FUNCTIONS A A function is a rel ation in wh ich each el ement of the d oma in (x value - in dependen t va ri a bl e) is pa ired w ith only one e leme nt of the nge (y value ­ d epe n de nt va riable) B A re lation can be tested to see if it is a funct ion by the vertical lin e test Draw a vertical line throug h a ny g raph, and if it hi t an x- value more than on ce, it is not a functi on (1-4) b b ­ 4ac = 0, exactly one rea l r ot c b - 4ac < 0, no real roots (two disti nct imaginary roots) I) Example l:f(x) =x2_4x+ I use f( x)= [!]E8rnrn FE A function Not a function A function A fu nction C Linear funct io ns ta ke the form: f(x) = mx+b, or y = m x+b where m = the lope, and b = the y-intercept Examp le: f(x ) = x-l , the s lo pe i 4/1 (ri e ove r run), and th e y-intercept is -I D he dis tance between two points o n a lin can be found j( X - u ing the distance form ul a, d = XI ) + ( Y2 - YI )2 X I) • Vd ) (Y + ' F The s tan dard form ofa linear functi o n is =Ax + By + C The lope is m = -A/B, and the y-intercept is -C/B G The zeros of a fun ct io n a re fo und by setting y to O and solvi ng for x I E xam ple 1: f(x) = 4x-1 (5) Exam p le 2: f(x) = 6, thi s func tio n has no zero, and is a h ri z ntal lin e th rough +6 o n the y-axi (6) 3, Exam ple 3: x = 4, th i i not a function, because the re is a verti ca l line through +4 on the x-axi , g ivi ng a n in fi n ite set of va lu es for y (7) X;~~5 ffi the discriminan t is > 0, the re are two rea l roots (15) 2) Example 2: f(x) = 2x2 + 2x + I us ing b - ac = -4, since the d isc rimin ant is < 0, there are two imaginary root (16) 3) Example 3: f(x) = x2 + 2x + I us ing b - 4ac = 0, · th ' I 17 ere IS one rea I' \ ( ) since t1le d ISCnl11l11ant IS = g(x) J RatIOnal fun ction s take the fo ri: f(x ) = b (x) I The pare nt fu nction i f(x) = X · ~2 x- + (21) E E / x= 1.2765957 y=-2 106855 Two real solutions E E a Find th e sum : (f + g)(x), x + + ~ ( x + ) (x - 4) + x X2 _ X _ x­ x- = x - ' and x f b Find the di fference: (f - g)(x), x + - ~ (x + 2) (x - ) - x x _ x _ x­ x- = x_ • an d x f Example 2: Gi ven f(x) = x + 2, g(x) = x: a Find the product: (fxg)(x), (x + 2)( x : ) = x + 2x -x=-;t an d x f b Find the quot ient: ( X - 4) Example 2: f(x) = 2xJ + x2 - 2x + 3, thi function has one rea l zero at x = -1.17, and two non -rea l roots (8) (x+2) (.-x- = t )(X), x;x+_2 = x2 -2x - S x x~930B511 , andx f O : + 2( 4) x ( x - 4) +2 Example: Gi en f(x)=x+2, g(x)= Find IfogJ(x): f ( x : ( _ x_ + ) x - +2 = + 2) = x + 5x - 16 , and x f x- M lnve r se function s : If og l(x) = Igo fj( x) - b ± jb - 4ac x=-1.010638 2a can y=2.9737903 Example: Given f(x)=2x - 4, g(x) = · +4 '-2X+4)_- (X-2+4) - -_x, Ifog l(x) -_f ( x 0265958 y~.OlO0806 A rationalfunction with asymptotes at the x & y axes F O E ~, be u ed to f ind the roots of a ll quadratic eq uations The a lue under th sq uare root symbo l is ca ll ed the d iscrim in an t It te ll s u the type of roots o f a quadratic equat ion ( 2X and IgO fj( x) = di st inct rea l roots - 4) x=-.5053192 y=.47379034 Two imaginal)! y=.00478157 L c om P.osition of functions:lfog l(x) = f(g(x» E !J Example : f( x) = x + I this fUllction has'two nonreal roots, (9) x= 10638298 l Qua dra tic funct ions take the form : f(x) = ax + bx + c y=1.0113173 I 111e graph ofa quadrati function i called a parabola (10) o E ome parabolas ar e qu adrat ic eq uation , but not t q uadratic fu nctions (II) Quadrati c functions or equat ions can have one real so luti on two rea l solutions, or no real so luti on (12-14) Th vertex of a parabo la i ca ll ed its critical p oi nt EE ra x=1.3829787 y=-2.671371 Two real roots, (.26,0) & (3.73,0) Example I: Given f(x) = x + 2, g ( x) == x: Example l: f( x) = 2x4 + x + X + 10, has a degree of 4, there are fo ur roots (so lutions) to th is po ly nomia l t lVO y=-.0100807 One real solullOn (f) EE tE a b - 4ac > 0, x = -(-4) f< - _ 4)/2 = 732, and - f< - - 4) /2 = 267 s ince K Oper ations of fu nctions: I Sum: (f + g)(x) = f(x) + g(x) Diffe rence: (f - g)(x) = f(x) - g(x) P rod uct: (f x g)(x) = f(x) x g(x) f( x) QuotIent: g (x) = g (x )' g( x) f No zeros + cx n-2 ••• + d x + e I When the hi g he t power o f the func ti on is a n odd intege r, there is at least one real zero When the h ighe t powe r is a n even integer, there may be no real zeros Both type can have imag inary roots of the form a + bi x=-1.170213 T h hi ghest p wer of a po lyno mi a l w ith o ne vari able is y=.0100806 ca ll ed it degr ee = x = -(-4) + Exa mple 3: f(x) = H Polyno mial function s take the form : f(x) = ax n + bx n- I ) The quadratic equat ion f(x) 2a Th e g raph of these functi ons cons ist of two pan s, one in quadrant I, and one in quad rant 3 The bran hes of rational functi on approac h line call ed asy mptotes (18) x x=-.9574468 Exa mple I: f(x) = x + (19) y=7.9737903 No real solutions Exam ple 2: f(x) = x (20) E The mid-point of a line segment can be found us ing the ( X2 + m id -po lllt for m ula, -b ± /b _ 4ac +4 = x x=.02659573 y=-.0100806 The asymptotes are the axes y1=x~2x+1 One real rOO! (-.93,0) N Families of functions : Graphs of fu nction fami lies Changes in va lues of the parent affect the appearance of the paren t g raph A par ent g p h is the basic grap h in a fam ily II the other fa mily m mbers move up, down, left right, o r turn based on changes in va lues I Polynomia l fun ctions 1: A bsolute value fu nctions: a f(x) = Ixl (38) a f(x) = x (22) b f( x) = 2x2 (23) b f(x) = -I xl (39) c f(x) = 5x2 (24) c f{x) = 12xl (40) _ d.( f(x) = 1.5xl (41) d.f(x) - -x (25) e f(x) = Ix + 21 (42) e f(x) = x + (26) f rex) = Ix - 21 (43) f f(x) = x ­ (27) g f(x) = Ixl + (44) g r(x) = (x + 2)2 (28) h f(x) = Ixl- (45) h.r(x) = (x - 2)2 (29) I>olynom ial function s 2: a f(x) = x3 (30) b f(x) = _x3 (31) c f{ x) = x3 + (32) d.f(x) = x3 - (33) e f( x) = 2x3 (34) f f( x) = 5x3 (35) g f( x) = (x + 2)3 (36) h.f(x) = (x - 2)3 (37) A Rectangula r coordinates arc o f the form (x,y), and arc pl otted on the Cartes ian coord in ate syste m B Poin ts are p lotted w ith two va lues, one the absci sa and the other th ordinate C T he absc issa i the x-va lue, call ed the domain and th e ord inate is the E S f] y- va lue, called the range /P D Many di ffe rent shape and func tion can be drawn on the Ca rt e ian system E Here is a g ive n ang le, orig inating fr the III -axi and rotating counter-clockwise Th is ang le is re prese nted by a P(4.6) li ne segment ori ginati ng at t.he o ri gin, and extend in g to a given poin t (P) (46) R Pola r coo rdinates are o f the fo rm P( r, 9), where r E B ,p = the I ( radiu , the di stance from the o ri g in (0,0) to I) (a g iven po int), and El = the magnitude o f an ang le I If r is pos iti ve, e is P(H,O) the meas ure o f any a ng le in sta nd ard pos ition th at has segm ent ,1' a its te rmina l s ide If r is negative, El is the measu re of any ang le that has ray o ppos ite segmen t O, P as its ter mina l side th (47&48) 22 EEJ P(-r,9) G Graphing w ith polar coordi nates : I Exa mp le I: 1'(4, 120 d egr ees) (49) E xa mp le 2: P ( ~ ) (50) H O ne angle graphed w ith polar coordinates can I' presen t seve l angle I If Pis a po int w ith polar coo rdinate (r, 9), then l' can also be g raphed by t he po lar coordi nates (-r, ffi P(4, 120) e + ( 2x + 1)1t) or (r, e + 2x1t), where x i any in teger E xa mlJle: S how fo ur differ nt pa ir ofp la r coo rd inate th at ca n be re prese nte d b the po int 1'(3, 60 r a n I degrees) (-r, e + (2 x + 1)1 80 d egrees) ~ (- (3), 60 + (1)180) (- (3), 60 - (1)1 80), (r, = 1'(-3, 240) or 1'(-3,120) P(4'3) e + 360x) ~ P(3, 60 + (1)360) or 1'(3, 60 + (2(360) = P(3, 780) hanging from rectangula r to pola r coordina tes : The fo ll owing fo rmulas are used to make thi s change: j( x f, + y2) , e = A rctan x > O = A rc ta n + 11:, X < 0, a nd fj = r adian s E xample I: Find the polar coordinates f, r 1'(-2,4) r = I r = = f j20 = 4.47 e = Arcta n j _ 2)2 + ( )2 _42 + 1t = 2.03, P(4.47, 2.03) Exam ple 2: Find the po lar co rd inates for P(3 ,5) r = 34 = 5.83, El = A rcta n "3 = 1.03 P(5.83, 1.03) ~_~ j()2+ 2) = hanging fro m polar to r ecta ng ular coordinates: The formulas used to make thi s change are : I x = r cos El e ~~a~Ple 1: v =r sin P ( ~), x = cos ( ~) = 2, an d s in (~) = 3.46 - P( 2, 3.4(h Exa mple 2: P(5, 60°), x = cos (60) = -4.76 Y = si n (60°) = -1.52 = P( -4 76, -1.52) K Grapbing imaginary numbers with po lar coord inate: The polar form of a complex number i x + yi = r(cos El + i s in e) Example: Graph the complex number -4, + 2i, and change to polar form r = ;;r+yL = j(_ + 2i 2) = /16 +4 = ,fiO = 4.47, El = rctan ( _24 ) + IT = 2.68,1)0Ia r fo rm = 1'( -4, i) = 4.47(cos 2.68 + i sin 2.68) A The notation P( n,n) = the num ber o f permutations of n objects taken all at one time B The notat ion P(n,r) represents the number of perm utation of n obj cts A Exponential properties: I Multiplication: x· xh = x + b E xample: x 2x = x6 take n r at a time P (n,r) Division : (;: ) = x·- b Example: Distribu tion with d ivisio n: ( r y y/ = ( ;: ) = x: 18·17·16·15·14·13·12·11·10·9·8·7·6·5· ·3·2·1 h 6.5.4.3.2.1 , not!cetatyou y can ca ncel 6!, leaving 18 ~ = 8.89 X 10 12 choices Example 3: A combination lock has four tumbl ers, and i num bered I ­ Power of a power: (X") b = x· b Exam ple: (x 2)3 = x6 Inverse power: x-I = { x"" = ~ x Root power: xI I = Example : XII2 20 on the dial How many combinations are possible P(20,4) j; = j; R bR B Logarithmic Properties and Logarithmic Fo rm: EE FE EB EfJ t§ Logarithmic Form: log.x = y, thi s is read a "the exponent of a to ge t the result x is y." Exa mple: log, 100 = 10, the exponent of x to get the res ult 100 = 10 or x lO = 100 Loga rithmic I)roperties : a M ultiplication: log.xy = log.x + log.y b Division: log y= logax - log.y * log.x = logaY' then c Power p rope rty: log.x b = b d rden tity property : If log.x x= y Change of Base p rope rty: I Lx, y and z are + nu mbers log ) z and x and }' are n t = I , then, log, z = -I - - og)x C Solving logarithmic equations: Exa mple I: Write log 1000 = in exponential form: 10" = 1000 Example 2: olve, log, Ii = i~ x l/4 =Ii 24 - 3x x = Example 4: log (2x + 8) - log (x + 2) = I +8 ~ ~ 8x = -12 (2x + 8) 10 = (x + 2) ~ Exam ple 5: log, = -* ~ log IOx +20= 2x E EJ x = - L5 ~ x- 1/3 = ~ x = - 1~5 Exa mple 3: y = 2 - - I (53) Exam ple 4: y = + I (54) Example 5: y = logzI""I~ and rttrloal ,,>W'tn uhuIK llIt,lIm P"TlfilU,,1fI r">fI the ~h Iwr 20111 l(IIJb K.r< harb l f 1t,,87 The ditference of two matrices.J - K is equa l t add ing J to th e add itive 1- by -I and add to r w I I0 I: 5I 010: -I i 00 : o I : -I I .J +KJ5 13 - 12 - II inverse of K .I = 26 ( ) + l( ) 139 56 [6 ( ) + t( ) C sing matn ce to solve sys tems of eq uat ions: If you have thr e systems of eq uation yo u can use an augmented matrix to find the solu tion et of the va ria ble You mu t fo llow the e guide li nes: I Any two rows Illa be interc hanged ny row may be repl aced by a non-zero multiple of that row Any row may be replaced by the um of that row and the multiple of another The goa l is to achieve an augmen ted matrix of the form; 110 : x,1 10 I : y I 100 I : z I, where x, y, z = the olution el Example : Solve x - 2y + Z = 3x + Y - z = 2x + 3y + 2z = u ing an augmented matrix Answer = 3x3 + x2 -5x + r 0 21 (9) + 4( )IJ 24 The aug mented matri i Answer = x + r -1 When th i sa me problem is performed with trad i­ ti onal divi sion, the an wer i the same tl 41 ,1 I I 13 I: I 13 I -I : 21 12 2: 71 Multiply row I by -3 and add 10 rO\ I I -2 I: I 107-4:-1 12 2: MUltiply row I by -2 and add to I' w I I -2 I: I 10 -4 :-19 10 0: -7 Mu ltiply row by - I and add trow II -2 I: I -4 : -1 I 100 : 12 1 -3 Answer = x + r - -5 - 81 J-K J -I 1\ free di wllloadS & h n red~ of t Itles t qUlc sluaY.COm 6111Ill~~ll~II~~ijI11111118