosing your topic and writing the draft and final versions of your exploration The final chapter in the book will support your involvement in the Theory of knowledge course It is a thought-provoking chapter that will stimulate you to think more deeply and critically about the nature of knowledge in mathematics and the relationship between mathematics and other areas of knowledge eBook Included with this textbook is an eBook that contains a digital copy of the textbook and additional high-quality enrichment materials to promote your understanding of a wide range of concepts and skills encountered throughout the course These materials include: * Interactive GeoGebra applets demonstrating key concepts * Worked solutions for all exercises and practice questions * Graphical display calculator (GDC) support To access the eBook, please follow the instructions located on the inside cover Information boxes As you read this textbook, you will encounter numerous boxes of different colours containing a wide range of helpful information Learning objectives You will find learning objectives at the start of each chapter They set out the content and aspects of learning covered in the chapter Learning objectives By the end of this chapter, you should be familiar with « different forms of equations of lines and their gradients and intercepts « parallel and perpendicular lines « different methods to solve a system of linear equations (maximum of three equations in three unknowns) Key facts Afunction is one-to-one highlighted for quick reference to help you it identify clear learning points '3"8:1 is the i'lmEC of exactlyi, one element x in Key facts are drawn from the main text and Hints Specific hints can be found alongside explanations, questions, exercises, and worked examples, providing insight into how to pies, p nsig’ analyse[answer a question They also identify common errors and pitfalls Notes Notes include general information or advice if each element y in the Ifyou use a graph to answer a question on an 1B mathematics exam, Joimsinchidea deac and well-labelled sketch in your working Quadratic equations will be covered in detail in Chapter Examples ‘Worked examples show you how to tackle questions and apply the concepts and skills you are studying Find x such that the distance between points (1, 2) and (x, —10) is 13 units Solution d=13=\x— D2+ (=10 — 22 = 13> = (x — 12 + (—12) =692l D RRIEEN 44— 20D =10 S x—6)(xt4)=0=>x—6=0o0rx+4=0 =x=6o0rx=—4 How to use this book This book is designed to be read by you — the student It is very important that you read this book carefully We have strived to write a readable book — and we hope that your teacher will routinely give you reading assignments from this textbook, thus giving you valuable time for productive explanations and discussions in the classroom Developing your ability to read and understand mathematical explanations will prove to be valuable to your long-term intellectual development, while also helping you to comprehend mathematical ideas and acquire vital skills to be successful in the Analysis and Approaches HL course Your goal should be understanding, not just remembering You should always read a chapter section thoroughly before attempting any of the exercises at the end of the section Our aim is to support genuine inquiry into mathematical concepts while maintaining a coherent and engaging approach We have included material to help you gain insight into appropriate and wise use of your GDC and an appreciation of the importance of proof as an essential skill in mathematics We endeavoured to write clear and thorough explanations supported by suitable worked examples, with the overall goal of presenting sound mathematics with sufficient rigour and detail at a level appropriate for a student of HL mathematics For over 10 years, we have been writing successful courses During that time, we have received many teachers and students If you have suggestions for free to write to us at globalschools@pearson.com mathematical endeavours Ibrahim Wazir and Tim Garry Vi textbooks for IB mathematics useful comments from both improving this textbook, please feel We wish you all the best in your Algebraand function basics l | \ Algebra and function basics Learning objectives By the end of this chapter, you should be familiar with different forms of equations of lines and their gradients and intercepts parallel and perpendicular lines different methods to solve a system of linear equations (maximum of three equations in three unknowns) the concept of a function and its domain, range and graph mathematical notation for functions composite functions characteristics of an inverse function and finding the inverse function f~'(x) self-inverse functions transformations of graphs and composite transformations of graphs the graphs of the functionsy = | f(x)|,y = f(|x|) andy =% & Equations and formulae Equations, identities and formulae You will encounter a wide variety of equations in this course Essentially, an equation is a statement equating two algebraic expressions that may be true or false depending upon the value(s) substituted for the variable(s) Values of the variables that make the equation true are called solutions or roots of the equation All of the solutions to an equation comprise the solution set of the equation An equation that is true for all possible values of the variable is called an identity Many equations are often referred to as a formula (plural: formulae) and typically contain more than one variable and, often, other symbols that represent specific constants or parameters (constants that may change in value but not alter the properties of the expression) Formulae with which you are familiar include: A = mr%,d = rt,d = [%, — )7 ¥ (7, — y)? and V = %m’} ‘Whereas most equations that we encounter will have numerical solutions, we can solve a formula for one variable in terms of other variables - often referred to as changing the subject of a formula (a) Solve for b in the formula ađ + b* =  (b) Solve for I in the formula T = 217‘/% (c) Solve for R in the formula M = nR (REEr Solution (a)fa2 b2 =c¥= b2 = c2=a2 =1h = =y c:=T72 If bis alength then b = Vc? — a2 I T N I s (b)szfl‘/;i‘/;’znig’zw#liw (0 I= LB TR+ p)l= nR = IR+ Ir = nR Bapip =IR—nR=—-Ir=RI—n)=—1Ir = R = Ir ) Note that factorisation was required in solving for R in part (c) Equations and grap Two important characteristics of any equation are the number of variables (unknowns) and the type of algebraic expressions it contains (e.g polynomials, rational expressions, trigonometric, exponential) Nearly all of the equations in this course will have either one or two variables In this chapter we will only discuss equations with algebraic expressions that are polynomials Solutions for equations with a single variable consist of individual numbers that can be graphed as points on a number line The graph of an equation is a visual representation of the equation’s solution set For example, the solution set of the one-variable equation containing quadratic and linear polynomials x2=2x + 8is x € {—2, 4} The graph of this one-variable equation (Figure 1.1) is depicted on a one-dimensional coordinate system, i.e the real number line ~& =¥ -1 =k & v Figure 1.1 Graph of the solution set for the equationx* = 2x + The solution set of a two-variable equation will be an ordered pair of numbers An ordered pair corresponds to a location indicated by a point on a two-dimensional coordinate system, i.e a coordinate plane For example, the solution set of the two-variable quadratic equation y = x? will be an infinite set of ordered pairs (x, y) that satisfy the equation Four ordered pairs in the solution set are shown in red in Figure 1.2 The graph of all the ordered pairs in the solution set forms a curve as shown in blue A one-variable linear equation in x can always be written in the form ax + b = 0, with a # 0, and it will have exactly one solution, namely x = *—2— An example Figure 1.2 Graph of the ;"i‘:fi’“ setof the equation Quadratic equations wil be covered in detail in Chapter2, The polynomial 6x* + 7x + ax + b has a remainder of 72 when divided by (x — 2) and is exactly divisible (i.e remainder is zero) by (x + 1) (a) Find the values of@ and b (b) Show that (2x — 1) is also a factor of the polynomial and, hence, find the third factor The polynomial p(x) = (ax + b)* leaves a remainder of —1 when divided by (x + 1), and a remainder of 27 when divided by (x — 2) Find the values of the real numbers a and b (x — 2) and (x + 2) are factors of x* + ax? + bx + ¢, and the remainder is 10 when divided by (x — 3) Find the values of 4, b and c When divided by (x + 2), the expression 5x* — 3x2 + ax + leaves a remainder of R When the expression 4x* + ax? + 7x — is divided by (x + 2) there is a remainder of 2R Find the value of the constant a Given that the roots of the equation x* — 19x> + bx — 216 = are consecutive terms in a geometric sequence, find the value of b and solve the equation (a) Prove that when a polynomial P(x) is divided by (ax — b) the remainder is P(%) (b) Hence find the remainder when 9x* — x + is divided by (3x + 2) Find the sum and product of the roots of the following equations () x4—§x3+3x2—2x+5:0 (b) x—2P3=x*-1 © 30 oxr—x Bk 2D C R If @, B and v are the three roots of the cubic equation ax® + bx® + cx + d = 0, show that a + ay + By =& One of the zeros of the equation x* — 63x + 162 = is twice that of one of the other zeros Find all three zeros 20 Find the three zeros of the equation x* — 6x? — 24x + 64 = given For question 20, let the zeros be represented by = that they are consecutive terms in a geometric sequence common ratio 22 Find the value of k if the roots of the equation x* + x? + 2x + k =0 o (=t P Find the value of k such that the zeros of the equation x* — 6x% + kx + 10 = are in an arithmetic progression; that is, they can be represented by @, a + d and a + 2d for some constant d Taar ‘where ris the For question 21, use the result from question 20 are in geometric progression 79 Functions Rational functions Another important category of functions is rational functions, which are feo functions in the form R(x) = @ where fand g are polynomials and the domain of the function R is the set of all real numbers except the real zeros of polynomial g in the denominator Some examples of rational functions are: -1 MR=% ) = X +2 G+ HE-1 e stk x2+1 The domain of p excludes x = 5, and the domain of g excludes x = —3 and x = The domain of r is all real numbers because the polynomial x> + has no real zeros Find the domain and range of h(x) Sketch the graph of h Solution A fraction is zero only if its numerator is zero Because the denominator is zero when x = 2, the domain of is all real numbers except x = 2, i.e x € R, x # Determining the range of the function is a little less straightforward It is clear that the function could never take on a value of zero because that will only occur if the numerator is zero And since the denominator can have any value except zero it seems that the function values of h could be any real number except zero To confirm this and to determine the behaviour of the function (and shape of the graph), some values of the domain and range (pairs of coordinates) are displayed in the tables x approaches from the left b —98 —0.01 x 102 h(x) 0.01 &8 01! 12 0.1 =05 0.5 =1 J125] =2 5] J1E0! =10 201 1.99 —100 2.01 100 2.001 1000 1.999 | —1000 80 x approaches from the right h(x) 10 The values in the tables provide clear evidence that the range of h is all real numbers except zero, i.e h(x) € R, h(x) # The values in the tables also show that as x — —00, h(x) — from below (sometimes written h(x) — 0~) and as x — +00, h(x) — from above (h(x) — 07) It follows that the line with equation y = (the x-axis) is a horizontal vertical o f):'mzpme : asymptote for the graph of h As x — from the left (sometimes written x h(x) appears to decrease without whereas as x — from the right h(x) appears to increase without — 27), bound, (x — 2%), bound This indicates that the graph of h will have a vertical asymptote at x = This behaviour is confirmed by the graph of h horizontal ?(SZ:::“;Z o | The line y = cis a horizontal asymptote of the graph of the function fifat least one of the following statements are true: « asx— +oo, then fix) — ¢+ « asx— + asx — +00, then fix) — ¢~ —oo, then fix) — ¢ * + asx— —oo, then fx) — ¢~ ‘The line x = d is a vertical asymptote of the graph of the function fif at least one of the following statements are true: « asx—d*, then fr) — +00 « asx—d*, then fr) — —oc « asx—d~, thenfw) — +o0 « asx—d-, thenfix) —» —oc0 Example 2.16 3= 12 g1 otch the graph of fand identify Consider the function fix) = x>+ 3x—4 any asymptotes and any x- or y-intercepts Use your sketch to confirm the domain and range of the function Solution First, completely factorise both the numerator and denominator S =1 Le e BGER2) e S2) The x-intercepts will occur where the numerator is zero Hence, the x-intercepts are (—2, 0) and (2, 0) A y-intercept will occur when x = flo) = s 02)(&2) D@ = 3, so the y-intercept is (0, 3) 81 Functions Any vertical asymptote will occur where the denominator is zero, that is, where the function is undefined From the factored form of f we see that the vertical asymptotes are x = and x = —4 We need to determine if the graph of f goes down (fix) — —oc) or goes up (f(x) — 00) on either side of each vertical asymptote It's easiest to this by simply analysing what the sign of h will be as x approaches and —4 from both the left and right For example, as x — 1~ we can use a test value close to and to the left of (e.g x = 0.9) to check whether f(x) is positive or negative to the left of The further the number n O fo = (00K =2 NG ) O ©09—-1D09+4 " (—X+H) the number Sisto0 Conversely, the closer the then f(x) — number s to 0, the inhettheniinber o can be IRy expressed aS e P +o0 (rises) As x — 1" we use a test value close to and to the right of (e.g x = 1.1) to check whether f(x) is positive or negative to the right of mathematically using the concept of a limit foo = 3011 +2)A1 expressed n limit Q1D+ notationas: lim =0 = —2) | (FX) D D = e == pose— il then f(x) — —oo (falls) andlim e = oo, Conducting similar analysis for the vertical asymptote of x = —4 produces: fo = SN2 GG (=4h 1l = IDf(=4s 1l 40) =) () (= ;:>f(x)>0:x4‘74’, then fix) — +o0 (rises) Joxo = SN2 (G308 (GOl ) (S IOR ) = &) I C=) (GO U TS then fix) — —oo (falls) A horizontal asymptote (if it exists) is the value that f(x) approaches as x — *o0 To find this value, we divide both the numerator and denominator by the highest power of x that appears in the denominator (x? for function f) w1 J Hx=1 = B Figure 2.11 GDC the solution to 82 £ 2092 ke screen for Example 2.16 x2 F xz x? Hence, the horizontal asymptote is y = As we know the behaviour (rising or falling) of the function on either side of each vertical asymptote and that the graph will approach the horizontal asymptote as x — *00, an accurate sketch can be made, as shown Because x = —4, analysis and x = the zeros of the polynomial in the denominator are x = and the domain of fis all real numbers except and —4 From our and from the sketch of the graph, it is clear that between x = —4 the function takes on all values from —oc to + 00, therefore the range of fis all real numbers We are in the habit of cancelling factors in algebraic expressions, such as =1 _ &+ x— D=1 =T =x+1 x2 However, the function f(x) = = : 11 and the function g(x) = x + are not the same function The difference occurs when x = A1) = 12 %, which 1.—1 is undefined, and g(1) = + = So, is not in the domain of fbut is in the domain of g As we might expect, the graphs of the two functions appear identical, but upon closer inspection it is clear that there is a ‘hole’ (point discontinuity) in the graph of fat the point (1, 2) Thus, fis a discontinuous function, but the polynomial function g is continuous f and g are different functions (see Figure 2.12) oy = ot £ ! hole ‘Try graphing =l y="—7onyour GDC and zooming in closely to the region around the point (1, 2) Can you see the hole? In working with rational functions, we often assume that every linear factor that appears in both the numerator and in the denominator has been cancelled () Therefore, for a rational function in the form g(_x)’ we can usually assume that the polynomial functions fand g have no common factors Example 2.17 A Find any asymptotes for the function p(x) = %49 Solution The denominator is zero when x = 4, thus the line with equation x = 4is a vertical asymptote 83 Functions Although the numerator x* — is not divisible byx — 4, it does have a larger degree Some insight into the behaviour of functionp may be gained by dividing x — into x> — Since the degree of the numerator is one greater than the degree of the denominator, the quotient will be a linear polynomial P(x) Recalling from the previous section that D R(x) = Qx) BD’ where Q and R are the quotient and remainder, we can rewrite p(x) as a linear polynomial plus a fraction Since the denominator is in the form (x — ¢) we can carry out the division efficiently by means of long division % A x +4 FEER0R=0) X2 — 4x - 4x—9 4x— Hence, p(x) = );7 = 16 As x — *oq, the fraction — This tells us about the end behaviour =il of function p, namely that the graph of p will get closer and closer to the liney = x + as the values of x get further away from the origin Symbolically, this can be expressed as follows: as x — * 00, p(x) — x + We can graph both the rational function p(x) and the line y = x + 4ona GDC to visually confirm our analysis Plotl Plot2 Plot3 \Y1E(X22-9)/(X-4 y=x+4 If a line is an asymptote of a graph but is neither horizontal nor vertical, it is called an oblique asymptote (sometimes called a slant asymptote) The graph of any rational function of the form g(_xx)) where the degree of function fis one more than the degree of function g will have an oblique asymptote Using Example 2.17 as a model, we can set out a general procedure for analysing a rational function, leading to producing a sketch of its graph and determining its domain and range 84 Analysing a rational function foo ) R(x) = ——, given functions fand g have no common factors 809 Factorise: Completely factorise both the numerator and denominator o Intercepts: A zero of f will be a zero ofR and hence, an x-intercept of the graph of R The y-intercept is found by evaluating R(0) W Vertical asymptotes: A zero of g will give the location of a vertical asymptote (if any) Then perform a sign analysis to see if R(x) — +00 or R(x) — —00 on either side of each vertical asymptote - Horizontal asymptotes: Find the horizontal asymptote (if any) by dividing both fand g by the highest power ofx that appears in g, and then letting x— *oo Oblique asymptotes: If the degree of g is one more than the degree of f, then the graph of R will have an oblique asymptote Divide g into fto find the quotient Q(x) and the remainder The oblique asymptote will be the line with equation y = Q(x) < o Sketch of graph: Start by drawing dashed lines where the asymptotes are located Use the information about the intercepts, whether R(x) falls or rises on either side of a vertical asymptote, and additional points as needed to make an accurate sketch Domain and range: The domain zeros of g You need to study the range Often, but not always (as at the horizontal asymptote will ofR will be all real numbers except the graph carefully in order to determine the in Example 2.15), the value of the function not be included in the range End behaviour of a rational function Let R be the rational function given by X ax"+a,_x" N+ e tax+a Rx) = fi— = om0 where functions 8%) b,x™+ b, xm 1+ e +bix + by fand g have no common factors Then the following holds true: If n < m, then the x-axis (liney = 0) is a horizontal asymptote for the graph of R If n = m, then the line y = ay™ is a horizontal asymptote for the graph of R by, If n > m, then the graph of R has no horizontal asymptote However, if the degree of fis one more than the degree of g, then the graph of R will have an oblique asymptote 85 Functions @ Sketch the graph of each rational function without the aid of your GDC On your sketch, clearly indicate any x- or y- intercepts and any asymptotes (vertical, horizontal or oblique) Use your GDC to verify your sketch @ foo S = ®) g0 =S2 (© hew=1=2% = (e) pex) SR e 2, IR ® fo== i (@ Rooy =%~ Xis O Dl (f)f) M) M) =1+ % s hi (h)h) heo = @ g0 _ Use your GDC to sketch a graph of each function, and state the domain and range of the function 2x @ fo=22 S 0)b goo DT T =1 _x2=2x+1 (c) hx)= Tfi ot o 2x =D ) Coo =22 o (d) reo) = -1 Use your GDC to sketch a graph of each function Clearly label any X-Or y- intercepts and any asymptotes @ fn = S ZE=2) B RO _ (b)b goo = 3x2 e - Pl i @ 509 = 3—x2 x4 Ifa, band careall positive, sketch the curve y = PR e R for each of the following conditions: G=bk—q A drugis given to a patient and the concentration of the drug in the bloodstream is carefully monitored At time ¢t = (in minutes after patient received the drug), the concentration, C, in milligrams per litre (@) a