Functions and transformations

Functions nd trnsformtions A Trnsformtions nd the prbol B The cubic function in power form C The power function (the hperbol) D The power function (the truncus) E The squre root function in power form F The bsolute vlue function G Trnsformtions with mtrices H Sum, difference nd product functions I Composite functions nd functionl equtions J Modelling AREAS OF STUDY The behviour of functions of single rel vrible, including ke fetures of their grphs such s is intercepts, sttionr points nd points of inflection, domin (including miml domin) nd rnge, smptotic behviour, smmetr nd the lgebr of functions, including composition of functions. The behviour of these functions is linked to pplictions in prcticl situtions. Identifiction of ke fetures of grphs of: power functions = n for n N nd n =,, nd trnsformtions of these to the form = ( + b) n + c where, b nd c R the modulus function = Trnsformtions from = f () to = Af (n( + b)) + c (where A, n, b nd c R nd f is one of the functions specified bove) nd the reltion between the originl function nd the grph of the trnsformed A Trnsformtions nd the prbol Trnsformtions function (including fmilies of trnsformed functions for single trnsformtion prmeter) Grphs of sum, difference, product nd composite functions of f nd g, where f nd g re functions of the tpes bove Recognition of the generl form of possible models for dt presented in grphicl or tbulr form, using polnomil nd power functions Applictions of simple combintions of the bove functions, nd interprettion of fetures of the grphs of these functions in modelling prcticl situtions The reltionship of f ( ± ), f () nd f to vlues of f () nd f () for different functions f Composition of functions, where f composition g is defined b f (g()), given rn g dom f ; for emple +55, + ebookplus Digitl doc Quick Questions In this chpter we consider the bsic grphs of the qudrtic nd cubic functions, the hperbol nd truncus, squre root nd bsolute vlue functions. The following trnsformtions of the bove grphs re discussed: diltion, reflection nd trnsltion. Diltion A diltion is the stretching or compressing of grph. Let the bsic grph be = f ( ). Chpter Functions nd trnsformtions 57

Diltion w from the -is: = f ( ). Stretches or compresses the grph f ( ) b fctor of from the -is.. Ech -vlue of the bsic grph is multiplied b fctor of, tht is (, ) (, ).. When >, the grph of f () is stretched nd becomes nrrower.. When < <, the grph of f () is compressed nd becomes wider. Diltion w from the -is: = f( n). Stretches or compresses the grph f ( ) b fctor of from the -is. n. Ech -vlue of the bsic grph is multiplied b fctor of n, tht is (, ) ( n, ).. When n >, the grph of f () is compressed from the -is nd becomes nrrower.. When < n <, the grph of f () is stretched from the -is nd becomes wider. Note: For the grphs we will be looking t in this chpter horizontl diltion cn be epressed s verticl diltion. For emple, ( + ) cn be written s ( + ) = 8 ( + ). So in this cse horizontl diltion from the -is b fctor of is the sme s verticl diltion from the -is b fctor of 8. This cn simplif the process of describing trnsformtions for these prticulr grphs. The concept of diltion is illustrted in the following digrm: Originl grph Originl grph dilted from the -is Originl grph dilted from the -is Ref lection Reflection provides mirror imge of grph. Reflection cn tke plce in one or both es. Let the bsic grph gin be = f ( ). Ref lection in the -is: = f( ). The mirror imge of the originl grph ppers cross the -is (the mirror line).. Ech -vlue is the negtive of the originl, the -vlue is unchnged, tht is (, ) (, ). Reflection in the -is: = f ( ). The mirror imge of the originl grph ppers cross the -is (the mirror line).. Ech -vlue is the negtive of the originl, the -vlue is unchnged, tht is (, ) (, ). Reflection in both es: = f( ). The bsic grph is reflected in the -is nd then the -is (or vice vers).. Both the - nd -vlues re the negtives of the originl, tht is (, ) (, ). The concept of reflection is shown in the digrm below. The red str is the originl grph. Reflection in the -is Reflection in the -is Reflection in both es 58 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

Trnsltion: = f ( b) + c A trnsltion slides the grph. Trnsltion cn be horizontl (to the right or left long the -is), or verticl (up or down long the -is). Consider our bsic grph = f ( ).. If = f ( b) the bsic grph is trnslted b units prllel to the -is: () in the positive direction (i.e. to the right) when b > (b) in the negtive direction (i.e. to the left) when b <. Ech -vlue hs b dded to it, tht is (, ) ( + b, ).. If = f ( ) + c, the bsic grph is trnslted c units prllel to the -is: () in the positive direction (i.e. up) when c >, (b) in the negtive direction (i.e. down) when c <. Ech -vlue hs c dded to it, tht is (, ) (, + c).. If = f ( b) + c the bsic grph is trnslted both horizontll nd verticll. Originl grph Verticl trnsltion up Verticl trnsltion down Horizontl trnsltion to the left Horizontl trnsltion to the right ebook plus s Digitl doc Spredsheet Trnsformtions Nturll, the grph cn be subject to combintion of two or more trnsformtions. Combintion of trnsformtions When describing trnsformtions tht hve been pplied to bsic grph f ( ), it is best to put the grph into the formt = f ( b) + c. The order of trnsformtions is importnt s diltions nd reflections re pplied before trnsltions, so ensure tht ou describe the trnsformtions in this order (remember D-R-T). In this chpter we shll consider grphs, derived from bsic curves, using single trnsformtions diltions, reflections or trnsltions s well s combintions of those. Modelling of dt will lso be considered. The qudrtic function in power form The grph of = is prbol with the turning point t the origin. The domin of the function is R nd the rnge is R + {}. Throughout this section we refer to the grph of = s the bsic prbol. Let us now consider the effect of vrious trnsformtions on the grph of this bsic prbol. Chpter Functions nd trnsformtions 59

Qudrtic functions re lso power functions. Power functions re functions of the form f ( ) = n, n R. The vlue of the power, n, determines the tpe of function. When n =, f ( ) =, nd the function is liner. When n =, f ( ) = nd the function is qudrtic. Other power functions will be discussed lter. Under sequence of trnsformtions of f () = n, n R, the = = = generl form of power function, is f () = ( b) n + c (where, b, c nd n R). All liner nd qudrtic polnomils re lso liner nd qudrtic power functions, becuse ll liner nd qudrtic functions re trnsformtions of f ( ) = nd f ( ) =, respectivel. When qudrtic function is written in turning point form it is written in power form. For emple, the qudrtic function = + + 6 cn lso be represented s the power function = ( + ) +. = Diltion In power form, is the diltion fctor. It diltes the grph in the direction. The lrger is, the thinner the grph of the prbol. If is proper frction, tht is, < <, the grph is wider thn the bsic prbol. Reflection If is negtive, the grph of the bsic prbol is reflected in the -is, tht is, the grph is flipped upside down. If is replced with, the grph of the bsic prbol is reflected in the -is, tht is, the grph is flipped sidews. Due to its smmetr, this effect cnnot be seen on the bsic prbol, but it is more obvious with prbol tht hs lred been trnslted. For emple, the grphs of = ( ) nd = ( ) re reflections of ech other cross the -is. = ( ) = ( ) (, 9) (, ) (, ) Trnsltion Horizontl trnsltion If b >, the grph of the bsic prbol is trnslted horizontll to the right, nd if b <, the grph of the bsic prbol is trnslted horizontll to the left. For emple, grph with the eqution = ( ) is bsic prbol tht hs been trnslted units to the right, nd grph with the eqution = ( + ) is bsic prbol tht hs been trnslted units to the left. If the coefficient of is not, the eqution must be rewritten in the form = ( b) + c in order to be ble to work out the vlue of b. For emple, = ( + ) is trnslted of unit to the left, since b = b = = ( b) = ( + ) = [ ( + )] = 6 ( + ) Verticl trnsltion If c >, the grph is trnslted verticll upwrd, nd if c <, the grph is trnslted verticll downwrd. For emple, the grph with eqution = + is bsic prbol tht hs been trnslted units up, nd the grph with eqution = is bsic prbol tht hs been trnslted unit down. = + c c = c = 6 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

ebook plus s Digitl doc Spredsheet 8 The qudrtic function in power form Combintion of trnsformtions The grph of = ( b) + c shows the combintion of the trnsformtions shown bove. The turning point of the grph is (b, c). The domin of the prbol is R nd the rnge is [c, ) if > or [, c) if <. = ( b) + c (b, c) WORKED EXAMPLE Stte the chnges required to trnsform the grph of = into the grph of = ( + ). THINK WRITE Write the generl formul for the prbol. = ( b) + c Identif the vlue of. = Stte the effect of on the grph. The grph of = is dilted b the fctor of from the -is. Identif the vlue of b. b = 5 Stte the effect of b on the grph. The grph is trnslted units to the left. 6 Identif the vlue of c. c = 7 Stte the effect of c on the grph. The grph is trnslted units down. We cn use trnsformtions to find the eqution of the function from its grph b first emining the new position of the turning point. WORKED EXAMPLE Use trnsformtions to find the eqution of this function. (, ) THINK WRITE Write the generl formul of the prbol. = ( b) + c From the grph stte the horizontl trnsltion nd hence the vlue of b. From the grph, stte the verticl trnsltion nd hence the vlue of c. Trnslted units to the right, so b =. Trnslted units up, so c =. Substitute the vlues of b nd c into the generl formul. = ( ) + 5 The grph of the prbol psses through the origin. Substitute = nd = into the formul. Using (, ): = ( ) + 6 Solve for, which is the diltion fctor. = 6 + 6 = = 6 8 = Chpter Functions nd trnsformtions 6

7 Substitute the vlue of into = ( ) + nd write our nswer. The eqution of the prbol shown is: = 8 ( ) + WORKED EXAMPLE Given the eqution = k, determine the effect on the grph =, when k = {,, }. Sketch the grphs. THINK On the Grph & Tb pge complete the function entr line s: = Then tick the bo nd tp!. WRITE/DISPLAY Complete the function entr line s: = = = Tick the boes of ech of the equtions nd tp!. Answer the question b describing the chnges in words. As the vlue of k increses the grph becomes thinner nd stretches w from the -is. REMEMBER. The grph of = is clled bsic prbol.. The grph of = ( b) + c is the bsic prbol, dilted b the fctor of from the -is, trnslted b units horizontll (to the right if b > or to the left if b < ) nd c units verticll (up if c > or down if c < ).. If is negtive, the grph is reflected in the -is.. If is replced with, the grph is reflected in the -is. 5. The turning point of the prbol is ( b, c). 6. The domin of the prbol is R. 7. The rnge is c if > or c if <. = ( b) + c (b, c) 6 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

EXERCISE A Trnsformtions nd the prbol WE Stte the chnges required to trnsform the grph of = into the grph of ech of the following. = b = c = d = 6 e = f = ( ) g = ( + ) h = ( ) i = ( + ) j = (.5) + k = ( + ) l = ( ) MC The eqution of prbol is given b + m ( ) ). Incresing m will result in the grph being: A trnslted further to the left B trnslted further up C thinner D wider E reflected in the -is Mtch the grphs of the prbols with the following equtions. = + b = ( ) c = ( + ) d ( ) e = ( + ) + WE Use trnsformtions to find the eqution of ech function. = iii i ii = iv v (, ) b (, ) c (, ) d (, ) 5 MC The eqution of the grph shown opposite is best given b: A = ( c) + d B = c ( b) C = ( + c) + b D = (c ) + d E = d ( c) d b c e 6 Find the eqution of the imge of = under ech of the following trnsformtions: diltion b the fctor of from the -is b reflection in the -is c trnsltion b units to the right nd unit down Chpter Functions nd trnsformtions 6

d diltion b the fctor of from the -is, followed b trnsltion of units down e reflection in the -is, followed b trnsltion of units to the left. 7 Find the equtions of these grphs. b 5 c d e 9 6 f 8 8 WE Find the eqution of = under the following sequentil trnsformtions (in order): diltion b fctor of from the -is b reflection in the -is c trnsltion of prllel to the -is d trnsltion of prllel to the -is. 9 Find the imge of the point (, ) under ech of the following trnsformtions: reflection in the -is b reflection in the -is c diltion b fctor of from the -is d diltion b fctor of from the -is e diltion b fctor of from the -is f trnsltion of units horizontll in the positive direction g trnsltion of unit prllel to the -is. The prbol hs turning point t (z, 8); it intersects the -is t = nd one of the -intercepts is = 5. Find: the vlue of z b the eqution of the prbol. For the prbol whose rnge is, whose -coordinte of the turning point is nd whose -intercept is = o o, find: the -coordinte of the turning point b the eqution of the prbol c the coordintes of the -intercepts. 6 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

The design shown in the digrm t right cn be obtined b tking the red portion of the prbol nd trnsforming it to form ech of the other 9 frgments. (One or more trnsformtions m be used to form ech frgment.) If the highlighted frgment is given b f ( ),, define the other 9 frgments in terms of f ( ) nd specif their domins. B The cubic function in power form The grph of the function = is shown t right: Both the domin nd rnge of the function re R. The function is constntl incresing nd hs sttionr point of inflection (where the grdient is ) t the origin (, ). Throughout this section we shll refer to the shpe of the grph of = s positive cubic, or bsic cubic curve. Cubic functions re lso power functions. Power functions re functions of the form f ( ) = n, n R. The vlue of the power, n, determines the tpe of function. When n =, f () = nd the function is liner. When n =, f () = nd the function is qudrtic. When n =, f () = nd the function is cubic. When n =, f () = nd the function is qurtic. Other power functions will be discussed lter. Under sequence of trnsformtions of f () = n, n R, the generl form of power function is f () = ( b) n + c (where, b, c nd n R). All liner nd qudrtic polnomils re lso liner nd qudrtic power functions, but this is not the cse for cubic functions (or qurtic functions). For emple, cubic power function in the form f ( ) = ( b) + c hs ectl one -intercept nd one sttionr point of inflection. A cubic polnomil in the form f () = + b + c + d cn hve one, two or three -intercepts nd is therefore not power function. All cubic power functions re lso cubic polnomils, but not ll cubic polnomils re cubic power functions. For emple, the cubic function = ( ) + is polnomil nd power function. It is the grph of = under sequence of trnsformtions. Diltion The vlue is the diltion fctor; it diltes the grph from the -is. The lrger is, the thinner the grph. = = = ebook plus s Digitl doc Spredsheet 5 Cubic function = ( b ) + c Reflection If is negtive, the grph of the bsic cubic is reflected in the -is, tht is, the grph is flipped upside down. If is replced with, the grph of the bsic cubic is reflected in the -is, tht is, the grph is flipped sidews. For emple, the grphs = ( ) nd = ( ) re reflections of ech other cross the -is. = = ( ) = ( ) (, ) (, ) (, ) Chpter Functions nd trnsformtions 65

Trnsltion Horizontl trnsltion If b >, the grph of the bsic cubic is trnslted horizontll to the right, nd if b <, the grph of the bsic cubic is trnslted horizontll to the left. For emple, the grph with eqution = ( ) is bsic cubic trnslted units to the right, nd the grph of = ( + ) is bsic cubic, trnslted units to the left, tht is, prllel to the -is in the negtive direction. If the coefficient of is not, the eqution must be rewritten in the form = ( b) + c in order to be ble to work out the vlue of b. For emple, the grph of = ( 5) is trnslted 5 units to the right, since = ( 5) 5 5 = [ ( )] = 8 ( ) Verticl trnsltion The vlue of c trnsltes the grph verticll or long the -is. If c >, the grph is trnslted verticll up, nd if c <, the grph is trnslted verticll down. The coordintes of the sttionr point of inflection re (b, c). For emple, if = is trnslted unit up, the eqution of the resulting grph is = + nd the point of inflection is (, ); if it is trnslted units down, the eqution of the resulting grph is = nd the point of inflection is (, ). Combintion of trnsformtions The grph of = ( b) + c shows the combintion of the trnsformtions described bove. Finll, the domin nd rnge of = ( b) + c re R (ll rel numbers). b = b = = ( b) c = c = = + c = ( b) + c (b, c) WORKED EXAMPLE Stte the chnges necessr to trnsform the grph of = into the grph of = ( + ). THINK WRITE Write the generl eqution of the cubic function. = ( b) + c Identif the vlue of. = Stte the effect of on the grph. The grph is dilted b the fctor of in the direction. Identif the vlue of b. b = 5 Stte the effect of b on the grph. The grph is trnslted unit to the left. 6 Identif the vlue of c. c = 7 Stte the effect of c on the grph. The grph is trnslted units down. 66 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

WORKED EXAMPLE 5 For ech of the following grphs: i stte the coordintes of the sttionr point of inflection ii find the - nd -intercepts iii sketch the grph iv stte the trnsformtions tht the grph of = hs undergone to form ech new eqution. = ( + ) b = ( ) + 6 THINK WRITE/DRAW Write the eqution. = ( + ) i Since the rule is of the form = ( b) + c, identif the vlues of b nd c nd hence write the coordintes of the sttionr point of inflection (b, c). ii Find the -intercept b letting =. Find the -intercept b letting =. iii To sketch the grph on set of lbelled es, mrk the sttionr point of inflection nd the - nd -intercepts, then sketch the positive cubic pssing through the points mrked. i b =, c = Sttionr point of inflection: (, ) ii -intercept: =, = ( + ) = 7 = 8 -intercept: = ( + ) = ( + ) = + = = iii (, ) (, ) iv Stte the kind of reflection nd the verticl nd horizontl trnsltions. b Write the eqution. b = ( ) + 6 i Since the rule is of the form = ( b) + c, identif the vlues of b nd c nd hence write the coordintes of the sttionr point of inflection (b, c). ii Find the -intercept b letting =. Find the -intercept b letting =. Note: Do not round off until the ver lst step; for grphing purposes, round off our finl nswer to deciml plce. iv The grph is reflected in the -is. There is horizontl trnsltion of units to the left nd verticl trnsltion of unit down. i b =, c = 6 Sttionr point of inflection: (, 6) ii -intercept: =, = ( ) + 6 = 6 + 6 = 7 -intercept: = ( ) + 6 = ( ) = 6 = 6 =+ 6 5.8 Chpter Functions nd trnsformtions 67

iii To sketch the grph on set of lbelled es, mrk the sttionr point of inflection nd the - nd -intercepts, then sketch the positive cubic pssing through the points mrked. iii (, 6) iv Stte the kind of reflection nd the verticl nd horizontl trnsltions. (5.8, ) iv The grph is reflected in the -is. There is horizontl trnsltion of units to the right nd verticl trnsltion of 6 units up. To find the eqution of the curve from given grph, we need to estblish ectl wht trnsformtions were pplied to the bsic cubic curve. This is best done b observing the shpe of the grph nd the position of the sttionr point of inflection. WORKED EXAMPLE 6 Find the eqution of the curve, if it is of the form = (( b) + c. 5 THINK WRITE/DISPLAY Write the generl eqution of the cubic function. = ( b) + c Write the coordintes of the sttionr point of inflection (b, c) nd hence stte the vlues of b nd c. Substitute the vlues of b nd c into the generl formul. The grph psses through the point (, 5) (-intercept). Substitute the coordintes of this point into the eqution. 5 On the Min pge, complete entr line b either tping the line s shown, or tpe the eqution: 5 = ( ) +. Highlight it nd tp: Interctive Advnced solve Chnge the vrible to nd tp: OK The sttionr point of inflection is (, ). So b =, c =. = ( ) + Using (, 5): 5 = ( ) + 6 Write the solution for the eqution. = 7 Substitute the vlue of into = ( ) +. = ( ) + 68 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

REMEMBER. The grph of = is bsic cubic curve.. The grph of = ( b) + c where > is the bsic cubic, dilted b the fctor of from the -is, trnslted b units long the -is (to the right if b >, or to the left if b < ) nd c units long the -is (up if c >, or down if c < ). = ( b) + c (b, c). If <, the grph is reflected in the -is.. The sttionr point of inflection is t ( b, c). 5. Both the domin nd rnge re R. EXERCISE B The cubic function in power form WE Stte the chnges necessr to trnsform the grph of = into the grph of ech of the following. = = 7 b c = + d = 6 e = ( ) f = ( + ) g = ( ) h = 6(7 ) i = ( + ) j =6 6 ( ) k ( + + 5) l = ( + ) = Which of these trnsformtions were pplied to the grph of = to obtin ech of the grphs below? i reflection in the -is ii trnsltion to the left iii trnsltion to the right iv trnsltion up v trnsltion down b c d e f Chpter Functions nd trnsformtions 69

WE5 For ech of the following grphs: i find the sttionr point of inflection ii find the - nd -intercepts iii stte the trnsformtion(s) tht the grph = hs undergone to produce the given grph iv sketch the grph. = b = EXAM TIP Be creful when sketching grphs use pproprite scles on the es, clerl drw the grph nd use the correct domin. Show t lest one scle point on ech is nd two coordinte points on ech curve (including intercepts). [Authors dvice] ebookplus s c = 6 d = ( ) e ( ) f = ( ) = g = ( ) + h = ( + ) i = ( + ) 6 Digitl doc Spredsheet 6 Function grpher Questions to 6 refer to the function = (m ). MC The coordintes of the sttionr point of inflection re: A (, ) m B (m, ) C (, ) m m 5 MC The grph of = is dilted in the direction b the fctor of: A B m C m D m D (, m ) m E (, m ) E m 6 MC If m >, incresing m will cuse the grph to become: A wider B thinner nd trnslted not s fr to the right C shifted further to the left D shifted further to the right E shifted further down 7 Find the eqution of the grph resulting from ech of the following trnsformtions of the grph of = : diltion b the fctor of from the -is b reflection in the -is nd trnsltion b 5 units to the left c trnsltion b units to the right nd unit down d diltion in the direction b the fctor of, followed b the verticl trnsltion of units e reflection in the -is, then trnsltion of unit to the left nd unit down. 8 Find the eqution of the grph resulting from the following sequentil trnsformtions of the grph of = : diltion b fctor of from the -is b reflection in the -is c trnsltion of in the positive direction prllel to the -is d trnsltion of in the negtive direction prllel to the -is. 9 WE6 Find the equtions of these curves, if the re of the form = ( ( b) + c. b c d e (, ) 9 7 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

MC The grph of = ( + ) + hs been reflected in the -is, shifted units to the right nd unit up. The eqution of the resulting grph is: A = ( ) + B = ( + ) + C = ( ) D = ( ) + E = The grph of cubic function of the form = ( ( b) + c hs sttionr point of inflection t (, ) nd cuts the -is t =. Find the eqution of the function. The grph of = (b ) + c hs sttionr point of inflection t (, ) nd psses through the point (, ). Find the eqution of the curve. b Stte the shpe of the curve (tht is, whether it is positive or negtive cubic). The grph of = ( ( b) + c cuts the -is t = nd the -is t = 8. If it is known tht the diltion fctor is equl to : find the position of the sttionr point of inflection b sketch the grph. C The power function (the hperbol) The grph shown t right is clled hperbol nd is given b the eqution =. Power functions re functions of the form f ( ) = n, n R. The vlue of the power, n, determines the tpe of function. We sw erlier tht when n =, f () = nd the function is liner. When n =, f () = nd the function is qudrtic. When n =, f () = nd the function is cubic. When n =, f () = nd the function is qurtic. The power function tht produces the grph of hperbol hs vlue of n =. Thus, the function f (= ) cn lso be epressed s the power function f ( ) =. The grph ehibits smptotic behviour. Tht is, s becomes ver lrge, the grph pproches the -is, but never touches it, nd s becomes ver smll (pproches ), the grph pproches the -is, but never touches it. So the line = (the -is) is verticl smptote nd the line = (the -is) is the horizontl smptote. Both the domin nd the rnge of the function re ll rel numbers, ecept ; tht is, R\{}. The grph of = cn be subject to number of trnsformtions. Consider = + c or = ( b) + c b Diltion The vlue is diltion fctor. It diltes the grph from the -is. Reflection If is negtive, the grph of the bsic hperbol is reflected in the -is. If is replced with, the grph of the bsic hperbol is reflected in the -is. = = = = = = = = Chpter Functions nd trnsformtions 7

For emple, the grphs of = nd = re reflections of ech other cross the -is. Trnsltion Horizontl trnsltion The vlue b trnsltes the grph b units horizontll, tht is, prllel to the -is. If b >, the grph is trnslted to the right, nd if b <, the grph is trnslted to the left. For emple, the grph with eqution = is bsic hperbol trnslted = units to the right. This grph hs verticl smptote of = nd domin R\{} (nd horizontl smptote = ). If bsic hperbol is trnslted units to the left, it becomes = +, with verticl smptote of = nd domin R\{ }. Hence, the eqution of the verticl smptote is = b nd the domin is R\{b}. The horizontl smptote nd the rnge remin the sme, = nd R \ {}, respectivel. Verticl trnsltion The vlue c trnsltes the grph c units verticll, tht is, prllel to the -is. If c >, the grph is trnslted upwrd, nd if c <, the grph is trnslted c units downwrd. The grph with eqution = + is bsic hperbol trnslted units up. This grph hs horizontl smptote of = nd rnge of R\{} (nd verticl smptote = ). If bsic hperbol is trnslted units down, it becomes =, with horizontl smptote of = nd rnge of R\{ } (nd verticl smptote = ). Hence the eqution of the horizontl smptote is = c nd the rnge is R\{c}. Alws drw the smptote s dotted line nd lbel it with its eqution (for emple, = ) t the end of the smptote. Ensure tht the grph continues to pproch the smptote getting closer but not touching or crossing the smptote or bouncing w from the smptote. = = = (, ) = ebook plus s Combintion of trnsformtions The grph of = + c shows the combintion of these trnsformtions. b = + c b Digitl doc Spredsheet 5 The hperbol = c c b = b Finll, if the coefficient of is number other thn, to obtin the vlue of h the eqution should be rerrnged first. For emple, = = + 6 ( (+ ) Therefore, b = (not 6 s it m seem t first); tht is, the grph is trnslted units to the left. 7 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

WORKED EXAMPLE 7 Stte the chnges tht should be mde to the grph of = in order to obtin the grph of = +. THINK WRITE Write the generl eqution of the hperbol. b Identif the vlue of. = Stte the chnges to =, cused b. The grph of = is dilted b the fctor of from the -is nd reflected in the -is. Identif the vlue of b. b = 5 Stte the effect of b on the grph. The grph is trnslted units to the left. 6 Identif the vlue of c. c = 7 Stte the chnges to the grph, cused b c. The grph is trnslted unit down. WORKED EXAMPLE 8 For the grph of = +, stte: the equtions of the smptotes c the rnge. b the domin THINK WRITE Write the generl eqution of the hperbol. = b + c Identif the vlues of b nd c nd hence write the equtions of the smptotes: Horizontl smptote: = c Verticl smptote: = b b =, c = Horizontl smptote: = Verticl smptote: = b Stte the domin of the hperbol: R\{b}. b Domin: R\{} c Stte the rnge of the hperbol: R\{c}. c Rnge: R\{} Sketching the grph of the hperbol b hnd cn be esil done b following these steps:. Find the position of the smptotes.. Find the vlues of the intercepts with the es.. Decide whether the hperbol is positive or negtive.. On the set of es drw the smptotes (using dotted lines) nd mrk the intercepts with the es. 5. Treting the smptotes s the new set of es, sketch either the positive or negtive hperbol, mking sure it psses through the intercepts tht hve been previousl mrked. Chpter Functions nd trnsformtions 7

WORKED EXAMPLE 9 Sketch the grph of = +, clerl showing the intercepts with the es nd the position of the smptotes. THINK Compre the given eqution with = b + c nd stte the vlues of, b nd c. WRITE/DRAW =, b =, c = ebook plus s Tutoril int-5 Worked emple 9 Write short sttement bout the effects of, b nd c on the grph of =. Write the equtions of the smptotes. The horizontl smptote is t = c. The verticl smptote is t = b. Find the vlue of the -intercept b letting =. 5 Find the vlue of the -intercept b mking =. 6 To sketch the grph: Drw the set of es nd lbel them. b Use dotted lines to drw the smptotes. The smptotes re = nd =. c d Mrk the intercepts with the es. The intercepts re = nd = Treting the smptotes s our new set of es, sketch the grph of the hperbol (s is positive, the grph is not reflected); mke sure the upper brnch psses through the - nd -intercepts previousl mrked.. The grph of = is dilted b the fctor of from the -is, trnslted units to the left nd units down. Asmptotes: = ; = -intercept: = = + = = Point (, ) -intercept: = = + + = = ( + ) = + 8 = 8 = 6 Point = 6, = ( ) (, ) = (, ) = The net emple shows how to find the eqution of the hperbol from its grph. 7 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

WORKED EXAMPLE Find the eqution of the grph shown. 6 THINK WRITE Write the generl eqution of the hperbol. = b + c From the grph, identif the vlues of b nd c Remember tht the eqution of the horizontl smptote is = c nd of the verticl smptote is = b. b =, c = Substitute the vlues of b nd c into the formul. = + Substitute the coordintes of n of the known Substitute (, ): points of intersection with the es into the formul (s, -intercept). = + 5 Solve for. = + = = 6 6 Substitute the vlue of into = +. 6 = + 7 Trnspose (optionl). 6 = REMEMBER. The grph of = is clled hperbol.. The grph of = + c is the grph of the bsic b hperbol, dilted b the fctor of from the -is, trnslted b units horizontll (to the right if b >, or to the left if b < ) nd c units verticll (up if c >, or down if c < ). If <, the grph is reflected in the -is. The equtions of the smptotes re: = b nd = c. The domin of the function is R\{b} nd its rnge is R\{c}. = b = + b c = c Chpter Functions nd trnsformtions 75

EXERCISE C The power function (the hperbol) WE 7 Stte the chnges tht should be mde to the grph of = in order to obtin the grph of ech of the following. = b = c = 6 d = e = + 7 f = 5 + g = + h = + 6 i = Which of the following trnsformtions were pplied to the grph of = to obtin ech of the grphs shown below? i trnsltion to the right ii trnsltion to the left iii trnsltion up iv trnsltion down v reflection in the -is b c d e f g h WE8 For ech of the following, stte: i the equtions of the smptotes ii the domin iii the rnge. d g = = = h + 6 b e = + 6 = + f = 5 + c = = i = n + m 5 76 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

For ech of the following grphs, stte: i the equtions of the smptotes ii the domin iii the rnge. b c d e f n m b ebook plus s Digitl doc Spredsheet 6 Function grpher 5 On the sme set of es sketch the grphs of =, =, = nd =. 6 WE9 Sketch ech of the following, clerl showing the position of the smptotes nd the intercepts with the es. Check our nswers, using CAS clcultor. = b = + + c = d = e + 5 = 6 f = + 6 g = h = + i = + 5 + + j = k = + l + = 7 MC The eqution of the grph shown is likel to be: A C E = + = = B D = + = 8 MC Which of the following is true sttement for the grph of = +? A The domin is R\{}. B The rnge is R\{}. C The eqution of the horizontl smptote is =. D The eqution of the verticl smptote is =. E None of the bove. Chpter Functions nd trnsformtions 77

9 WE Find the eqution for ech of the following hperbols, if the re of the form = + c. b b c d e f 5 5 ebook plus s Digitl docs WorkSHEET. Histor of mthemtics The histor of some mjor curves D If function is given b f( ) =, sketch ech of the following, lbelling the smptotes nd the intercepts with the es. f ( + ) b f () c f ( ) d f ( ) + e f( ) f f ( ) Sketch the grph of + =, nd stte its domin nd rnge. (Hint: First trnspose the eqution to mke the subject.) The power function (the truncus) The grph shown t right is known s truncus. The eqution of the grph is given b: EXAM TIP When ou re sked to give the domin nd rnge in question, be sure to demonstrte clerl which nswer is which. For emple, for =, dom f : R, rn f : [, ]. [Authors dvice] = Power functions re functions of the form f ( ) = n, n R. The vlue of the power, n, determines the tpe of function. We sw erlier tht when n =, f () = nd the function is liner. When n =, f () = nd the function is qudrtic. When n =, f ( ) = nd the function is cubic. When n =, f () = nd = = the function is qurtic. When n =, f ( ) = nd the power function produces the grph of hperbol. The power function tht produces the grph of truncus hs vlue of n =. Thus, the function f( ) = cn lso be epressed s the power function f () =. The function is undefined for =. Hence, the eqution of the verticl smptote is = nd the domin of the function is R\{}. 78 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

We cn lso observe tht the grph pproches the -is ver closel, but never touches it. So = is the horizontl smptote. Since the whole grph of the truncus is bove the -is, its rnge is R + (tht is, ll positive rel numbers). Similr to the grphs of the functions, discussed in the previous sections, the grph of = cn undergo vrious trnsformtions. Consider the generl formul = ( b) + c, or = ( b) + c. Diltion The vlue is the diltion fctor. It diltes the grph from = the -is. The diltion fctor does not ffect the domin, = rnge or smptotes. = = = Reflection If is negtive, the grph of bsic truncus is reflected in the -is. The rnge becomes R (tht is, ll negtive rel numbers). = = If is replced with, the grph of the bsic truncus is reflected in the -is. The effect of this reflection cnnot be seen in the bsic grph, but it becomes more obvious if the grph hs been trnslted horizontll first. For emple, the grphs of = nd = ( ) ( ) re reflections cross the -is. The verticl smptote chnges from = to = nd the domin chnges from R \{} to R \{ }. = = ( ) = = = ( ) = = Trnsltion Horizontl trnsltion The vlue b trnsltes the grph b units horizontll. If b >, the grph is trnslted to the right, nd if b <, the grph is trnslted left. For emple, the grph of the eqution = results from trnslting bsic truncus units ( ) to the right. The verticl smptote is = = b = b = = ( b) Chpter Functions nd trnsformtions 79

nd the domin is R\{}. If bsic truncus is trnslted units to the left, it becomes = ( + ), where the verticl smptote is = nd the domin is R \{ }. Hence, the eqution of the verticl smptote is = b nd the domin is R \{b}. The rnge is still R + nd the eqution of the horizontl smptote is =. Verticl trnsltion The vlue c trnsltes the grph c units verticll. If c > the grph is trnslted upwrd, nd if c <, the grph is trnslted c units downwrd. For emple, the grph with eqution = + results when bsic truncus is trnslted unit upwrd. The horizontl smptote is = nd the rnge is (, ). If bsic truncus is trnslted unit down, it becomes =, with = s the horizontl smptote nd (, ) s the rnge. Hence the eqution of the horizontl smptote is = c nd the rnge is (c, ). Note: If is positive (see grph below), the whole grph of the truncus is bove the line = c (the horizontl smptote) nd hence its rnge is > c, (c, ). If is negtive, the whole grph is below its horizontl smptote nd therefore the rnge is <, or (, c). = = + c c = = c = = = = The grph of = ( b) + c shows the combintion of these trnsformtions. c = + c ( b) = c b = b WORKED EXAMPLE Stte the trnsformtions required to chnge the grph of = into the grph of =. ( ) THINK WRITE Write the generl formul for the truncus. ( b) + c Identif the vlue of. = 8 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

Stte the effect of on the grph. The grph of Identif the vlue of b. b = = is reflected in the -is. 5 Stte the effect of b on the grph. The grph is trnslted units to the right. 6 Identif the vlue of c. c = 7 Stte the effect of c on the grph. The grph is trnslted unit down. WORKED EXAMPLE For the function = ( + ), stte: the equtions of the smptotes b the domin c the rnge. THINK WRITE Write the generl formul for the truncus. ( b) + c b Write the generl equtions of the smptotes. Verticl smptote: = b Horizontl smptote: = c Identif the vlues of b nd c. b =, c = Stte the equtions of the smptotes b substituting the vlues of b nd c into corresponding formuls. Write the domin of the truncus, which is R\{b}. c Check whether is positive or negtive. c > Asmptotes: = nd = b Domin: R\{ } Write the rnge (which for > is > c). Rnge: > To sketch the grph of truncus, first put it in the form = ( b) + c, then compre the given eqution with the generl formul to see wht chnges should be mde to the bsic curve (the grph of = ) to trnsform it to the one ou wnt. This should give ou n ide of how the grph will look. The following lgorithm cn then be used:. Find the position of the smptotes.. Find the intercepts with the es.. On the set of es, drw the smptotes (using dotted lines), lbel with the eqution nd mrk the - nd -intercepts.. Treting the smptotes s the new set of es, sketch the bsic truncus curve. > < 5. Mke sure the curve psses through the points mrked on the es. Chpter Functions nd trnsformtions 8

WORKED EXAMPLE Sketch the grph of =, clerl showing the position ( + ) of the smptotes nd the intercepts with the es (correct to deciml plce). THINK WRITE/DRAW Write the generl formul for the truncus. ( b) + c Identif the vlues of, b nd c. =, b =, c = Write short sttement bout the trnsformtions the grph of = should undergo in order to be chnged into the one in question. Write the equtions of the smptotes ( = c nd = b). The grph of = is reflected in the -is, trnslted unit to the left nd units up. Asmptotes: = nd = 5 Find the -intercept (round off to deciml plce). -intercept: = = ( + ) ( + ) = ( + ) = + = ± = ± = 6 Find the -intercept. -intercept: = 7 To sketch the grph: drw the set of es nd lbel them; use dotted lines to drw smptotes; mrk the - nd -intercepts; treting the smptotes s the new set of es, drw the bsic truncus curve upside down (since is negtive); mke sure it intersects the es in the right plces. or =... 7. = ( + ) = = (.7, ) (., ) = (, ) = In the bove emple we hve considered sketching the grph from the given eqution. Sometimes the opposite tsk is required; tht is, the eqution of the function should be estblished from its grph. 8 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

WORKED EXAMPLE Find the eqution of the curve shown in this digrm. = THINK WRITE Write the generl eqution of the truncus. ( b) + c Compred to =, the grph is shifted units to b = nd c = the right. (There is no shift long the -is.) Substitute the vlues of b nd c into the formul. = + ( ) = ( ) Substitute the coordintes of the -intercept into the formul. Using (, ) : = ( ) = = ( ) 5 Solve for. = 6 Substitute for in the eqution. = ( ) REMEMBER. The grph of = is clled truncus.. The grph of = ( b) + c is the bsic truncus curve, dilted b fctor of from the -is nd trnslted b units long the -is (to the right if b > or to the left if b < ) nd c units long the -is (up if c > or down if c < ). If is negtive, the grph is reflected in the -is.. The verticl smptote is = b.. The horizontl smptote is = c. 5. The domin is R\{b}. 6. The rnge is > c if >, or < c if <. c b = ( b) + c Chpter Functions nd trnsformtions 8

EXERCISE D The power function (the truncus) WE Stte the trnsformtions required to chnge the grph of = into the grph of ech of the following: = b = c = ( + ) d = e 5 ( ) = f ( + ) = + 6 g = h = + ( ) MC To obtin the grph shown, the grph of = ws: A reflected in the -is nd trnslted units down B trnslted units to the left C reflected in the -is nd trnslted units to the left D reflected in the -is nd trnslted units to the right E reflected in the -is nd trnslted units up i = 5 ( + ) = MC Which of the following trnsltions took plce, so tht the grph of = ws chnged into the one shown t right? A m units to the left nd p units up B m units to the right nd p units up C m units to the left nd n units up D m units to the right nd n units up E m units to the left nd n units down WE For ech of the following stte: i the equtions of the smptotes ii the domin iii the rnge. = b = c = ( ) d = e ( + ) = 5 f ( + ) = g = + h = i = + 5 ( ) = m p = n Questions 5 tο 7 refer to the following digrms. i ii iii 8 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

iv v vi ebook plus s Digitl doc Spredsheet 6 Function grpher 5 MC Which of the bove functions hs the domin R\{ }? A ii onl B iii onl C ii nd iii D ii, iii nd vi E v nd vi 6 MC Which of the bove functions hs the rnge <? A i, ii nd iv B iii, v nd vi C iv onl D v onl E iii nd vi 7 MC Which of the grphs hs smptotes = nd =? A i B ii C iii D iv E v 8 WE Sketch ech of the following, clerl showing the position of the smptotes nd the intercepts with the es (correct to deciml plce where pproprite). = b = c = 5 ( ) ( + ) d = e ( ) = f = g = ( ) ( + ) j = h = ( + ) k = + ( ) i = + ( ) l = ( ) 9 Find the equtions for ech of the following curves, if it is known tht ll of them re of the form = ( b) + c nd is either + or. EXAM TIP Alws check ou hve met the Alws check ou hve met the requirements of the nswer formt, for emple, ensure tht ou hve rounded correctl nd show the correct number of deciml plces. [Assessment report 6 VCAA] b c n q p s r m d e f t c d b f e g h Chpter Functions nd trnsformtions 85

g h i j l k [ VCAA 6] WE Find the eqution for ech of the following. b c = = 5 = = = = d 7 = e.5 = f 7 = = = = E The domin of truncus is R\{ }; its rnge is > nd its grph cuts the -is t nd =. Find the eqution of the function. The domin of truncus is R\{}; its rnge is (, ) nd its grph cuts the -is t = 5. Find the eqution of the function. The squre root function in power form The squre root function is given b = (or = ). Power functions re functions of the form f ( ) = n, n R. The vlue of the power, n, determines the tpe of function. We sw erlier tht when n =, f ( ) = nd the function is liner. When n =, f ( ) = nd the function is qudrtic. When n =, f () = nd the function is cubic. When n =, f () = nd the function is qurtic. When n =, f () = nd the power function produces the grph of hperbol. When n =, f ( ) = nd the power function produces the grph of truncus. The power function tht produces the grph of the squre root function hs vlue of n =. Thus, the function f( ) = cn lso be epressed s the power function f( ) = f () =. The function is defined for ; tht is, the domin is R + {}, or [, ). 86 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

As cn be seen from the grph, the rnge of the squre root function is lso R + {}, or [, ). Throughout this section we will refer to the grph of = s the bsic squre root curve. Let us now investigte the effects of vrious trnsformtions n on the bsic squre root curve. Consider the function = b+ c, or =( ( b) + c. Diltion The vlue is diltion fctor; it diltes the grph from the -is. The domin is still [, ). = = = Reflection If is negtive, the grph of bsic squre root curve is reflected in the -is. The rnge becomes (, ]. The domin is still [, ). = = = (, ) (, ) (, ) = If is replced with, the grph is reflected in the -is. For emple, the grphs with equtions = nd = re reflected cross the -is. The domin becomes (, ] nd the rnge is [, ). = (, ) (, ) = Trnsltion Horizontl trnsltion The vlue b trnsltes the grph horizontll. If b >, the grph is trnslted to the right, nd if b <, the grph is trnslted to the left. The grph with the eqution = results when the bsic curve is trnslted units to the right. This trnslted grph hs domin [, ) nd rnge [, ). If the bsic curve is trnslted units to the left, it becomes = + nd hs domin [, ) nd rnge [, ). The domin of squre root function fter trnsltion is given b [b, ). Verticl trnsltion The vlue c trnsltes the grph verticll. If c >, the grph is trnslted verticll up, nd if c <, the grph is trnslted verticll down. If = is trnslted units verticll up, the grph obtined is = +, with domin [, ) nd rnge [, ). If the bsic curve is trnslted units down, it becomes =, with domin [, ) nd rnge [, ). The rnge of the squre root function is [c, ) for >. (, ) (, ) (, ) (, ) (, ) b = c = b = = b c = = + c Chpter Functions nd trnsformtions 87

Combintion of trnsformtions The grph of = b + c shows the combintion of these trnsformtions. The point (b, c) is the end point of the squre root curve. For emple, the end point of = + is (, ). It is lws good prctice to lbel the end point with its coordintes. Mke sure it is n open circle if the -vlue is not in the required domin nd closed circle if its -vlue is within the function s domin. Consider the function = b + c. The grph of = + b + c hs ( b, c) s its end point. If this function is reflected in the -is, it becomes = + b + c with end point (b, c). The eqution = + b + c cn then be rewritten s = b + c. For emple, the grph of = + cn be rewritten s = + +, which hs n end point of (, ) nd bends to the left. The domin is (, ] nd the rnge is [, ). The eqution = + + results in = + + when it is reflected in the -is. The domin chnges from [, ) to (, ] nd the rnge remins [, ). The eqution = + cn be rewritten s = ( + ) ; the domin is [, ) nd the rnge is [, ). ) = b + c (b, c) = b + c (b, c) = + b + c (b, c) WORKED EXAMPLE 5 Stte the trnsformtions required to chnge = to = + + 5 +. THINK WRITE Write the generl formul for the squre root curve. = b + c Identif the vlue of. = Stte the effect of on the grph. The grph is dilted b fctor of from the -is nd reflected in the -is. Identif the vlue of b. b = 5 5 Stte the effect of b on the grph. The grph is trnslted 5 units to the left. 6 Identif the vlue of c. c = 7 Stte the effect of c on the grph. The grph is trnslted units up. WORKED EXAMPLE 6 For ech of the following functions find the domin nd rnge. = + b = + c = + ebook plus s Tutoril int-5 Worked emple 6 88 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

THINK WRITE Write the generl formul. = b + c Write the question. = + Identif the vlues of b nd c. b =, c = Stte the domin b. The domin is [, ). Stte the rnge ( c for > ). The rnge is [, ). b Write the question. b = + Fctorise the epression under the squre root sign. = (+ ) Stte the domin. The domin is [, ) ). Identif the vlue of c nd check whether is positive or negtive. c =, < 5 Stte the rnge. The rnge is (, ]. c Write the question. c = + Identif the vlues of b nd c. b =, c = Since the function is of the form = b + c, the domin is b. The domin is (, ]. Stte the rnge ( c). The rnge is [, ). To sketch the grph of the squre root function, we need to compre the given formul with > = b + c > = b + c = b + c. This will give us n ide of the chnges required to trnsform the bsic squre root < (b, c) < curve into the one we wnt. It will lso let us know the w the curve will look. The digrm bove illustrtes the ide. Once the coordintes of the end point nd the direction of the curve re known, the intercepts with the es (if n) should be found before sketching. WORKED EXAMPLE 7 Sketch the grph of = +, clerl mrking intercepts nd the end points. THINK WRITE/DRAW Write the eqution. = + Write the coordintes of the end point. End point: (, ) Stte the shpe of the grph. Shpe: Chpter Functions nd trnsformtions 89

Find the -intercept b letting =. -intercept: = = = + = = ( ) = = 5 Find the -intercept if there is one. There is no -intercept. 6 Sketch the grph b plotting the end point, mrking the -intercept, nd drwing the curve so tht it strts (, ) t the end point nd psses through the -intercept. (, ) WORKED EXAMPLE 8 Given f : [, ) R, where f( ) = nd g(( ) = f ( ) + b, where nd b re positive rel constnts, consider the effect on g(( ) s nd b increse individull. THINK On the Grph & Tb pge, tpe the eqution s shown. Press the dimond t the top of the screen nd select Dnmic Grph. Complete the tble s: Dnmic: v Dnmic: w Strt: Strt: End: End: Step:. Step:. Use the rrow kes to djust the vlue of the vrible. Left nd right will djust v; up nd down will djust w. WRITE Write our description in words. As increses, the grph is dilted w from the -is, with the grph stretched further from the -is. As b increses, the grph is trnslted up prllel to the -is. REMEMBER. The grph of the function = b + c is the grph of =, dilted b the fctor of from the -is nd trnslted b units long the -is nd c units long the -is. = b + c. If <, the bsic grph is reflected in the -is. (b, c). The end point of the grph is ( b, c).. The domin is b. 5. The rnge is c for >, or c for <. 6. If = b + c, the domin is b; the grph of = is reflected in the -is. 9 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

EXERCISE E The squre root function in power form WE5 Stte the trnsformtions required to chnge = to ech of the following. = b = c = d = + e = f = g = + h = 6 + i = + For ech of the functions in question write the coordintes of the end point. MC The grph shown below ws obtined b trnslting the grph of = : A units up nd 9 units to the right B units down nd 9 units to the right C 9 units up nd units to the right D units down nd 9 units to the right E none of the bove MC To obtin the grph in the digrm t right, the grph of = ws: A trnslted units to the right nd units up B trnslted units to the right nd units up C trnslted units to the right, units up nd reflected in the -is D reflected in the -is, trnslted units up nd units to the right E reflected in the -is, trnslted units up nd units to the left. 5 WE6 Find the domin nd rnge for ech of the following functions. = + b = c = d = + + e = 5 f = + g = + + h = + i = + 5 9 (, ) j = 7 k = 6+ + l = Questions 6 tο 7 refer to the digrm t right. 6 MC The eqution of the grph is of the form: A = b+c, c,> B = b+c, c,< C = b +c, c,> D = b +c, c,< E could be either B or C 7 MC The domin nd rnge (in tht order) of the function re: A (, ] nd (, ] B (, ) nd (, ) C (, } nd {, + ) D (, ] nd (, ] E (, ] nd (, ] 8 WE7 Sketch the grph of ech of the following, clerl mrking intercepts nd end points. = + b = + c = d = 6+ e = + + + f = + g = h = 6+ + + i = [ VCAA 7] (, ) EXAM TIP Grphs should be drwn showing correct fetures, such s smoothness nd end points. [Assessment report 7] Chpter Functions nd trnsformtions 9

ebook plus s Digitl doc Spredsheet 6 Function grpher 9 WE8 MC The eqution of the grph shown t right is: A = B = C = D = + E = (, ) The grph of = ws dilted b the fctor of from the -is nd trnslted m units to the right nd units down. It intersects with the -is t = 5. Find: the vlue of m b the eqution of the curve. The end point of the squre root curve is t (, ) nd its -intercept is 9. Sketch the grph of the curve nd hence estblish its eqution. The grph of = ws dilted b the fctor of from the -is, reflected in the -is, trnslted unit to the left nd p units up. Find: the vlue of p, if the grph cuts the -is t = b the eqution of the curve c the -intercept d the domin e f the rnge. Hence, sketch the grph, showing the coordintes of the end points nd the intercepts with the es. F The bsolute vlue function The function f ( ) = is clled n bsolute vlue function or modulus function. The domin of this function is R nd its rnge is R + {}. Its grph is smmetricl in the -is nd hs cusp ( shrp point) t the origin. The smbol represents the mgnitude of, (tht is, the size of ), regrdless of its sign. Therefore, =, if, if < Compre the grphs of = nd =. For, the grphs of the two functions re identicl, while for < the grph of = is the reflection of = in the -is. In generl, n grph of the form = f() is clled n bsolute vlue function. To sketch the grph of = f ( ), we need to sketch the grph of = f () first nd then reflect in the -is the portion of the grph which is below the -is. = = = WORKED EXAMPLE 9 Sketch the grph of =. THINK We first need to sketch the grph of =. Stte the shpe of this grph. WRITE/DRAW Let = Shpe: positive prbol, trnslted unit down 9 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

Stte the coordintes of the turning point (note tht it is lso the -intercept). Turning point: (, ) Find the -intercept b letting =. -intercept: = = = =± = + or Sketch the grph of the prbol (Figure A). 5 Reflect the portion of the prbol for < < in the -is; mrk the new -intercept (Figure B). (, ) (, ) (, ) (, ) (, ) (, ) = Figure A Figure B Similr to the grphs discussed in the previous sections, the grph of the bsolute vlue function cn be trnsformed through diltions, trnsltions nd reflections. If = f ( ) + c, is the diltion fctor. It diltes the grph from the -is. The lrger is, the thinner the grph. If <, the grph is reflected in the -is. = = = The vlue c trnsltes the grph long the -is. If c >, the grph is moved c units up nd if c <, it is moved c units down. c = c = = + c WORKED EXAMPLE Sketch the grph of = +. THINK Compre the given function with = f() + c, nd write short comment. To sketch the required shpe we first need to sketch = ( stright line). Find the -intercept b letting =. WRITE/DRAW The grph of = is trnslted unit up. Let =. -intercept: = = = ebook plus s Tutoril int-5 Worked emple Find the -intercept b letting =. -intercept: = = = Chpter Functions nd trnsformtions 9

Sketch the line. (, ) (, ) 5 Reflect the portion below the -is in the -is; mrk the new -intercept. (, ) (, ) 6 Move the grph unit up; mrk the new -intercept nd the coordintes of the cusp. (, ) (, ) Absolute vlue functions s hbrid functions An bsolute vlue epression cn be thought of s two seprte epressions, depending on whether it is negtive or positive. + cn be written s ( + ) or ( + ) depending upon the vlue tht tkes. To determine these prticulr vlues of, we need to solve the two inequlities + > nd + <, giving > nd <, respectivel. This gives us domin for the two epressions bove, so we cn write representtion for the bsolute vlue epression s: +, wher ere + = ( +), wher ere < WORKED EXAMPLE This is useful process when used to rewrite n bsolute vlue function tht is to be grphed, becuse it gives us rule for ech prt of the grph in the form of hbrid function. It is lso importnt when needing to differentite function of this tpe in lter chpter. Epress f( ) = 55 s hbrid function, defining the domin of ech prt nd grphing the function. THINK Brek the function into two prts: negtive nd positive prt. Simplif the domin nd function for ech. WRITE/DRAW 5, wher ere 5 f( ) = 5 = ( 5 ), where 5 < < First function: 5 First domin: 5 5 Second function: (5 ) = 5 + Second domin: 5< < ebook plus s Tutoril int-55 Worked emple 5 9 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

Rewrite the function in hbrid form with the two rules with their respective domins. Grph the two functions for the specific domins. 5 The bsolute vlue function cn esil be drwn using CAS clcultor. On the Grph & Tb pge, complete the function entr line s: = bs(5 ) Tick the bo nd tp!., 5 f( ) = 5 + +, wher ere wher ere 5 (, ) 5 < 5 (, ) 5 f() = 5 REMEMBER. The smbol denotes the mgnitude of.. =, if =, if <. To sketch the grph of = f() : () sketch the grph of = f ( ) (b) reflect the portion of the grph which is below the -is in the -is. = EXERCISE F The bsolute vlue function WE9 Sketch the grph of ech of the following, showing ect vlues of intercepts of es. = b = c = 6 d = 6 e = f = ( ) g = h = ( + ) i = [ VCAA 7] EXAM TIP Alws give nswers in ect form unless the question sks for numericl pproimtion (for emple, give our nswer to deciml plces ). [Assessment report 7] Chpter Functions nd trnsformtions 95

ebook plus s Digitl doc Spredsheet Absolute vlue function MC Which of the following functions best describes this grph? A = ( ) B = ( + ) C = + D = E = ( + ) + For ech of the following functions stte the domin nd rnge. = b = + c = d = e = + + f WE Sketch the grphs of ech of the following. = b = + 5 6 c = + = d = + e = f = ( + ) g = h = + i 6 = ebookplus Digitl doc WorkSHEET. ebookplus Interctivit int-7 Trnsformtions with mtrices G j l = = + 8 k = + EXAM TIP Use ruler for liner nd bsolute vlue grphs. [Assessment report 7 VCAA] 5 WE Given the function f ( ) = : rewrite the function s hbrid function with pproprite domins b find f () nd f () c sketch the grph, lbelling n significnt points. 6 Given the function f ( ) = + : rewrite the function s hbrid function with pproprite domins b find f ( ) nd f () c sketch the grph, lbelling n significnt points. 7 The design shown t right is to be embroidered on the outer side of pir of children s socks. The totl length of the design is cm nd its width is 8 cm. If we drw the set of es through the centre of the design, the 6 red section cn be thought of s the bsolute vlue function on restricted domin. Find the rule for the red section nd specif the domin. b Using our knowledge of the trnsformtions, nd the rule for the red section, find the rules for the blue, green nd ellow sections of the design. c Using grphics clcultor, sketch the functions tht were obtined in nd b. Hve ou obtined the right design? 6 Trnsformtions with mtrices Trnsformtions of grphs cn be described using mtrices s n lterntive to function nottion. The trnsformtions tht hve been considered so fr (diltions, reflections nd trnsltions) cn be represented in mtri form. This describes how prticulr point on grph will be moved (or mpped) to resultnt loction b the ppliction of diltion, reflection or trnsltion, or combintion of the three. Remember the definition of trnsformtion is rule tht links ech point in the Crtesin plne to nother point. So the mtri cn be used for n 96 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

97 Chpter Functions nd trnsformtions point on curve, or in fct used to determine the new rule for function tht hs undergone one or series of trnsformtions. The new points or rules re termed imges of the originl. Reflections nd diltions We cn summrise the use of mtrices to mp these trnsformtions, T, of points on curve s follows: (Let ' be the trnsformed vlue of, nd ' be the trnsformed vlue of.) T = = ' ' = represents reflection in the -is. T = = ' ' = represents reflection in the -is. T = = ' ' = represents diltion of fctor of from the -is. T = = ' ' = represents diltion of fctor of from the -is. These opertions cn be combined to represent more thn one trnsformtion, for emple T = = ' ' = represents reflection in the -is, diltion of fctor of from the -is, nd diltion of fctor of from the -is. WORKED EXAMPLE Using mtrices, find the loction of the point ( (, ) under the following trnsformtions of the point (, ): diltion b fctor of from the -is reflection in the -is. THINK WRITE/DISPLAY Construct the correct mtri tht represents the trnsformtions described. T = = ' ' On the Min pge, tp: ) 8 Enter the vlue into the mtri nd multipl the mtri b tpping: 7 then entering the mtri vlues nd pressing: E

98 Write the mtri eqution nd interpret this to nswer the question. = The imge of the point (, ) is t (, ). Mtri opertions cn be done using CAS clcultor, but s the mtri multipliction required here is simple, it is recommended it be done b hnd. The difficult prt is to correctl identif the trnsformtion mtri. Once ou hve done tht it is mtter of performing mtri multipliction. Trnsltions Trnsltions require slightl different process. The trnsformtion mtri is mtri, nd finding the new imge requires ddition of the mtrices rther thn multipliction. Mtrices describing trnsltions re of the form b c. This represents: trnsltion of b units in the positive direction of the -is nd trnsltion of c units in the positive direction of the -is. Note b nd c > : If either of the terms is negtive, the trnsltion is in the negtive direction. A zero entr indictes there is no trnsltion in prticulr direction. So trnsltion of point (, ) cn be described s follows: T b c = = + ' ' = + b + c + WORKED EXAMPLE Find the loction of the point (' (, ') under the following trnsformtions of the point (, ): trnsltion of units in the direction trnsltion of 5 units in the direction. THINK WRITE Construct the correct mtri tht represents the trnsformtions described. T = = + ' ' 5 5 Construct nd solve the pproprite mtri eqution. T = = + ' ' 5 + = 5 Interpret this to nswer the question. + = 5 The imge of the point (, ) is t (, ). Mtri opertions cn be done using CAS clcultor, but s the mtri ddition required here is simple, it is recommended it be done b hnd. Mths Quest Mthemticl Methods CAS for the Csio ClssPd

99 Chpter Functions nd trnsformtions Note it is ver importnt to use the correct mthemticl lnguge, both for the size nd the direction of the trnsformtion. For emple, diltions re b fctor of, in other words multiple of the originl vlue, s distinct from trnsltions, which re described s of n units, which is set distnce. In terms of direction, the epression from the -is cn lso be epressed s prllel to the -is, in the direction or even horizontll. The sme pplies for from the -is. Putting it ll together The forml nottion often used to describe trnsformtion begins s T : T R R, which is sing the trnsformtion tht mps point (, ) to nother point (, ) is. nd then the trnsformtion is described. For emple, trnsformtion involving diltion b fctor of from the -is followed b trnsltion of in the direction nd in the direction could be defined s T : T R R, T(, ) = (, + ), or simpl (, ) (, + ). When more thn one trnsformtion is described, it is known s composition of the trnsformtions. When series of trnsformtions re described, the need to be done in the correct order s stted in the question. WORKED EXAMPLE Find the loction of the point (' (, ') under the following trnsformtion of the point (, ): diltion b fctor of from the -is diltion b fctor of from the -is reflection in the -is trnsltion of units in the direction. THINK WRITE Perform the first three trnsformtions together. Construct the correct mtri tht represents these trnsformtions described. T = = ' ' Construct nd solve the pproprite mtri eqution. T = = ' ' 6 = Now perform the trnsltion. Construct nd solve the pproprite mtri eqution. Note we re using the product mtri from step. T = = ' ' + 9 6 6 + = Interpret this to nswer the question. The imge of the point (, ) is t 9 6, ( ) Remember tht trnsformtion mps n point on curve to nother b the sme rule. Rther thn mpping series of individul points on the sme curve, we cn simpl find new rule under trnsformtion (or series of them) nd use this new rule to determine the loction of n points from the curve described b the originl rule.

WORKED EXAMPLE 5 Write the resultnt eqution from the following trnsformtions of the curve described b = : diltion b fctor of from the -is reflection in the -is trnsltion of units in the negtive direction. THINK Perform the first two trnsformtions together. Construct the correct mtri tht represents these trnsformtions described. T = ' = ' Construct nd solve the pproprite mtri eqution. Now perform the trnsltion. Construct the correct mtri tht represents this trnsformtion described. T = ' = ' + Construct nd solve the pproprite mtri eqution. Note we re using the product mtri from our first eqution in step. WRITE = ' = ' + = 5 The trnsformed vlues of nd re nd. Therefore ' = nd ' =. ebookplus Tutoril int-56 Worked emple 5 6 Epress ' nd ' in terms of nd. 7 Stte the resultnt eqution. = ' nd = ' + = ' = So for n point on the grph of the originl function, =, we cn mp the corresponding point under the trnsformtions bove b substituting the vlues into this trnsformed eqution. Let s hve look t nother emple. This time we will complete reflections/diltions nd trnsltions in the one step. WORKED EXAMPLE 6 Find the imge of the curve with eqution = fter reflection in the -is, followed b diltion of fctor of from the -is, nd then trnsltion b + in the direction. THINK Construct the correct mtrices tht represent the trnsformtions described. T = ' = ' + WRITE nd Mths Quest Mthemticl Methods CAS for the Csio ClssPd

Chpter Functions nd trnsformtions Construct nd solve the pproprite mtri eqution. ' ' + = + The trnsformed vlues of nd re + nd. So ' = + nd ' =. Epress ' nd ' in terms of nd. = ' nd = ' 5 Substitute new - nd -terms for =, nd drop the primes. = = = 6 Stte the resultnt eqution showing ll trnsformtions. = b. represents diltion of from the -is nd diltion of b from the -is.. represents reflection in the -is, nd represents reflection in the -is. b c. represents horizontl trnsltion of b nd verticl trnsltion of c. REMEMBER Trnsformtions with mtrices Identif the diltions nd/or reflections described b the following mtrices. i ii iii iv WE WE Find the imge of the point (, 5) under the bove trnsformtions in question. Find the imge of the grphs of the following equtions under the trnsformtions in i nd ii. = b = 5 c = Identif the trnsltions described b the following mtrices. i ii iii 5 EXERCISE G

5 WE Find the imge of the point (, ) under the trnsformtions given in question. 6 Find the imge of the following equtions under ech of the trnsformtions defined in questions i nd ii. = b = 7 The trnsformtion T : R R which mps the curve with the eqution = to the curve with the eqution = ( 6) +, could hve: A T = + 6 B T = + 6 C T = + D T = + E T = + 6 For the following trnsformtions, where T : R R, stte wht the trnsformtion T represents nd determine the imge of the eqution f( ) =. T = + 6 b T = + c T = + 9 A function g(( ) is mpped to the curve h(( ) = g((( + )) +. Crete mtri eqution tht will mp g(( ) to h(( ). WE5 The following trnsformtions re pplied, in order, to the grph of = : diltion b fctor of from the -is reflection in the -is trnsltion of unit in the direction. Use mtrices to determine the imge eqution under these trnsformtions. b Find the imge of the point (, ) nd check whether this point lies on the curve of the eqution from. WE6 If f( ) = nd g ( ) = +, describe, in order, the trnsformtions ( ) performed to the grph of f ( ) to give g() nd crete mtri eqution which would mp f ( ) to g(). If f ( ) = g((( + )) + nd g ( ) =, find f ( ) in terms of onl, using: n lgebric method without the use of CAS clcultor b mtrices nd CAS clcultor. If f ( ) = g( ) nd g( ( ) =, find f ( ) in terms of onl, using: n lgebric method without the use of CAS clcultor b mtrices nd CAS clcultor. If = h+ ( + ) + 6, find h( ( ) using mtri methods. Mths Quest Mthemticl Methods CAS for the Csio ClssPd

H Sum, difference nd product functions Sum functions A sum function is of the form = f () + g(), or lterntivel = = (f + g)(). Mn functions include two or more terms dded (or = subtrcted) together. For emple the function = + cn be thought of s the sum of the functions = nd =. These grphs cn be drwn b sketching the two individul functions on the sme set of es then dding the -vlues (ordintes) for ech -vlue nd plotting the resulting points. This is useful method when we know the bsic shpe of the individul functions but do not recognise the whole function. We would not use this method for fmilir function such s = +, s we hve lernt ws of sketching this without breking it up into prts. Using the emple in the first prgrph, = +, we = do not recognise the shpe of this function, but we know the two individul functions re the bsic positive prbol nd the hperbol. We could therefore sketch the grph of the = + = prbol nd the hperbol nd dd the -vlues together for corresponding -vlues to obtin points on the curve of the sum function which cn be joined together to obtin the grph of the sum function. Note tht the domin of the hperbol is restricted to R \ {} so the -vlue t = is undefined. As ou cnnot dd n undefined number, this -vlue is lso undefined for the sum function. A generl rule is tht the sum function is onl defined for the domin over which both of the individul functions re defined. The domin of the sum function is, therefore, the intersection of the domins of the individul functions. If h() = f () + g(), then the domin h() = domin f ( ) domin g(). Difference functions A difference function is of the form = f ( ) g(), or lterntivel = (f g)(). It is essentill the sme s sum function ecept tht one of the individul functions is subtrcted from the other. So = could be sketched b the sme method s described bove but insted of dding the -ordintes, we would subtrct one from the other. The domin of difference function is determined in the sme w s sum function. We could etend our rule bove to include difference functions. If h() = f () ± g(), then the domin h() = domin f () domin g(). We cn lso think of difference function s dding negtive nd it could be written = +. With this in mind, n lterntive method of sketching the grph of difference function is to reflect the grph of the second function = = (in this emple, ) in the -is nd then dd the ordintes s for sum function. = Chpter Functions nd trnsformtions

When sketching grphs of sum/difference functions, there re ke points tht cn be found on either individul function to esil identif the vlue of the ordinte of the sum or difference function. These re the -intercepts nd n point of intersection of the individul functions. The -intercept is where the ordinte of tht prticulr function is zero, so the grph of the sum or difference function is ctull the ordinte of the other function for tht vlue of..5 At the intersection, the ordinte of the sum function will be double tht of the two individul functions. For difference function, n intersection of the two.5 individul functions corresponds to on -intercept ( = ) of the difference function. Another useful -vlue to look for is where the.5 grphs of individul functions hve -vlues tht re of the sme mgnitude but one is positive nd one.5.5.5.5 is negtive. This point is n -intercept of the sum.5 function. WORKED EXAMPLE 7 Using ddition of ordintes, sketch the grph of f( ) = =lo log e ( + +)+ +, [, ]. THINK WRITE/DRAW ebook plus s Tutoril int-57 Worked emple 7 Sketch the grphs of = log e ( + ) nd = on the sme set of es over the required domin [, ]. (, ).5 = log e ( + ) (, ) (, log e ()) =.5.5.5.5 Moving from left to right, dd the -coordintes of the two grphs for the ke points nd plot the resultnt points. The ke points re the: end points -intercepts points of intersection. The new points on the grph re mrked b n sterisk. (, ) = log e ( + ) (, ) (, log e ()) =.5.5.5.5.5 Left end points (, ) nd (, ), so the new point will be t (, ). Right end points (, log e ()) nd (, ), so the new point will be t (, + log e ()). -intercepts (, ) nd (, log e ()). Points of intersection (.,.) nd (.5,.5), so the new points will be (.,.88) nd (.5,.). Mths Quest Mthemticl Methods CAS for the Csio ClssPd

Join these points with smooth curve to crete f (). (, ).5 f() = log e ( + ) + (, ) = = loge ( + ) (, log e ()).5.5.5.5 Remove the two individul grphs to leve the sum function. (, + log e ()) (, ) f() = log e ( + ) +.5.5.5.5.5 Product functions A product function is of the form = f ( ) g(), or lterntivel = (fg )(). When grphing product functions, it is useful to grph the individul functions, f nd g, nd for n relevnt vlues of, to identif the -vlues, or ordintes nd multipl these together to obtin the -vlue of the product function. If the -vlue is undefined t prticulr vlue of for either of the individul functions, then the product function is undefined for tht vlue. We cnnot multipl b n undefined number. If h() = f ()g(), then the domin h() = domin f () domin g(). When emining the grph of the two individul functions, it is useful to look t -intercepts nd points where the vlue of either function is ±. The product function will lso hve n -intercept t point where either individul function hs n intercept (s multipling b zero gives zero). At point where function = ±, the product function will hve vlue equl to the vlue of the other function, or its negtive. It is lso useful to observe tht where the individul functions re both bove the -is, or both below the -is, the vlue of the product function will be positive, tht is, bove the -is. This is becuse the product of two positive numbers or two negtive numbers is positive. Alterntivel, where one function is bove nd one below the -is, the vlue of the product function will be negtive, tht is, below the -is. WORKED EXAMPLE 8 If f ( ) = nd g ( ) = +, sketch the grph of f( ) g ( ) = +. THINK WRITE/DRAW Sketch the grphs of f () nd g(). (, ) (, ) = = + Chpter Functions nd trnsformtions 5

Find the domin of f ( ) nd the domin of g(). Dom f = R nd dom g = [, ) Find the domin of f ( ) g(). Dom fg = [, ) Find the -intercepts of both f nd g nd hence find the -intercepts of the product fg. 5 Find the vlues of for which the product is negtive. 6 Find the vlues of for which the product is positive. 7 Find the turning point using CAS clcultor. Round the nswer to deciml plces s pproprite. -intercept for f () is when = nd f ( ) = -intercept for g() is when = nd g() = Hence, the -intercepts for the product re when = nd =. f () is negtive nd g() is positive for (, ), so fg is negtive for (, ). f () nd g() re both positive for (, ), so fg is positive for (, ). The turning point is (,.77). 8 Sketch the grph of the product. = + (, ) (, ) (,.77) REMEMBER. For the sum/difference function, dom(f () ± g()) = dom f () dom g (). The grph of the sum/difference function cn be obtined b using the ddition of ordintes method.. For the product function, dom (f () g()) = dom f () dom g(). Some fetures of the grph of the product function re s follows: () the -intercepts of f () g() occur where either f () or g() hve their -intercepts (b) f() g() is bove the -is where f () nd g() re either both positive or both negtive (c) f() g() is below the -is where one of the functions f () or g() is positive nd the other is negtive. EXERCISE H Sum, difference nd product functions Sketch the grphs of f ( ) = g(( ) + h(( ) using ddition of ordintes, given the following functions g() nd h(). Stte the domin of f () in ech cse. g() =, h() = b g ( ) =, h ( ) = c g() =, h() = d g ( ) =, h h( ( ) = + 6 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

Determine the eqution of g(( ) h(( ) in ech of the following cses then, using ddition of ordintes, sketch the grph of g() h(). g ( ) = +, h ( ) = b g() =, h() = + For ech of the following, find the domin of f ( )g( ). f( ) =,g( g ( ) = b f () = +, g() = + c f( ) =, g ( ) = d f () =, g() = e f( ) =,g( g ( ) = + WE 7 Sketch the grph of f( ) = + for, b the ddition of ordintes, showing the generl shpe nd n smptotes. 5 Two functions re defined s f ( ) = nd g ( ) =. Sketch the grph of ech on the sme set of es for. b Find the smllest possible vlue of given tht the domin of the function h, where h() = (f + g)(), is. c Find f () nd g(), nd hence find h(). d Find f () nd g(), nd hence find h(). e Find f () nd g(), nd hence find h(). f Using this informtion, sketch the grph of h(( ) (on the sme set of es s in ). 6 Given the functions f ( ) =, g(( ) = nd h(( ) = f ( ) + g(( ): Sketch the grph of ech on the sme set of es for. b Find f ( )nd g ( ), nd hence find h ( ). c Find f () nd g(), nd hence find h(). d Find f () nd g(), nd hence find h(). e Find f () nd g(), nd hence find h(). f Wht is the rnge of the function h (in ect form). g Using this informtion, sketch the grph of h() (on the sme set of es s in ). 7 WE 8 Two functions re defined s f ( ) = nd g ( ) =.Le. Let h ( )= f ( g )g ( ). Find the domin of h. b Sketch the grph of ech on the sme set of es. c Find f () nd g(), nd hence find h(). d Find f () nd g(), nd hence find h(). e Find f () nd g(), nd hence find h(). f Wht is the rnge of the function h (in ect form)? g Using this informtion, sketch the grph of h() (on the sme set of es s in ). 8 Sketch the grphs of the functions f( ) = + 5 nd g ( ) = 8 nd use these to find the domin of the function h(( ) = f ( ) +g ( ). On the sme es, sketch the grph of h(( ), including the coordintes of n end points. 9 Use CAS clcultor to view nd sketch the grphs of f ( ) = + nd g ( ) =, for.5.5. Then, without using the clcultor, use these grphs to sketch the grph of h, if h() = (f g)() on the sme set of es. Using the clcultor, check the shpe of the grph ou hve drwn nd use it to identif n significnt points such s intercepts nd cusp points to deciml plces. (You m need to djust our window settings in order to clerl identif these points.) Chpter Functions nd trnsformtions 7

I Composite functions nd functionl equtions Composite functions A composite function is formed from two functions in the following w. If f () = + 5 nd g() = re two functions, then we combine the two functions to form the composite function g(f ( ()) = f () = ( + 5). Tht is, f () replces in the function g(). The composite function reds g of f nd cn be written g f. Another composite function is f (g()) = g() + 5 = + 5. In this cse, g() replces in f (). This composite function reds f of g nd cn be written f g. For the composition function f (g()) to be defined, the rnge of g must be subset of (or equl to) the domin of f,, tht is rn g dom f.. It is esiest to list the domin nd function of both f () nd g() first when deling with composite function problems. For emple: f () = nd g ( ) = : f ( ) g(( ) Domin R [, ) Rnge [, ) [, ) Composite functions cn be rther comple to grph b hnd, so CAS clcultor cn be used for ssistnce when sketching. WORKED EXAMPLE 9 For the pir of functions f ( ) = + nd g ( ) = : show tht f (g ( )) is defined b find f(g( ( )) c stte its domin. THINK Crete tble showing the domin nd rnge of both functions. WRITE f ( ) g(( ) Domin R \{ } R + Rnge R \{} R + For f ( g()) to eist the rnge of g must be subset of f. b Form the composition function f ( g()) b substituting g() into f ( ). c The domin of f ( g()) must be the sme s the domin of g(). Since the domin of g() is R +, it is the domin of f ( g()). rn g() dom f () f (g()) is defined. f(g g( )) = f( ) f(g g( )) = + Domin of f ( g()) = R + 8 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

Functionl equtions Sometimes we re required to solve or nlse equtions tht re in terms of unknown functions, for emple, f ( ) or f (), rther thn being in terms of unknown vribles, for emple or. An emple of the tpe of problem ou might encounter is to find function tht stisfies f ( + ) = f ( ) + f (). Alterntivel, ou might be required to determine if prticulr function stisfies the rule f( ) = f (). Equtions such s f ( + ) = f ( ) + f () re clled functionl equtions. There re generll two ws to solve these tpes of problems: lgebricll or using CAS clcultor. WORKED EXAMPLE Determine if f ( ) = stisfies the eqution f ( + ) = f ( ) f ( ). THINK WRITE Substitute the function into the LHS nd RHS of the eqution seprtel. LHS = f ( + ) = + RHS = f ( ) f () = Simplif the LHS of the eqution to determine LHS = f ( + ) = + if it equls the RHS of the eqution. LHS = f ( + ) = LHS = RHS Answer the question. f () = stisfies the eqution f ( + ) = f () f () WORKED EXAMPLE Determine if g(( ) = stisfies the eqution g( ) = g ( ). THINK On the Min pge, define the eqution b entering the eqution s: Highlight it nd tp: Interctive Define Set: Func nme: g Vrible/s: OK WRITE/DISPLAY To clculte g() nd g(), complete the entr lines s: g() g() Press E fter ech entr. Emine the solutions to determine if the epressions re equl. LHS = g( ) = RHS = g() = Chpter Functions nd trnsformtions 9

Check tht the eqution holds true for ll vlues of b completing the entr line s: g() = g() Then press E. Since the LHS = RHS, the sttement is true nd the eqution holds for ll vlues of. Write the solutions for g( ) nd g() nd nswer the question. g( ) = g() = g( ) = g() When g() = it stisfies the eqution g( ) = g(). If we consider the sme eqution f( ) = f f ( ) for different function, for emple, f( ) =, we obtin two different equtions, f( ) = nd f ( ) =, which re not equl. However, if we define this function on CAS clcultor nd enter the sttement f ( ) = f (), the result is =. This mens this eqution holds true when = but not for n other vlues of. REMEMBER. For the composite function f (g()) to be defined, the rnge of g must be subset of the domin of f.. Furthermore, if f (g()) is defined, the domin of f (g()) equls the domin of g().. For equtions involving lgebr of functions, determine if the eqution is true for the prticulr function b considering the LHS nd RHS of the eqution seprtel to test if the eqution holds true for ll vlues of. EXERCISE I Composite functions nd functionl equtions WE 9 For ech of the following pirs of functions: i show tht f ( g()) is defined ii find f ( g()) nd stte its domin. f () = nd g ( ) = + b f ( ) = + nd g() = + c f () = ( ) nd g() = d f () = nd g() = e f () = ( + )( + ) nd g() = Show tht the function f ( ) = stisfies + f f the eqution f = ( ) + ( ). WE Show tht f( ) = [ VCAA 7] stisfies the eqution f ( ) + f ( ) = + +. f( ) EXAM TIP Show tht questions require detiled working out to be given, so provide s mn steps s possible to ttin full mrks. Write s required t the end of our solution once ou hve shown the problem is correct. [Assessment report 7] Mths Quest Mthemticl Methods CAS for the Csio ClssPd

WE Determine which of i v hold for the following functions: f( ) i f ( ) = f () f () ii f( ) = f( ) iii f () + f () = ( + ) f () v f () = f () f () iv f f = () f( ) f( ) = b f () = c f( ) = d f( ) = e f () = f f () = 5 f() = ( + ) nd g ( ) =, determine the vlues of such tht f (g(( )) eists. 6 If f: R R, where f( ) = nd g:r { } R, where g ( ) = + +, determine if f g nd g f eist nd, if so, find the composition functions. 7 If f: R R, where f( ) = nd g:r R, where g(( ) =, show tht f g is not defined. B restricting the domin of g, find function h such tht f h is defined. 8 Given w(( ) = +, > nd v(( ) =, R +, stte the domin nd rnge of ech function. Hence, find if w v nd v w eist nd, if so, stte their rules including their domins. 9 Show tht the eqution g(( ) = stisfies the eqution g( ) = g(( ). Show tht this sttement is true for ll functions of the form g() = n, where n is n odd nturl number. Show tht g(( ) = stisfies the eqution g(( ) = g(( )g( ). Show tht this eqution is true for ll functions of the form g() = n, where n is nturl number. Given the two functions: f:[, 6] R, f( )= ( ( 6) g: [, 6] R, g() = + 6 find the mimum vlue of f () g() for 6, correct to deciml plces, nd determine the vlues of in this intervl for which the mimum occurs. [ VCAA 7] EXAM TIP Be creful to nswer ll spects of the question (this question requires the -vlue nd the mimum). Alws re-red the question to check ou hve met ll requirements before moving on to the net question. [Assessment report 7] J Consider f:[, ] ] R, f( ) = nd g: R R, g(( ) =. Wht trnsformtions re required to obtin f (g(( )) from f ( )? Modelling People such s scientists, finncil dvisers, business nlsts, economists, sttisticins nd others often hve to del with lrge nd smll sets of dt. Once the dt re collected, we re often interested in finding the rules tht link fetures of the dt. The process of finding such rule is clled modelling nd the rule itself is known s the mthemticl model. When finding the model, the best w to strt is to plot the dt, s the shpe of the grph might suggest the tpe of reltionship between the vribles. Tpes of grphs B recognising the shpe of grph, it is possible to find the rule or mthemticl model tht describes it. Throughout this chpter, severl tpes of grphs hve been investigted. Chpter Functions nd trnsformtions

The prbol: = The grph of cubic function: = The hperbol: = = = The truncus: = The grph of squre root function: = = = Reflections nd trnsltions cn be pplied to ech of these grphs, but the bsic shpe of ech grph remins the sme. WORKED EXAMPLE Mtch ech of the following grphs with the pproprite model. i = ii = iii = iv = v = b c d e THINK Mtch the grphs using the informtion in the summr bove. WRITE i is prbol; it mtches grph b. ii is cubic; it mtches grph e. iii is hperbol (the grph is in opposite qudrnts); it mtches grph c. iv is truncus (the grph is in djcent qudrnts); it mtches grph. v is squre root function; it mtches grph d. Mths Quest Mthemticl Methods CAS for the Csio ClssPd

WORKED EXAMPLE 5.5.5. 5 5.59 The dt in the bove tble ectl fit one of these rules: = =, =,, = or =. Plot the vlues of ginst. b Select the pproprite rule nd stte the vlue of. THINK WRITE/DRAW Plot the vlues of ginst. 5 5 b Stud the grph. It ppers to be squre root curve. Write the pproprite rule. To find the vlue of : select n pir of corresponding vlues of nd. (Since we need to tke squre root, the best to choose is the one where is perfect squre.) Substitute selected vlues into the rule nd solve for. We need to mke sure tht the selected rule is the right one. Replce with.5 in the rule. 5 Substitute the vlues of from the tble into the formul nd check if ou will obtin the correct vlues of. 6 As the vlues of obtined b using the rule mtch those in the tble, the choice of model is correct. b Assume tht =. Using (,.5): 5. = = =.5 Verifing: = 5. (, ): = 5. = (,.5): = 5. =.5 (,.): = 5. =. (, 5): = 5. = 5 (5, 5.59): = 5. 5 = 5.59 The rule tht fits the dt is =, where =.5. Chpter Functions nd trnsformtions

The process of fitting stright line to set of points is often referred to s regression. Sttisticl dt is esiest to del with in liner form. If the dt is not liner, then liner reltionship cn still be found b trnsforming the scle. A regression line cn then be fitted. For emple, = m + c is hperbol. However, if we substitute X for, the rule becomes liner: = mx + c. The grph of versus X will be stright line with grdient of m nd -intercept of c. These vlues (m nd c) cn then be estblished from the grph nd thus the hperbolic model cn be determined. Note: In qudrtic reltionship, X is substituted for ; in cubic reltionship, X is substituted for. WORKED EXAMPLE It is believed tht, for the dt in the tble below, the reltionship between nd cn be modelled b = + +b+ + c. 5 5. 8.6.8. Plot the vlues of ginst. b Clculte the vlues of, b nd c (correct to deciml plces) nd write the equtions. THINK WRITE/DISPLAY On the Sttistics pge, enter -vlues into column A nd -vlues into column B. Lbel ech column nd respectivel. To grph the dt, tp: SetGrph Setting Set: Tpe: Sctter XList: min\ YList: min\ Freq: Mrk: squre Set Use 6 if the window needs djusting. Mths Quest Mthemticl Methods CAS for the Csio ClssPd

b Tp: Clc Qudrtic Reg Set: XList: min\ YList: min\ OK Interpret the vribles given on the screen. =.5 b =. c =.96 =.5. +.96 Correct to deciml plces. If the reltionship between the vribles is not given, we hve to mke n ssumption of model from the grph of the dt. We then hve to trnsform the dt ccording to our ssumption. If the ssumption ws correct, the trnsformed dt, when plotted, will produce perfectl stright, or nerl stright, line. Note: In this section we will consider onl the rules of the tpe = + b, = + b, = + b nd so on (we will not llow for horizontl trnsltion), so tht the pproprite substitution cn be mde. WORKED EXAMPLE 5 5 6 5 6 Estblish the rule connecting nd tht fits these dt. THINK Using either grph pper or CAS clcultor, plot ginst. WRITE/DRAW 5 5 5 5 5 6 The sctterplot ppers to be hperbol. Write the pproprite formul (remember tht we do not consider horizontl trnsltions in this section). Assumption: = +b Chpter Functions nd trnsformtions 5

Check our ssumption: prepre new tble b replcing vlues of with (leve the vlues of unchnged)..5..5..7 5 6 Plot ginst. 5 5 5 5...6.8 5 Comment on the shpe of the grph. The grph is ver close to stright line, therefore the ssumption of hperbolic model is correct. 6 If we replce with X, the rule becomes = X + b, which is the eqution of the stright line, where is the grdient nd b is the -intercept. These ( nd b) cn be found from the grph s follows: drw in the line of best fit. 5 5 5 5 7 Write the formul for the grdient. m =...6.8 X 8 Select n points on the line. Using (.7, ) nd (, 5): 9 Substitute the coordintes of the points into the formul nd evlute. 5 m =. 7 = 5 8. =. Write the vlue of. Since is the grdient, = m =.. Write the generl eqution of the stright line. Substitute the vlue of m nd the coordintes of n of the points, s (, 5) into the eqution. Solve for c. (Alterntivel, red the -intercept directl from the grph.) = m + c 5 =. + c 5 =. + c c = 5. =.88 Stte the vlue of b. Since b is the -intercept, b = c = 88..88 5 Substitute the vlues of nd b into = + b to obtin the rule tht fits the given dt. The rule for the given dt is: =. + 88. 6 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

REMEMBER. Modelling is the process of finding the rule tht fits the given dt.. The rule itself is clled mthemticl model.. The best w to strt modelling is to produce sctterplot of the originl dt.. Use the sctterplot of the dt to mke n ssumption of the model of the reltionship. It should be of the tpe = + b, = + b, = + b nd so on. To test the ssumption, trnsform the dt ccordingl. If the ssumption is correct, the trnsformed dt when plotted will produce stright, or nerl stright, line. 5. To find the vlues of nd b in the model, drw line of best fit; is the grdient of the line nd b is the -intercept. EXERCISE J Modelling WE Mtch ech of the grphs with the pproprite model: i = + b ii = + b iii = + b iv = +b v = + b b c d e ebook plus s Digitl doc Spredsheet 76 Modelling WE The dt in ech of the tbles below ectl fit one of these rules: =, =, =, = or =. For ech set of dt, plot the vlues of ginst nd drw the grph. Select the most pproprite rule, nd find the vlue of. 8..... 8. b 6 6 5 Chpter Functions nd trnsformtions 7

c 5 5.8.5.5.8 d.5.5..6.96.6 e 5 5.5.5.5 f.5.5.5 MC Which of the grphs below could be modelled b = +b? i ii iii iv v A i onl B i, ii nd iii C iv nd v D i, ii nd iv E i, iv, nd v WE It is believed tht for the dt in the tble below, the reltionship between nd cn be modelled b = + b. 5..9.5 9 6.8 Plot the vlues of ginst. b Plot the vlues of ginst nd drw the line of best fit. c Find the vlues of nd b nd hence the eqution describing the originl dt. 5 The tble below shows the vlues of vribles, nd. 6 8.5.5. Estblish the mthemticl model of the reltionship between the vribles, if it is known tht it is of the form = + b. 6 The tble below shows the results, obtined from n eperiment, investigting the frequenc of sound, f,, nd the length of the sound wve, λ. λ..5 5 8 f 68 7 5 8 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

Plot f ginst λ. b From the following reltionships select the one which ou think is suggested b the plot: f = λ, f =, f = λ. λ c Bsed on our choice in prt b, plot f ginst either λ, or λ, drw in the line of best λ fit nd use it to find the rule tht connects the vribles. 7 For her science ssignment, Rchel hd to find the reltionship between the intensit of the light, I,, nd the distnce between the observer nd the source of light, d. From the eperiments she obtined the following results. d.5.5.5 I 7 68 7 Use grphics clcultor to plot the vlues of I ginst d. Wht form of reltionship does the grph suggest? b Nthn (Rchel s older brother) is phsics student. He tells Rchel tht from his studies he is certin tht the reltionship is of the tpe I = d. Use this informtion to help Rchel to find the model for the required reltionship. 8 WE 5 The tble below gives the vlues of vribles, nd. Estblish the rule, connecting nd, tht fits these dt. 5 7 9 7 9 9 Joseph is finncil dviser. He is studing the prices of shres of prticulr compn over the lst months. Months 5 6 7 8 9 Price, $ 6. 6.8 7.5 8. 8.5 8.9 9. 9.65.. Represent the informtion grphicll. b Estblish suitble mthemticl model, which reltes the shre price, P, nd the number of the month, m. c Use our model to help Joseph predict the shre price for the net months. ebook plus s Digitl doc Investigtion Gol ccurc Chpter Functions nd trnsformtions 9

SUMMARY Grphs of the power functions Nme Eqution Bsic shpe Domin Rnge Specil feture Prbol = ( b) + c (b, c) R If > c If < c Turning point t (b, c) Cubic = ( b) + c (b, c) R R Sttionr point of inflection t (b, c) Hperbol = b + c or = ( b) + c c b R \{b} R \{c} Horizontl smptote = c, verticl smptote = b Truncus = ( b) + c or = ( b) + c c b R \{b} If > > c If < < c Horizontl smptote = c, verticl smptote = b Squre root = b + c or = ( b) + c (b, c) b If > c If < c End point t (b, c) The eqution for n grph = f () bove cn be written in the generl form: = f ( b) + c. This form cn be used to describe trnsformtions of ll of the functions considered. For ll of the bove functions:. is the diltion fctor: it diltes the grph from the -is.. When n eqution for these tpes of grphs is put into its generl form of = f ( b) + c, the horizontl diltion cn be described in terms of verticl diltion.. If <, the bsic grph is reflected in the -is.. f (b ) or f ( + b) is the reflection of f ( + b) in the -is. 5. b trnsltes the grph b units long the -is (to the right if b >, or to the left if b < ). 6. c trnsltes the grph c units long the -is (up if c >, or down if c < ). Mths Quest Mthemticl Methods CAS for the Csio ClssPd

Chpter Functions nd trnsformtions To put equtions into generl form: If the coefficient of 7. is number other thn, to find the vlue of b nd, the eqution should be trnsposed to mke the coefficient of equl to. For emple, = ( + 5) + =[( [ )] + + ( ( ( )] 5 )] = + 5 ( ) ( ) + ( + 5 =9 9 + 5 ( ) ( ) + ( + 5 Hence, = 9, b = 5. The bsolute vlue function = mens =, if nd =, if < = Domin: R Rnge: R + {} To sketch = f ( f ( f ) : Sketch the grph of. = f ( f ( f ). Reflect the portion of the grph tht is below the. -is in the -is. Or: Epress the function in hbrid form with specific domins where the bsolute vlue epression is positive. nd negtive. Sketch ech rule for the specified domin.. For functions of the form = f ( f ( f ) + c, nd c hve the sme impct on the grph of the bsolute vlue function, s on the grphs of ll other functions discussed in this section. Trnsformtions with mtrices The use of mtrices to mp trnsformtions of points nd equtions cn be summrised s follows, where (, ) is the imge of the point (, ) under the trnsformtion. T = = ' ' = represents reflection in the -is. T = = ' ' = represents reflection in the -is. T = = ' ' = represents diltion of fctor of from the -is. T = = ' ' = represents diltion of fctor of from the -is. Trnsformtions cn be combined to represent more thn one trnsformtion. For emple, ' ' = + = + + = + = +

describes the following: diltion b fctor of from the -is, diltion b fctor of from the -is, reflection in the -is, horizontl trnsltion of + nd verticl trnsltion of +. Sum nd difference of functions For the grph of the sum/difference function, dom (f () ± g()) = dom f () dom g(). The grph of the sum/difference function cn be obtined b using the ddition of ordintes method. For the product function, dom (f ()g()) = dom f () dom g(). Some fetures of the grph of the product function re s follows: the -intercepts of f ()g() occur where either f () or g() hve their -intercepts f ()g() is bove the -is where f () nd g() re either both positive or both negtive f ()g() is below the -is where one of the functions f () or g() is positive nd the other is negtive. Composite functions nd functionl equtions For the composite function f (g()) to be defined, the rnge of g must be subset of the domin of f. Furthermore, if f (g()) is defined, the domin of f (g()) equls the domin of g(). Equtions involving lgebr of functions, for emple f ( ) = f (), re generll tested to determine if the re true for prticulr functions. To determine if n eqution is true for prticulr function, consider the LHS nd RHS of the eqution seprtel to determine if the eqution holds true for ll vlues of. Alterntivel, ou m find prticulr -vlue for which the eqution does not work; tht is, counteremple. These tpes of equtions cn be investigted b defining the functions on the CAS clcultor nd then testing the lgebric function eqution. Modelling Modelling is the process of finding the rule (mthemticl model) tht fits the given dt. To model:. Plot the originl dt on grph pper or use CAS clcultor.. Mke n ssumption of the model.. Trnsform the dt in ccordnce with our ssumption.. Check the ssumption b plotting the trnsformed dt (if correct, the grph will be stright or nerl stright line). 5. Drw in line of best fit. 6. Find the eqution of the line ( = m + c). 7. Replce in the eqution with the trnsformed vrible (for emple,, ). Mths Quest Mthemticl Methods CAS for the Csio ClssPd

CHAPTER REVIEW SHORT ANSWER For the function ) : stte the coordintes of the turning point b stte the domin nd rnge c sketch the grph. The grph of = + c hs verticl b smptote t = nd horizontl smptote t =. Find the vlues of b nd c. This grph then undergoes the following trnsformtions: reflection in the -is diltion b fctor of from the -is horizontl shift of units right. b If the intersection of the two grphs is t (m, ), find the vlue of m. c Hence find the eqution of the trnsformed grph. Epress the function f( ) = 6 5 in the form + b +. Hence describe the trnsformtions + c required to produce this curve from the grph of =. The grph of cubic function hs sttionr point of inflection t (, ). It cuts the -is t =. Find the eqution of the grph. 5 The grph of = ws dilted b the fctor of from the -is, reflected in the -is nd then trnslted units to the left nd unit down. Stte the eqution of the smptotes. b Stte the domin nd rnge. c Stte the eqution of the new grph. d Sketch the grph. 6 Stte the chnges necessr to trnsform the grph of = into the one shown. b Find the eqution of the grph. 7 The domin of truncus is R\{}, the rnge is (, ) nd the grph cuts the -is t =. Find the eqution of the function. 8 The bsic squre root curve ws reflected in both es nd then trnslted so tht its intercepts t the es were (, ) nd ( 5, ). Find the size nd the direction of the trnsltions; hence, find the eqution of the new grph. 9 Sketch the grph of = ( + ), clerl showing the coordintes of the cusps, the intercepts with the es nd the position of the smptotes. b Stte the domin nd rnge of the grph in. The grph of f: : [ 5, ] R where f ( ) = + 6 + 9 is shown. On the sme set of es sketch the grph of = / /f ()/. b Stte the rnge of the function with the rule = / /f ()/ nd domin [ 5, ]. [ VCAA 6] The point (, ) undergoes trnsltion given b the mtri to (, ). Find nd b nd describe b the trnsformtions involved. The point (, ) undergoes series of trnsformtions given b the mtrices b nd then to ( 7, ). Find the vlues of nd b. b Find the imge under the trnsformtions of: i = ii = + A point on curve (, ) undergoes trnsformtion descibed b to (', '), where is rel constnt such tht >. Chpter Functions nd trnsformtions

Find the vlues of nd in terms of, ' nd '. b If the point is on the curve =, find the imge of the curve in terms of under this trnsformtion. c If the point (, 6) is on the trnsformed curve, find the vlue(s) of nd hence the rule of this imge. The grph of the function f : (, ) R, f ( ) = + is shown below. (, ) (, ) f() = +.5.5.5 Let g() = f () +, nd sketch this grph on the sme set of es. Hence, sketch (f + g)(). 5 The grph of the function f:( :, ), f ( ) = / / is shown below. f() = (, ) (, ) Let g() = f () + nd sketch the grph of this function on the sme set of es. Hence, sketch f (g()). 6 Let f ( ) = determine which of the following reltionships re true. f () f ( ) = f () f () b f () f () = f () + f () c f ( ) + f ( ) = f () + f () d f () f ( ) = f () + f () 7 The dt in the tble below ectl fit one of these models: =, = or =. 5 5 5 6.5.5.6 Plot the vlues of ginst nd use the sctterplot to choose suitble model. b Plot the vlues of ginst either, or (depending on our choice in prt ). Did ou choose the right model? Eplin our nswer. c Find the vlue of. MULTIPLE CHOICE The eqution of prbol is given b = m ( + ), where m >. The increse in m will result in: A the grph being thinner B the grph being wider C the increse of the domin D the increse of the rnge E the grph being shifted further to the right The coordintes of the turning point of the prbol = ( + 6) re: A (6, ) B ( 6, ) C (, ) D (, ) E (, ) The grph of = (b b ) + is dilted in the direction b the fctor of: A B b C D b E b The grph of = ( + ) hs sttionr point of inflection t: A (, ) B (, ) C (, ) D (, ) E (, ) 5 If f( ) = +, then f ( ) + will hve: A the horizontl smptote = B the horizontl smptote = C the horizontl smptote = D the verticl smptote = E the verticl smptote = 6 The eqution of the grph shown is likel to be: A = B = + C = + D = + E = + 7 If the grph of = is reflected in the -is, trnslted units to the right nd units up, the resulting grph would hve the eqution: A = + B = + b Mths Quest Mthemticl Methods CAS for the Csio ClssPd

C = D = + E = + 8 Which of the following is not true for the grph t right? A The verticl smptote is =. B The horizontl smptote is =. C The domin is R\{}. D The rnge is R\{ }. E The vlue of the -intercept is greter thn. 9 To obtin the grph shown, we need to: A trnslte the grph of = one unit to the left nd reflect in the -is B trnslte the grph of l = l one unit to the left nd reflect in the -is C trnslte the grph of = left nd reflect in both es one unit to the D trnslte the grph of = one unit to the left nd dilte it in the direction E none of the bove The eqution of this grph is of the form: A = m+ n, > B = m + n, > C = m+ n, < D = m + n, < E = + m + n, < The eqution of this grph could be: A =d d B = + d C =d d D =c c E = c (m, n) d c b The domin of the function f( ) = + is: A [, ) B [, ) C [.5, ) D [, ) E [, ) The rnge of the function = is: A (, ) B (, 5] C [, ) D (, ) E (, ] The eqution of the grph shown in the digrm below is best described b: A = + + B = + C = + D = E = + 5 The vlue(s) of k for which k + = k + re: A onl B C or D or E or [ VCAA 6] 6 Under the trnsformtion T : R R of the plne defined b T = +, the imge of the curve = hs the eqution: A = + B = + C = + D = + E = 7 If g:[, ] R, where g(( ) = ( ), nd h:(, ] R, where h(( ) =, then the function f ( ) such tht f ( ) = g(( ) h(( ) is defined b the rule: A f: R R, where f () = ( ) B f: : (, ] R, where f () = ( ) C f: : [, ] R, where f () = ( ) D f: : (, ] R, where f () = ( ) E f: : [, ] R, where f () = ( ) ( + ) 8 Given the function f( ) = then f h could eist if h(( ) ws defined s: A h() = + B h() = C h() = D h() = + E h ( ) = Chpter Functions nd trnsformtions 5

9 The dt in the following tble ectl fit one of these models: =, = or =... 8. 9. The vlue of is: A. B. C.7 D.9 E. For certin dt the vlues of were plotted ginst nd the line of best fit ws drwn s seen on the digrm below. The model tht reltes the vribles nd is: A = B = 9 + (, 9) C = 9 D = (., ) E = EXTENDED RESPONSE The grph of = f ( ) is shown t right. Sketch the grph of ech of the following functions on the sme set of es with the originl grph nd give the coordintes of the points A, B, C nd D. i = f () ii = f ( ) iii = f ( ) iv = f () + v = f () vi = f ( + ) b M, fbric designer, wishes to use the curve of = f () (red) to crete wv pttern s shown in the digrm t right. If she wnts the wves to be units prt verticll, suggest the best w she could lter the eqution of = f (). (Remember fbric hs fied width!) Consider the function f:r R, f() = ( ) ( ) +. The coordintes of the turning points of the grph of = f() re (, ) nd (b, 5 7 ). Find the vlues of nd b. b Find the rel vlues of p for which the eqution f() = p hs ectl one solution. c For the following, k is positive rel number. i Describe sequence of trnsformtions tht mps the grph of = f () onto the grph of ii Find the -is intercepts of the grph of = f in terms of k. k d Find the rel vlues of h for which onl one of the solutions of the eqution f ( f + h) = is positive. [ VCAA ] The grph of the function f: (,.5) (.5, ) R, f( ) = is shown below. (.5, ) (.5, ) f() = (,.5) (,.5).5.5.5.5.5.5 D(, 6) B A C(, ) 7 units prt = f. k EXAM TIP Students must ensure tht the show their working if question is worth more thn one mrk, students risk losing ll vilble mrks if onl the nswer is given nd it is incorrect. The instructions t the beginning of the pper stte tht if more thn one mrk is vilble for question, pproprite working must be shown. When students present working nd develop their solutions, the should use conventionl mthemticl epressions, smbols, nottion nd terminolog. [Assessment reports nd 7 VCAA] 6 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

The grph of = f () is to be trnsformed to become the grph of = f ( ) +. Describe these two trnsformtions. b Crete mtrices to represent these trnsformtions. c Use these mtrices to find the imges of the points (, ) nd (,.5) under these trnsformtions, nd use these vlues to deduce the imges of the points (, ) nd (,.5). d On the sme es, sketch this trnsformed function, showing the coordintes of the four points from c bove. e Using n method, find rule for this trnsformed function. A proposed section of ride t n musement prk is to be modelled on the curve = ( 6 + 5 ) 5, where is the height (in metres) of the ride bove ground level nd is the horizontl distnce (in metres). The -is represents ground level. It will trvel through tunnel from A to C; B is the lowest point in the tunnel nd D is the highest point on the ride. D A B C Find the horizontl distnce from A to E. b Find the gretest depth below ground level nd the mimum height bove ground level tht the rollercoster will rech in this section (correct to deciml plces). c Describe the impct tht diltion b fctor greter thn from the -is would hve on: i the mimum depth nd mimum height from b ii the point t which the rollercoster would emerge from the tunnel iii the grdient of the slope t this point. 5 Len nd Ale re plnning to bu new house. The ve been wtching the prices of -bedroom houses in specific re, where the wnt to live, for the whole er. During ech month the collected the dt nd then, t the end of the month, the clculted the verge price for tht month. The results of their clcultions re shown in the tble below. (The prices given re in thousnds of dollrs.) E Month 5 6 7 8 9 Price 8 55 6 66 7 7 7 75 7 7 7 Plot the prices ginst the months. Wht model does the grph suggest? b If the model of the form = ( b) + c is to be used for these dt, wht is (judging from the grph) the most suitble vlue for h? c Plot the vlues of (the prices) ginst ( b), where b is the vlue ou ve selected in prt b. Comment on the shpe of the grph. d Drw line of best fit nd find its eqution. Hence, stte the vlues of nd c in the model. e Write the eqution of the model. f According to the Rel Estte Institute, the propert mrket is on sted rise (tht is, the prices re going up nd re likel to rise further). Do the dt collected b Len nd Ale support this theor? g Use the model to predict the verge price for the net months. Chpter Functions nd trnsformtions 7

h Len nd Ale were plnning to spend no more thn 5 for their new house. Severl months go the prices were in their rnge, but the could not find wht the wnted. If the prices re going to behve ccording to our model, how long do the hve to wit until the prices fll bck into their rnge? 6 An egle sors from the top of cliff tht is 8. metres bove the ground nd then descends towrds unsuspecting pre below. The egle s height, h metres bove the ground, t time t seconds cn be modelled b the eqution h = 5 +, where t < 5 nd is constnt. t 5 Find the vlue of. b Find the egle s height bove the ground fter i 5 seconds ii seconds. c After how mn seconds will the egle rech the ground? d Comment on the chnges in speed during the egle s descent. e Sketch the grph of the eqution. After seconds, the egle becomes distrcted b nother bird nd reches the ground ectl seconds lter. For this second prt of the journe, the reltionship between h nd t cn be modelled b the eqution h = (t ) + c. f Find the vlues of nd c. g Full define the hbrid function tht describes the descent of the egle from the top of the cliff to the ground below. ebook plus s Digitl doc Test Yourself Chpter 8 Mths Quest Mthemticl Methods CAS for the Csio ClssPd

ebookplus ACTIVITIES Chpter opener Digitl doc Quick Questions: Wrm up with ten quick questions on functions nd trnsformtions. (pge 57) A Trnsformtions of the prbol Digitl docs Spredsheet : Investigte trnsformtions. (pge 59) Spredsheet 8: Investigte the qudrtic function in power form. (pge 6) B The cubic function in power form Digitl docs Spredsheet 5: Investigte the cubic function in power form. (pge 65) Spredsheet 6: Investigte grphs of functions. (pge 7) C The power function (the hperbol) Tutoril WE 9 int-5: Wtch worked emple on sketching the grph of hperbol. (pge 7) Digitl docs Spredsheet 5: Investigte the hperbol. (pge 7) Spredsheet 6: Investigte grphs of functions. (pge 7) WorkSHEET.: Find the domin, rnge, coordintes of turning points nd equtions of smptotes of vrious grphs. (pge 78) Histor of Mthemtics: Investigte the histor of mjor curves. (pge 78) D The power function (the truncus) Digitl doc Spredsheet 6: Investigte grphs of functions. (pge 85) E The squre root function in power form Tutoril WE 6 int-5: Wtch worked emple on implied domin nd rnge. (pge 88) Digitl doc Spredsheet 6: Investigte grphs of functions. (pge 9) F The bsolute vlue function Tutorils WE int-5: Wtch worked emple on sketching the grph of n bsolute vlue function. (pge 9) WE int-55: Wtch worked emple on epressing n bsolute vlue function s hbrid function. (pge 9) Digitl docs WorkSHEET.: Identif trnsformtions, stte domin nd rnge, sketch grphs of power functions nd bsolute vlue functions. (pge 96) Spredsheet : Investigte grphs of bsolute vlue functions. (pge 96) G Trnsformtions with mtrices Interctivit Trnsformtions with mtrices int-7: Consolidte our understnding of using mtrices to trnsform functions. (pge 96) Tutoril WE 5 int-56: Wtch how to use mtrices to determine the resultnt eqution fter trnsformtions. (pge ) H Sum, difference nd product functions Tutoril WE 7 int-57: Wtch how to use ddition of ordintes to sketch the sum of two functions. (pge ) J Modelling Digitl docs Spredsheet 76: Investigte modelling with functions. (pge 7) Investigtion: Gol ccurc. (pge 9) Chpter review Digitl doc Test Yourself Chpter : Tke the end-of-chpter test to test our progress. (pge 8) To ccess ebookplus ctivities, log on to www.jcplus.com.u Chpter Functions nd trnsformtions 9