A relation can be tested to see if it is a function by the vertical line test.. Draw a vertical line thro gh any graph, and if it hit an x-value more than once, it is not a function.. 3
Trang 1FUNCTIONS
A A function is relation in which each element of the domain (x value
- independent va a ble) is paired with only one element of the ran e
(y value d ependent va riable)
B A relation can be tested to see if it is a function by the vertical line test
Draw a vertical line thro gh any graph, and if it hit an x-value more than
once, it is not a function (1-4)
b b2 4ac = 0, exactly one rea l r o
c b2 - 4ac < 0 no real roots (two distinct imaginary roots) I) Example l:f(x) =x2_4x+ I use
-b ± /b2 4ac
x = -(-4) + f< - 4 2
Afunction Not a fuction Afunction A function
C Linear m x+b where m functions = take the thloe pe, form: f(x) and b = = mx+b, the y-intercepor y = FE
Example: f(x) = 4 -l, the slope i 4/1 (ri e over run),
the y-intercept is
D he distance between two points on a lin can be found
u ing the distance formula, d = j(X 2 - XI )2+ (Y2 - YI )2 X;~~5
E The mid-point of a line segment can be found using the
( X + X I) (Y2+ Vd )
mid-polllt formula, 2 • 2 '
F The standard form ofa linear function is 0 =Ax + By +
C The lope is m = -A/B, and the y-intercept is -C/B
G The zeros solving for of x a function are found by setting y to O and ffiNo zeros
I Example 1: f(x) = 4x-1 (5)
2 Example 2: f(x) = 6, this functi n has no zero, and i
a h riz ntal line through +6 on the y-axi
3 Example 3: x = 4 thi i n t a function, because ther
is a vertical line through +4 on the x-axi , givi ng a
in fi nite set of values for y
H Polynomial functions take the form: f(x) = axn + bxn- I
+ cxn- 2 •• + dx + e
I When the hig e t power of the functi n is an o
integer, there is at least one real zero
2 When the highe t power is an even integer, there m
3 Both type can have imaginary roots of the form a + bi x=-1.170213
4 Th highest p wer ofa poly omial wih o e variable is y=.0100806
called it degree
Example 4 there are fol: f(x) ur r= oots 2x 4 + x(solutions) to 2 + X + 10, thas is poa degree ly omiaof l tE9
Example 2: f(x) = 2xJ + x2 -2x + 3, thi function h
one real zero at x = -1.17, and two non-real roots
Example 3 f x) = x2 + I this fUllction has'two
l Quadratic functions take the form : f(x) = ax2 + bx + c y=1.0113173
I 111e graph ofa quadrati function i called a parabola (10)
2 ome parabolas are quadratic equation , but not t Eo
q adratic functions (II)
3 Quadratic functions or equations can have one rea
solution two real solutions, or no real solution
4 Th vertex of a parabola i c ll d its critical point
- b ± jb2- 4ac x=-1.010638 ) The quadratic equation f(x) = 2a can y=2.9737903
be u ed to find the roots of all quadratic equations
6 The alue under th square root symbol is called t
d iscriminant It tells u the type ofroots of a quadrat
equation
a 2 - 4ac > 0, t lVO distnct real roots
_ 4)/2 = 3.732, and
x = -(-4) - f< - 4 - 4) /2 = 267 since the discriminant is > 0, there are two real roots (15) 2) Example 2: f(x) = 2x2 + 2x + I using b 2- 4ac = -4, since the discriminant is < 0, there are two imaginary root (16)
3) Example 3: f(x) = x2 + 2x + I using b2 - 4ac = 0
1 d · 0 th ' I 17 since t le ISCnl11l11ant IS = ere IS one rea I' \ ( )
J RatIOnal functions take the fori: f(x) = b (x)
I he parent function i f(x) = X ·
2 The graph of these functi ns consist of two pans, one
in quadrant I, and one in quadrant
3 The bran hes of rational function approach lne called asymptotes
4 Example I: f(x) = x + 3
5 Example 2: f(x) = x (20)
6 Example 3: f(x) = ~2 x - + 3 (21)
K Operations of functions:
I Sum: (f + g)(x) = f(x) + g(x)
2 Difference: (f -g)(x) = f(x) -g(x)
3 Product: (f x g)(x) = f(x) x g(x)
4 QuotIent: g (x) = g (x)' g(x) f 0
5 Example I: Given f(x) = x + 2 g (x) == x: 4
a Find the sum: (f + g)(x), x + 2 + ~ ( x + 2) (x - 4) + x X 2 X _ 8 x
x - 4 = x - 4 ' and x f 4
b Find the di fference: (f - g)(x), x + 2 - ~
( x + 2 ) (x - 4) - x x _ 3x _ 8 x
x - 4 = x _ 4 • and x f 4
6 Example 2: Given f(x) = x + 2, g(x) = x: 4
a Find the product: (fxg)(x), (x + 2)( x : 4 ) =
x 2 + 2x -x=-;t and x f 4
b Find the quotient: ( t)(X), x; x +_24 = ( X- 4) x2-2x - S (x+2) - x - = x , andx f O
L comP.osition of functions:lfogl(x) = f(g(x» E ! J8 Example: Gi en f(x)=x+2, g(x)= : 4 +2
Find IfogJ(x): f ( x : 4 + 2) =
2 ( 4) x 0265958 ( _ x_ + 2 ) + 2 = x + x - + 2 y~.OlO0806
with asymptotes
5x - 16 at the x&y axes ,and x f 4
4
x-M.lnverse functions: Ifogl(x) = Igofj(x) F E O
· + 4
Example: Given f(x)=2x -4, g(x) = ~, _ ( ' X + 4)_ (X + 4) _ Ifogl(x) - f - 2 - - 2 - 2 - -4 - x,
x=.02659573 ( 2 4 ) 4
and I gO fj( x) = 2 = x The asymptotes
are the axes
y=-.0100807 One real solullOn
x=1.276595 y=-2 1 Two real
x=-.9574468 y=7.9737903
No real solutions
x=1 y=-2.671 Two real roots (.26,0) &(3.73,
x=-.5053192 y=.47379034 Two imaginal)!
x~930B511 y=.00478157 y1=x ~2x +1 One real rOO! (-.93,0)
1
Trang 222
N Families of functions: Graphs of function families Changes in values of
the parent affect the appearance of the parent raph A parent graph is
the basic graph in a fam ily II the other family m mbers move up, down,
left right, or turn based on changes in values
I Polynomial functions 1:
a f(x) = x2 (22)
b f(x) = 2x2 (23)
c f(x) = .5x
_
d.f(x) - -x (25)
e f(x) = x2 + 2
x
f f(x) = 2 (27)
g r(x) = (x + 2)2 (28)
h.r(x) = (x - 2)2 (29)
2 I>olynomial functions 2:
a f(x) = x3 (30)
bf(x) = _3 (31)
c f{ x) = x3 + 2
d.f(x) = x3 -2
e f(x) = 2 3 (34)
f f( x) = 5x3 (35)
g.f(x) = (x + 2)3 (36)
h.f(x) = (x -2)3 (37)
3 Absolute value functions:
a f(x) = Ixl (38)
bf(x) = -I l (39)
c f{x) = 12xl d.(f(x) = 1.5x
e.f(x) = Ix + 21
f rex) = Ix - 21 (43)
g f(x) = Ix + 2 (44)
h f(x) = Ixl- 2 (45)
A Rectangular coordinates arc of the form (x,y), and arc plotted on the
Cartesian coordinate system
B Points are plotted with tw values, one the absci sa and the other th
o rdinate
C The abscissa i the x-value, called the domain and the ordinate is the
y-value, called the range
D Many different shape and function c n be drawn on the
Carte ian system
E Here is a given an le, originating fr III the -axi and rotating counter-clockwise This angle is represented by a
li e segment originating at t.he origin, and extending to a given point (P) (46)
R Polar coordinates are o the form P( r, 9), where r = the
radiu , the distance from the origin (0,0) to I) (a given
point), and El = the magnitude of an angle
I If r is positive, e is the measure of any an le in
standard position that as segment 0,1' a its terminal
side
2 If r is negative, El is the measure of any angle that as
th ray opposite segment O,P as its terminal side
(47&48)
G Graphing with polar coordinates:
I Example I: 1'(4, 120 degrees) (49)
2 Example 2: P ( 4 ~) (50)
H One angle graphed with polar coordinates can I' present
several an le
I If Pis p int with polar coordinate (r, 9), then l' can
/P
P (4 6)
E B 7 , p
I (
P(H ,O)
P (-r , 9)
P (4, 1 20)
2
also be graphed by the polar coordinates (-r, e + ( 2x + 1)1t) or
(r, e+ 2x1t), where x i any integer
2 ExamlJle: Show fo r differ nt pair ofp lar coordi- r n a
nate that can be represented b the p int 1'(3, 60 I
degrees)
3 (-r, e+ (2x + 1)180 degrees) ~(-(3), 60 + (1)180) (-(3), 60 -(1)1 80), = 1'(-3, 40) or 1'(-3,120) P ( 4 '3)
4 (r, e+ 360x) ~ P(3, 60 + (1)360) or 1'(3, 60 + (2(360) = P(3, 780) hanging from rectangular to polar coordinates: The following
formulas are used to make this change:
I r = j( x 2 + y2), e =Arctan f , x > O
2 9 =Arctan f + 11: , X < 0, a nd fj = radians
3 Example I: Find the polar coordinates f, r 1'(-2,4) r= j _ 2)2+ (4 )2
= j20 = 4.47 e= Arctan _42 + 1t = 2.03, P(4.47, 2.03) ~_~
4 Example 2: Find the polar co rdinates for P(3,5) r = j() 2 + 52) =
34 = 5.83, El = Arctan "35 = 1.03 P(5.83, 1.03)
hangin from p lar to rectan ular coordinates: The formulas used to
make this change are:
I x = r cos El
2 v = r sin e
3 ~~ a~Ple 1: P ( 4 ~), x = 4 cos ( ~) = 2, and 4 sin (~) = 3.46 - P(2, 3.4(h
4 Example 2: P(5, 60°), x =5 cos (60) =-4.76 Y =5 sin (60°) = -1.52 = P( -4 7 , -1.52)
K Grapbing imaginary numbers with p lar coordinate: The polar
form of a complex number i x + yi = r(cos El + i sin e) Example: Graph the complex number -4, + 2i, and change to polar form
r = ;;r+yL = j ( _ 42 + 2i2) = /16 + 4 = ,fiO = 4.47, El = rctan ( _4 ) + IT = 2.68,1)0Iar form = 1'( -4, i) = 4.47(cos 2.68 + i sin 2.68)
Trang 3A Exponential properties:
I Multiplication: x· h = x + b
Example: x2x4 = x6
2 Division: (;: ) = x·-b
Example: (:: 64)= x lO
3 Distribution with multiplication: (xy)· = x·y
Example: (xy)s = xSy
4 Distribution with division: ( y r= x:
Example: y / = ( ;: ) y
5 Power of a power: (X")b = x
Example: (x2)3 = x
6 Inverse power: x-I =
Example: x"" =
x
7 Root power: xII =
Example: XII2 =
8 Rational power: x/ b = b
Example: x312 =
B Logarithmic Properties and Logarithmic Form:
1
1 Logarithmic Form: log.x = y, this is read a
exponent of a to get the result x is EE
Example: log, 100 = 10, the exponent of x to get
result 100 = 10 or xlO = 1
2 Logarithmic I)roperties:
a Multiplication: log.xy = log.x + log.y
b Division: log y= logax - log.y FE
c Power property: log.xb = b * log.x
d rdentity property: If log.x = logaY' then x = y
3 Change of Base property: I Lx y and z are + numbers
log) and x and }' are n t = I, the , log,z = -I og)x - - EB
C Solving logarithmic equations:
Example I: Write log 1000 = 3 in exponental form:
10" = 100
Example (xI/4)4 = 2: (1i)4 olve, ~lx og, = 4 Ii = i ~ xl4 = Ii EfJ
Exampl24 - 3xe 3: lx = ogi6 2x -6) = log4(24 - 3x) 2x - 6 t§
Example 4: log (2x + 8) - log (x + 2) = I log
2 + 8) (2x + 8)
(x+2) = I ~ 10 1 = (x+ 2) ~ IOx+20= 2x E E J6
+ 8 ~8x = -12 ~x = - L5
x-1 /3
Example 5: log, 5 = -* ~ = 5 ~x = - 1~5
D Graphing Expon ntial and Logarithmic Equato s:
Example I: y = 3', and (1/3)' on the same graph
Example 2: y = 92 + 1
Example 4: y = 22> -1 + I (54)
Example 5: y = logz<x + 2) (55)
Example 6: y = log2(x - 2) (56)
Example 7: y = log2(x - 2) + 2 (57) ttJ
A The n tation P(n,n) = the number of permutations of n objects taken all
at one time
B The notation P(n,r) represents the number of permutation of n obj cts
I
taken r at a time P(n,r) = !, P(n,r) = ( 11.) I'
Il - r
C The notation n! is read as n -factorial
Example I: How many wa s can five cartons of cereal be arranged?
P(5,5) ~5! = 5432 I = 120
Example 2: If 18 people show up to serve on ajury, hO\ many 12 person
18!
juries c n be chosen? P(18,12)
( 18 -12 )! 18·17·16·15·14·13·12·11·10·9·8·7·6·5·4 ·3·2·1 h
6.5.4.3.2.1 , ot!cetatyou
can c ncel 6!, leaving 18 ~7 = 8.89 X 1012 choices Example 3: A combination lock has four tumblers, and i numbered I
20 on the dial How many combinatio s are possible P(20,4) ~ ;gi
_ 20·19·18·17·1 6 ·15·1 4 ·13·12·11·10·9 ·8 · · 6·5· 4 ·3·2 ·1 _
- 16·15·14· 13·12 ·11·1O·9·8·7·6·5·4.J.2·1
116,280
Example 4: If26 people enter a 1 -mile race and all ofthem finish, h w many
possible orders of finishing are there? P(26,26) = 26! 26! = 4.032 x 1026
D Permutations that contain repetitions or are placed in a Circular Pattern
Repetitions: The number of n objects of which x & yare alike = ~ .y
Circular Patterns = (n-I)!
Example 1: How many word arrangements can be made from the word
5!
radar? n = 5, x&y = a&r = 2 2!2! = 30
Example 2: How many word pattern can be formed from Mi issippi? ll!
2!2!2!4! = 207,900
Example 3: There are 24 children who are going to play dodge ball 10
of them will start out in the circle Ilow many ways could the
remaining children form circular combinations? (n-I)! = 1 ! =
6,227,020,800
E Combinations: Differ from permutations in that order i not a consider
ation The number of n objects taken r at a time is C(n,r) = ( II! )
11- r !r!
Example 1: A book club ha selected eight books to read, how many four
group combinations are pos ible? ~qn,r) = ( 8! ) = 70
8 - 4 !4!
Example 2: How many five card hands an be dealt from a regular deck
52'
of c rds? qn,r) ~= (52 _ 5 ) !5! = 2598960
A Synthetc division is a method used to make long division of p Iynomial
les cumbersome
I It is mainly used when you have a very long numeral r
2 It can only be lIsed with a di i or in the form (x -n)
3 You can convert the form (3x + n) to (x + 1/3n) and then use synthetic
division
B Synthetic Division of the form x2 -3x -54.;-x -9
I Example I: The standard way of dividing polynomials
x - 9 ) X 2 - 3x - 54
x+
-_(x2
-6x (6x -5
o Answer = x + 6 r 0
2 Using synthetic division, we use the econd part of the divi or, but change
the sign (+9, trom the above example) We then Ii I the coefficients of the
dividend From the example above, th y \\ould be I, -3, -54
Then bring down the coefficients one at a time, mUltiply them b} the
Trang 4divisor (+9), and add them to the next coefficient This gives the same
result From the example on pg 3:
2 1 1 -3
-9
1 6 0 An wer = x +
Note: The power of x always begins one lower in the a swe
Example 2: 3x3 + x2 + 5x - 2 -;.- x +1
::.1 13 1 5
-3 2
3 -2 7 -9 nswer = 3x2 -2x + 7 r
Example 3: In the following example notice that there mu t be
place-hold r for missing powers of the variable Z 4 + Oz3 + O
Oz + 16 -;.- z +
-21 0 0 0
-2 4 -8 1
I -2 4 -8 32 Answer = z3 _2z2 + 4z -8
C S nthetic divi ion of the form 3x2 + lOx - 9 + 3x~2 In thi example all
number are first divided by 3 then at the end of the process all fractions
are c anged by multiplication
Exam pie 1: 3x - 2 ) ( 3x 2 + lOx - 9) become x - t ) ( X 2 + lj) x
-10 - 9
tl "3
2
3
Answer = x + 4 r - t ' now multiply all fractions by 3
Answer = x + 4 r -1 When thi same problem is performed with trad i
ti nal division, the an wer i the same
Example 2: 2x - I )( 6 x 4 - x 3- Ilx2+ 9x - 2), now divide all parts
by 2, and use s nthetic division
1 I
t l 3 - 2 - 2 2 -I
3 I
-3 -5 2 0 Answer = 3x3 + x2 -5x + 2 r 0
Again, when standard long division is perform d, the same answer i
attained Note: I I' there w re any fracti ons in a ny part of the an wer
they wou ld be removed by multiplicatio
MATRICES
A Matrix: rectangular array of elements in columns and rows Rows are
named before columns, therefore a 2 4matrix has two rows and four colwnn
B Properties of Matrices:
I A mat x with onl one row is call d a row matrix
2 matri that ha onl one olum ll i ca lled a column matrix
3 Two matrices are equal if and only if they have the same dimensions
and contain the same identical eleme ts
4 The sum or a 2x3 and a 2 3 matrix is a 2x3 matrix in which the
elements are added to the corresponding elements
9 Exam ple: Find.1 + K ifJ = 1 2 4 -61 and K =
1-3 0 21 -3 -'i'
.J+KJ5 13 - 4
12 - 3 I
5 The ditference of two matrices.J -K is equa l t adding J to the add itive
inverse of K .I = 1 and K
1- 3 0 I 3
J-KJ - I -5
-1-8 3
6 The product ofa scalar (x), (a value with magn itude, but no direction),
and a mat x (.I) is d, with each element f 1 multiplied time the
va ue x IX tImes the matrIX.I =
1-18 0 12
7 The product of 2 -two by t\\'o matrices i a two b two matrix The
procedure isa follow: J = 12 41 ,16 91
JxK='2(6)+ 4(3) 2(9) +4(2) I J24 26
[6 ( 6 ) + t( 3 ) 6 ( 9 ) + l( 2 ) 1 139 56
C sing matnce to solve systems of equations: If you have thr e systems
of equ tion you c n use an augmented matrix to find the solution et of
the vari ble You mu t follow the e g ide nes:
I Any two rows Illa be interchanged
2 ny row may be replaced by a non-zero multiple of that row
3 Any row may be replaced by the um of that row and the
another The goal is to achieve an augmented matrix of
1100
10 I 0 : y
100 I : z I , where x, y, z = the olution Example: Solve x - 2y + Z =
3x + Y -z =
2x + 3y + 2z = 7 u ing an a gmented matri
The augmented matri i I I -2 I: 7
13 I -I :
12 3 2:
Multiply row I by -3 and add 10 rO\ 2 I I -2 I: 7
107-4:-19
12 3 2: 7 MUltiply row I by -2 and add to I' w 3 I I -2 I: 7
10 7-4 :-19
10 7 0: -7
Mul ply row 2 by -I and add trow 3 II -2 I: 7
I 0 7 -4 : -19
100 4: 12
Add row 3 to I' W 2 I 1 -2 I : 7
107 0:-7
10 0 ~ : 12
Multiply row 2 by 117 -2 1 I: 7
01 0: -I
004 : 12
Multiply row by 1/4 I -2 I : 7
o 0 I: 3
Multiply row 2 by 2 and add to row I I 0 I 5
001 : 3 Multiply row by -I and add to r w I 100: 2
o I 0 : -I
00 1 : 3
The solution set =
ISBN-1 : 978-14
2320248-ISB -10: 142320248
U S 4.95 CAN $7 50 Author : S Orcutt
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