POWER GUIDEMATH

YEARS

16DOING OUR BEST, SO YOU CAN DO YOURS

AUTHORSJulia Ma & Steven Zhu

CONTRIBUTIONS & REVISIONSMichael Nagel

EDITORSDean Scha�er & Sophy Lee

ALPACA-IN-CHIEFDaniel Berdichevsky

2 0 1 02 0 1 1E D I T I O N

AlgebrathroughCalculus

M A T H

®

the World Scholar’s Cup®

DemiDec, The World Scholar’s Cup, Power Guide, and Cram Kit are registered trademarks of the DemiDec Corporation. Academic Decathlon and USAD are registered trademarks of the United States Academic Decathlon Association.

DemiDec is not affiliated with the United States Academic Decathlon.

MATH POWER GUIDE®

I. WHAT IS A POWER GUIDE?........................................................ 2 II. CURRICULUM OVERVIEW............................................................3 III. GENERAL MATH..............................................................................4 IV. ALGEBRA…........................................................................................ 7 V. GEOMETRY………….......................................................................... 37 VI. TRIGONOMETRY.............................................................................46 VII. POWER LISTS................................................................................... 54 VIII. POWER TABLE..................................................................................59 IX. POWER STRATEGIES…………………………………………………………60 X. ABOUT THE AUTHORS..................................................................61

BY

JULIA MA CALTECH

ALTA HIGH SCHOOL

STEVEN ZHU HARVARD UNIVERSITY FRISCO HIGH SCHOOL

EDITED BY

DEAN SCHAFFER STANFORD UNIVERSITY

TAFT HIGH SCHOOL

SOPHY LEE HARVARD UNIVERSITY PEARLAND HIGH SCHOOL

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n this case would’ve hadmes as many e rational roo

find the sum

um of the ro

s the coefficiem of the roots

ots is –47=

4-7

m of the roots

11

ore variables, p your typin

heorem, and

termine all

ynomials in nt—the num

and all the facof C, and we’

e rational re

the various faots of 36 + 2xe correct coefefficient for t

er all the fact

14±,

13±,

12±

d to list all thpossible ratioots, we can u

m or produc

oots is ab- f

ent of the secs of 4x2 – 7x

47

s of x3 + 3x2 –

storing –3 ang because t

d the remaind

the possible

the form of Amber in front

ctors of C ’ll use p to re

eal roots can

factors into thx3 + x4 – 11x2

fficient for Athat term is 1

tors of A 1±,

19±,

16±,

he factors of onal roots use the facto

ct of the roo

for all polyno

cond-highest+ 5

– 4x – 12

as a variable then you wo

der equals 0,

e rational ro

Axn + Bxn-1 +t of the term

epresent all th

n be found

he above exp2 – 12x

A 1

13±,

118±,

112

C over 1, 2,

or theorem to

ots, but not

omials, wher

t degree term

may help on’t need

then (x –

oots of a

…C m with the

he factors

with the

ression

16

3, and 6,

o find the

the roots

re a is the

m

Solving In

L

Th

an

th

g Inequalitinequality: a An inequa

Examp Examp

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Be car Examp

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x <

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nd ac for eve

e constant Example: F a = 5 c = –2

The pr

ies definition

ality states thple: 4 + 5 < 1ple: 4x + 2 > uadratic a linear inequreful to flip thple: –3x + 7

We subtract 7 3x > –2 hen we divid

< 32

Mat

ll use ab-

um of the roo

to find the p

en-numbered

Find the prod

roduct of the

hat two expre12 3y – 4

uality, treat thhe sign if you> 5 from both si

de both sides

th Power Guide

ots is 3=13

roduct of th

d polynomial

duct of the ro

e roots is 52-

essions are no

he inequalityu multiply or

ides to get th

by -3 and fli

| 12

he roots is ac-

ls, where a i

oots of 5x2 +

ot equal

y as an equatir divide by a

he term with

ip the sign

ac for odd-nu

is the leading

8x – 2

ion and isolanegative num

x by itself

umbered pol

g coefficient,

ate the variabmber

lynomials

, and c is

ble

Th

Linear

Ex

In the abov

to the left o

he answer ma

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If the inequ

The darken

r inequalities

xample: y ≤

To graph tit At the line

-3

-3

Math

ve graph, the

of, but not in

ay also appea

circle means

uality were x

ned circle me

can have mo

2+x31

this inequalit

, y is equal to

-1 -2

-1 -2

h Power Guide |

e inequality

ncluding, x =

ar in the form

that the valu

x ≤ 32 , then t

eans that the

ore than one

ty, we must

o the functio

0

0

13

is true in the

= 32

m of a numbe

ue 32 is not i

the number l

solution inc

variable

plot the line

on

1 2

1 2

e shaded area

er line

ncluded in th

line would lo

cludes 32

e and then sh

3

3

a, which is t

he solution

ook like this:

hade the regi

he region

ion above

Ex

We shade a

xample: y < –Now, we s

Usually in“greater th Someti

dotted But rea

Mat

above the lin

–2x + 1 hade below t

n graphs, “lean” looks theimes when yrather than s

ally, the shad

th Power Guide

ne because y c

the line becau

ss than” looe same as “gry is not equasolid

ding is the im

| 14

can also be g

use y is less t

oks the samereater than oral to the fun

mportant part

greater than t

than the func

e as “less thr equal to”

nction, the li

t

he function

ction

an or equal

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to,” and

unction is

6 If you’ve e

To solve a After f The ro Test n

Th Examp

Fi (x O W

N W Le

W W W O

wh

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ever seen The B

a quadratic infinding the roots will partnumbers in eahe ones that ple: x2 + 6x –rst, we factor+ 7)(x – 1) <ur roots are -

We will place t

otice that theWe will choose

et’s use -8, 0 Rule of thuvery easy to

When we plugWhen we plugWhen we plug

n the numbhere the ineq

hus, the soluratic inequali

Boondock Saints

Math

nequality, treoots, place thtition the numach region make the ine

– 7 < 0 r < 0 -7 and 1 these roots o

e roots dividee a number iand 2 umb6: wheneo plug into thg -8 into the ig 0 into the ing 2 into the inber line, we wquality is true

tion to the qities may hav

, the line at the

-7

-7

h Power Guide |

eat it like an ehem on a nummber line int

equality true

n a number l

e the line intn each region

ever you can he inequalityinequality, wnequality, wenequality, wewill place ane

quadratic ineqve an x and a

e beginning abo

15

equation andmber line to different r

will be part

line

to three region to test the

choose 0 as y we get 9 < 0, we get –7 < 0, e get 9 < 0, wn x where th

quality is –7 a y

out “rule of thu

1

1

d solve for th

regions

of the solutio

ons: x < –7, –inequality

a test value,

which is falsewhich is tru

which is false he inequality

< x < 1

mb” is most ex

he roots

on

–7 < x < 1, an

, do so, as it

e e

y is false and

xcellent. – Steve

nd x > 1

is usually

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en

A

Ex

Absolute valu A number

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When theto two opp Examp

x – x – x =

When we the inequa Examp

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we x

xample: y ≤ xWe can rep

Since y is and the are

ue r’s absolute vlways non-neple: 4=45-ple: 45=0e expression iposite valuesple: =3 -x – 3 = 2 – 3 = –2 = 1 or x = 5 have inequa

ality symbol ple: 6 +x ≥or the first in

6+x ≥ 7 or the seconde change the

6+x ≥ –7 The value less than or

Mat

2+x 2 present this i

less than or ea includes th

value is essenegative 45 inside the ab 2

alities with ab

≥ 7 nequality, we

d inequality, sign of the v

to the right r equal to

th Power Guide

nequality gra

\ equal to the he curve itsel

tially its dista

bsolute value

bsolute value

just remove

however, wevalue to the r

of the inequ

| 16

aphically

function, wlf

ance from 0

e signs is a fu

es, we have t

the absolute

e have to flipright of the sy

uality is now

we shade the

on a number

unction, we s

to be careful

e value signs

p the symbol ymbol

negative 7, a

area below t

r line

set the funct

with the dir

of inequality

and the symb

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y because

bol is now

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ons: RationFunctions For each p

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hus, x ≥ 1 or ple: 7 -2x nce an absolever be true

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nal, Expon

possible valuof the depen

u plug in a vaxample: y = f

When x = f(x) denote

than one valxample: y = f

f(2) = 4 f(–2) = 4 In this case

alue of x, howxample: y = f

f(9) = 9 f(9) = 9 In this case

uation to be tical line placone point line intersect

We can place e graph at on

Math

x ≤ 3 – 13 < –3 lute value ca

trick questio

nential, and

ue of a givenndent variablelue for x, you

f(x) = x2 2, f(2) = (2)2

es that y is a flue of x may f(x) = x2

e, y was 4 whwever, may hf(x) = x = 3 = –3

e, y may not a function, iced anywher

ts the graph a

a vertical linne point, whi

h Power Guide |

an never ma

ons like this o

d Logarith

independene, y u should get

2 = 4 function of xhave the sam

hen x was 2 aave more tha

equal both 3t must pass tre on the gra

at more than

ne anywhere ich means th

17

ake an expres

one

hmic

nt variable, x

one value fo

x me value for y

and when x wan one value

3 and -3 the vertical liaph of a func

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on the graphat the graph

ssion negativ

x, of a functio

r y

y

was -2 for y

ine test ction can cro

then it isn’t a

ph above, anh represents a

ve, the inequ

on, there can

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a function

d it would oa function

uality can

n be only

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domai Possib

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vertical line wThis graph

ain of a functvalue of x thain ble limitationer, and takine of a functioare several li

quare roots ansymptotes ca

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negativ The fu Thus, -In addition The cu

site function ), sometimes he above is reomposite fun

Mat

will cross theh, therefore, dtion is all posat causes a m

ns on domainng the logariton is all possiimitations onnd exponentan graphically

ve graph, y =urves approave infinity unction will n-1 is not in thn, the line x =urves will app

is the result denoted (f

ead as “f of gnctions f(g(x)

th Power Guide

e graph abovdoes NOT ressible values

mathematical

n include divihm of a nonble values of n range ial functionsy illustrate ra

= –1 is a horizach the line

never actuallyhe range of t= 0 is a verticproach, but nof combining)(x), is a typ

g of x” )) and g(f(x))

| 18

ve at two placepresent a fuof x (the inderror in the

iding by 0, tan-positive numf y (the depen

only give noange limitatio

zontal asympy = –1 whe

y reach y = –1the functioncal asymptotenever touch, tng two or mopical exampl

) are not nece

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dependent vafunction is N

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ptote en x stretche

1

e the line x = 0

ore functions e of a compo

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0

ariable) NOT includ

uare root of a

le)

values, for ex

es out to inf

0 at once

osite function

ame

ded in the

a negative

xample

finity and

n

In

Ex

nverses An invers

input The invers f(f–1(x)) = The doma The range Not all inv

The g

If the O Ex

on

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xample: Give

f(g(x)) = 3(

g(f(x)) = 3

The only t

se is the “un

se of the fun(f f–1)(x) = x

ain of a functe of a functioverses are funraph of an in

function f is ne-to-one mxample: the nce

The inverse-to-one funcpasses the honction in at m

Math

en f(x) = 3x +

(x1 ) + 2

2+x31

ime the abov

ndo” of a fun

ction f(x) is dx tion is the ran

on is the domnctions nverse is the m

one-to-one, eans that no inverse of f(

e of f(x) is f-1

ction (whichorizontal linemost one po

Power Guide |

+ 2 and g(x) =

ve two comp

nction: it tak

denoted as f-

nge of its invmain of its inv

mirror image

then its invevalues in thex) = x is a f

1(x) = x h, as a functe test—any hint

19

= x1

osite functio

kes the outp

1(x)

verse verse

e of the funct

erse is a funce function’s rfunction bec

tion, must byhorizontal li

ons are equal

put of a func

tion across y

ction range appearcause f only

y definition ne placed on

is when x = –

ction and re

= x (see grap

r more than omaps to eac

pass the vern the graph

–1

eturns the

ph below)

once ch y-value

rtical line intersects

R

The above It doe

Yo To find th

Examp Le To

Somet Examp

f-1(7)? Th

Rational func The doma

To deequals Th D An

In

e graph is f(xs not, howevou could alsohe inverse of ple: f(x) = x3 et y = x3 o find the inv

x = y3 Solving forThus, f-1(x)

times we do nple: Given th

he answer is ction ain of a functetermine thes 0 hese values wivision by zerny division b

Removable If a fac

each otNon-remo If a fac

n both cases, c

Mat

x) = x2, whichver, pass the ho say that its ia function, l

verse, we swi

r y, we get th) = 3 x not need to ahat f(x) is a

simply 3, sin

tion includese domain of

will be the onro causes a m

by zero produe discontinuictor (x – c) isther out, provable discon

ctor (x – c) is c must be a r

th Power Guide

h is a functionhorizontal-lininverse does et y = f(x), ch

itch the varia

he inverse fun

actually find one-to-one f

nce all we do

s all of its posa function,

nly ones exclumathematical uces either a rities are “holes in both theducing a “hotinuities are only in the d

real number

| 20

n because it pne test, so itsnot pass the hange all x’s

ables

nction y = 3

the inverse efunction, if f

in an inverse

ssible x-valuefind the x-v

uded from theerror

removable ores” in the grae numerator ole” at x = c asymptotesdenominator

passes the ves inverse is novertical-line to y’s and all

x

equation f(3) = 7, wha

e is switch th

es values at wh

e domain

r a non-remoaph and denomi

r, an asympto

ertical-line tesot a functiontest l y’s to x’s

at is f(3) = 7

he x and the y

hich the den

ovable discon

inator, the tw

ote exists at x

st n

7, what is

y

nominator

ntinuity

wo cancel

x = c

Examp

W

If

The range The ra If the

there i

Ex

If the

asymp

c ide

Ex

If the exists

ple: What is

We must factox2 + x – 6 =The denomThus, the d

we factor the

y = 2)( -x(4)( -x(

Both the n Thus, - At x = The other We can

e of a functioange of rationexponent de

is no horizon

xample: y = x

The degreeThe degreeThe degreethere is no The range,

degree of th

ptote exists at

is the leadinenominator

xample: Wha

Both the nc = 3 and b Remem

highest In the

The horizo

Note that

because f( -

degree of that y = 0 (the

Math

the domain o

or the denom= (x – 2)(x + minator has tdomain inclue numerator,

3)+(x3)+(x

numerator an-3 is the loca–3, a hole exroot, x = 2, innot cancel (

on includes alnal functionsegree of the ntal asymptot

1-x2+x+x3

e of the nume of the denoe of the numhorizontal a

, therefore, inhe numerato

t y = bc

g coefficient

at is the horiz

numerator anb = –1 mber that tht power denominator

ontal asympto

even though

)187- = –3

he numeratorx-axis)

h Power Guide |

of y = x+x

x-x2

2

minator 3)

two roots, 2 audes all value, we can find

nd the denomation of a remxists in the gris the location(x – 2) out ofll of its possibs can be limitnumerator iste

erator is 3 ominator is 1merator is grasymptote ncludes all reor is the sam

t of the num

zontal asymp

nd the denom

he leading co

r, the term w

ote is y = 1-3

h y = –3 is a

r is less than

21

6-x12-x ?

and -3 es of x exceptd out more de

minator have movable discoraph n of a verticaf the expressible y-valuested by a horis greater tha

reater than th

eal numbersme as the deg

merator, and

ptote of y = 3

minator have

oefficient com

with the high

or y = –3

a horizontal

the degree o

t 2 and -3 etails about t

a factor of (xontinuity

al asymptote ion

izontal asympan the degree

he degree of

gree of the d

b is the lead

43

24

x-x6x7+x3 ?

a degree of 4

mes before

est power is –

asymptote,

of the denom

the graph

x + 3)

ptote e of the deno

f the denomi

denominator

ding coefficie

4

the variable

–x4

it is still in t

minator, an a

ominator,

inator, so

, then an

ent of the

with the

the range

asymptote

E

Asbe

The invers

Given

Wde

Exponential f

The indep The g

The base o The doma The range A horizon The invers Regardless

coefficient

s x increases, ecause the de

Eventually

which effecse of a ration

n a rational fu

We must findependent varifunction

pendent variaeneral form oof an exponeain is all real e is all positivntal asymptotse of an expos of its baset of 1 becaus

Mat

the numeratnominator h

y, the ratio o

ctively equalsnal function i

unction Q(x)

)x(P

d the inversiable (as befo

able in an expof this type oential functionumbers

ve numbers te exists at y =onential funce, an exponee a0 = 1

th Power Guide

tor will increhas a higher e

of the numer

s 0 is not necessa

)) , the inverse

se by intercore)

ponential funof function ison, a, must be

= 0 ction is a logaential functio

| 22

ease at a slowexponent

rator to the

arily a functio

e is NOT sim

changing var

nction is an s ax e positive

arithmic funcon will con

wer rate than

denominato

on

mply P(x)

)x(Q

riables and

exponent

ction tain the poi

the denomin

r will approa

solving for

int (0, 1) if

nator will

ach ∞

1- ,

the new

f it has a

L

Logarithmic

The indep The g

Lo

You m Ln

Ln Yo

Logarithmargument Examp

Th So 23 x =

Logarithm Examp

W Th

Examp

32

functions

pendent variaeneral form iog stands for

Log(x) is thmay also see Ln stands for t

e is a conste = 2.7182e is import Conseq

n(x) is the samou will need

ms are used t

ple: log2(8) =he argument olve for x if 2

= 8, so the p= 3

ms and exponple: log7(72) =

We can rewritehus, 2 = x ple: ln(e 3

2

) =

x=

Math

able of a logais log(x) a logarithm

he same as loLn(x) the natural lotant like pi 28182846…tant because lquently, the lme as loge(x)to know wheto find the p

= x is 8, and the

2x = 8 power is 3

nential expres= x e this equatio

x

Power Guide |

arithmic func

taken on basog10(x)

ogarithm, tak

lots of naturalogarithm ba

ere the Log apower to wh

e base is 2

ssions cancel

on as 72 = 7x

23

ction is in th

se 10

ken on base e

al phenomenased on e is c

and Ln functihich a base

each other o

he argument o

e

na are based ocalled the nat

ions are on yis taken to p

out when the

of a logarithm

on e tural logarith

your calculatoproduce the

e bases are the

m

hm

or resulting

e same

Comple D

Special ru When

of the Ex Ex

Whenlogarit Ex Ex

By thcombi Ex

Ex

The th To fin

Lo

Ex

The base o The doma The range A vertical The invers Regardless

argument’

ex NumberDefinitions A complex

a and i =

All pure rwith b = 0

les exist for on the entire a

logarithm xample 1: logxample 2: log

We cannoterms, not

n two logaritthm with thexample 1: logxample 2: loghe same tokeine them intoxample 1: log

xample 2: log

hree rules abond a logarithm

ogbased(argume

xample: FindSince most

and plug in

The answe If you

of a logarithmain is all posie is all real nuasymptote exse of a logaris of the base’s coefficient

rs

x number is b are real nu

= 1- or i2 =real numbers0 and a = 0, r

Mat

operations onargument has

g A2 = 2log Ag(x3 + 7x2 – 5ot move the

the entire arthms of the e arguments mg A + log B =g5(x + 2) + loen, when two one logarithg A – log B =

g6(x + 2) – lo

ove can also m in a base o

ent) = log

log(ar

d log6(43) t calculators

n )6log()43log(

er is about 2.0plug in 6 to mic functiontive numbers

umbers xists at x = 0thmic functie, a logarithmis 1, because

any number umbers, and i= -1 s and all purespectively

th Power Guide

n logarithmss an exponen

A 5)2 = 2log(x3

other exponrgument

same base multiplied to

= log AB g5(x – 6) = lo

wo logarithmhm with the

= log A/B

g5(x – 6) = lo

be used in reother than 10

)baseg()umentrg

don’t have

0992 the 2.0992 p must be poss

ion is an expomic functione log(1) = 0

in the form i is the imagi

ure imaginary

| 24

nt, we can tu

+ 7x2 – 5) nents becaus

are added, ogether

og6(x+2)(x –ms of the sa

first argume

og5(6-x2+x )

everse 0 or e, use the

a base-6 log

power, you gsitive

onential funcn will contai

a + bi inary unit

y numbers a

urn the expon

se they only

we can com

6) = log5(x2 –ame base areent divided b

e following f

garithm func

get 43

ction in the point

are technical

nent into a c

y apply to i

mbine them

– 4x – 12) e subtracted

by the second

formula

tion, use the

t (1, 0), prov

lly complex

coefficient

ndividual

into one

d, we can d

e formula

vided the

numbers,

O

Operations w We can si

Examp W

W

N

Treat i as Use the di

In the Complex

The co The co The co

A fraction Fix th

of the

Examp

W W

(8(4

with complexmplify higheple: Find the

We need to fin

We can see thaThis observ 1. Divi 2. Tak 3. Rais 4. i3 = Let’s try i71

1. Divi 2. Tak 3. Rais 4. i1 =

otice that theWith this rWe can als We kn 0 + i53

a variable whistributive pr

e end, simplifconjugates aomplex conjuomplex conjuomplex conju

n with an imahis by multip

denominato

ple: 2i - 83i+4

We need to geWe will multip

2i)+2i)(8 - 82i)+3i)(8+4

Math

x numbers er powers of ie value of i75

nd the pattern

at the patternvation gives ide 75 by 4 e the remainse i to the powi75= -i 13 ide 713 by 4e the remainse i to the powi713 = i e sum of everrule, we easilso find i + i2 +

now that i + i2

+ i54 = 0 + i +hen adding aroperty whenfy i2 = -1 are pairs of cougate of i is -ugate of 2 – ugate of 4 is aginary exprelying both th

or

t rid of the i ply top and b

=1-16i+ 642+i8+32

Power Guide |

i

n to the pow

i1 = i

i2 = –1

i3 = –i

i4 = 1

i5 = i

n repeats everus easy short

der, ¾, and iwer that you

der, ¼, and iwer that you

ry four termsy find that i3

+ i3 + … + i5

2 + i3 + … + + (-1) = i – 1and subtractinn multiplying

omplex numb-i 3i is 2 + 3i 4 because th

ession in the he numerato

in the denombottom by th

4+16i6-24i =

6+26

25

wers of i

ry 4 powerstcut to solve

ignore the 4 u found in Ste

ignore the 4 u found in Ste

s is 0 34 + i35 + i36 +53 + i54 i52 = 0 ng (combineg two comple

bers that com

here is no imadenominator

or and denom

minator he conjugate

6832i+ =

341+13

i75

in the denomep 2

in the denomep 2

+ i37 = 0

e like terms) ex numbers

me in the for

aginary part r needs to be

minator by th

of the denomi6

minator

minator

m a + bi and

e simplified he complex c

minator, 8 +

d a – bi

conjugate

2i

C

Reading L

Complex num Any polyn

numbers All compl

Examp Since com

odd numb For a qua

quadratic If the

Th

If the

Th

If the In

im

g Graphs oLinear Funct Linear fun

These

By rea Fi

In

W

mbers as roonomial with

ex roots (thaple: If a poly

mplex roots mber of real rooadratic equatiformula: b2 –discriminant

he roots are

discriminant

he root is 2ab-

discriminantn the quadratmaginary unit

of Functionions nctions (linefunctions ha

ading the grarst, we look f

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b = –1 We still need t

We can rea

Using the f

m = –21

Mat

ots of equatiodegree n wi

at have nonzeynomial has amust come in

ots ion, the natu– 4ac t is positive, b

2adiscrim±b-

t is 0, the roo

ab

t is negative, tic equation, t

ns

ar equationsave x raised t

ph, we can fifor the y-intercept in the g

cept form, wh

to find m, thead two point

formula for s

th Power Guide

ons ll have n (po

ero imaginarya root of 7 + 2

pairs, then a

ure of the ro

both roots arntmina

ots are real an

the roots aretaking a squ

) are straightto the first po

figure out theercept, the vagraph above hich is y = m

e slope ts from the gr

slope, m = xy

| 26

ossibly nond

y parts) come2i, it must ha polynomial

oots is determ

re real

nd identical

e complex couare root of

t-line graphsower

e equation it alue of y wheis -1

mx + b, the y-i

raph, (–2,0)

12

12

x-xy-y , we ha

distinct) roots

e in conjugatave another rl with an odd

mined by the

onjugates a negative di

represents ere the line cr

intercept is b

and (0, –1)

ave m = ( -0--1

s among the

te pairs root, 7 – 2id degree mus

e discrimina

iscriminant c

rosses the y-a

b

(-2)0-

e complex

st have an

ant of the

creates an

axis

Q

H

Th

Quadratic Fu Quadratic

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The ab St

tu To fin

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W –1

a =

The g

Higher order Higher or

If the same s

herefore, the

unctions c functions (qfunctions ha

bove graph sandard formrning point o

nd the equatince the parabhe lowest poing the vertex nd a, we need

We can read fr1 = a(0 – 2)2

= 21

raph above r

r functions rder functionorder (degre

side of the y-

Math

above graph

quadratic eqave x raised t

hows a parabm is also know

of the paraboon of the parbola opens uint is (2, -3)into the stan

d to plug in arom the grap– 3

epresents y =

ns (higher oree of the high-axis

Power Guide |

h represents y

quations) areto the second

bola that follwn as vertex fola rabola, we mpwards, we l

ndard form eqanother pointh the point (

= 21 (x – 2)2 –

der equationhest exponen

27

y = –21 x – 1

U-shaped grd power

lows the stanform because

must find the look for the l

quation, we t (0, -1)

– 3

ns) fall into twnt) is even, th

raphs

ndard form y e the point (h

vertex first lowest point

have y = a(x

wo general tyhe graph will

= a(x – h)2 +h, k) is the v

– 2)2 – 3

ypes of graphl start and en

+ k vertex, the

hs nd on the

Thwh

Th

If the

he above grahich means t

he graph sho

If a test queliminate t In this Then, grapmatches Alternativeand see if t

order is odd,

Mat

aph starts anthe order is ev

ws the funct

uestion ever the answer chcase, we wou

ph the remai

ely, you can they solve cor, the graph w

th Power Guide

nd ends on tven

ion y = 21 x4

asks you to hoices whose uld eliminateining choices

plug points rrectly

will start and

| 28

the same side

+ x3 – 2x2

find the equ orders canne all the chois on your ca

from the gra

end on oppo

e, the positiv

uation from ot possibly bices with oddalculator to f

aph into the

osite sides of

ve side, of th

a graph like be correct d orders find the equa

e remaining e

f the y-axis

he y-axis,

this one,

ation that

equations

E

L

The abwhich

The gExponential f Exponenti

The abLogarithmic Logarithm

bove graph sh means the oraph shows tfunctions ial functions

bove graph sfunctions

mic functions

Math

tarts on the norder is oddthe function

create curve

hows y = ex,

s create curve

Power Guide | 2

negative side

x5 + 2x4

s that have a

and the asym

es that have a

29

e of the y-axis

a horizontal a

mptote is y =

a vertical asym

s and ends on

asymptote

0

mptote

n the positivee side,

Sequenc A

A

The ab Notice

Flgr

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Examp Here,

8 – 11

To nth te Th It

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n will Examp

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bove graph se that this gripping the eaph

, and Meanequences ic sequences ple: 2, 5, 8, 1the differenc– 5 = 3

1 – 8 = 3 find the rm = first terhe 8th term inmakes sense

eries metic series is

ormula to fin

2t last + term t

yield the totple: Find therst, we must

he formula to

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hows y = ln(xraph is the inexponential g

ns

are patterns 11… ce d between

nth term rm + d(n – 1n the examplthat 7 “gaps

s the sum of a

nd the sum o

term) gives t

tal sum e sum of the afind n, the n

o find the nu

th Power Guide

x), the naturverse of the egraph on the

of numbers t

n consecutive

of an ) le sequence a” exist betwe

an arithmetic

f the first n t

the average o

arithmetic prnumber of te

umber of term

| 30

ral logarithm exponential ge x = y line

that have a c

terms is 3

arithmetic

above would een the first a

c sequence

terms is n(fir

of all the term

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ms is n = last

of x, and the

graph e will yield t

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31, 34, 37…9eries

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e asymptote

the above log

ference d

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1) = 23 th terms

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iplying the a

94, 97

1+term

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garithmic

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mmation proGreek letter si

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ple: Find ∑12

7=n

sting the termrom m = 7 to

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We can solve trms

5+9+7+5

n alternative The term iIn this caseIf the probmiddle two

iven a set ofot work quence c sequences ple: 1, 2, 4, 8the common

2=

Math

k of the termnumber of poe number of st post and d

e difference d

know n = 23,

14=2

97)+(31

oblems will uigma is ∑

e sum of the

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∑ m3

ms, we find to m = 12, we

use the summ

s the average the arithmetthis problem

9=13+11

strategy is toin the middlee, the middleblem had an o terms to finf unrelated n

are patterns o8, 16… n ratio r is 2

h Power Guide |

ms as fence poosts by addinf gaps then, wdivide that di

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, we can use

472

use sigma not

numbers 1 t

he lower bound limits of sumas 1 + 2 + 3

that they are have 6 term

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of two or motic mean of 5by adding u

o recognize the will equal te of the five teven numbend the averagnumbers, of c

of numbers w

31

osts separatedng one to thewe must take istance by the

ur set-up is n

the summati

tation

to 10 as ∑10

1=k

k

und, and inc

mmation + 4 + 5 + 6

21, 24, 27, 3ms (12 – 7 + 1

mula, sum = 6

ore numbers5, 7, 9, 11, anup all the term

he terms as athe average terms is 9 er of terms, wge of the whocourse, the a

with a comm

d by uniforme number of g

the distancee length of a

n = +3

31 - 97

ion formula t

k

creases by 1 f

+ 7 + 8 + 9 +

30, 33, 36 1 = 6)

1=2

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s nd 13? ms and divid

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we would onlole sequence arithmetic se

mon ratio r

m gaps, then wgaps e between thegap

23=1

to find the su

for each term

+ 10

171

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ly need to av

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e last post

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number of

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24

To fin Examp

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2=

nd the nth terple: What is

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1024512 =

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512256 =

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he 13th term

ries ric series is th

formula to fi

mon ratio

ple: Find ∑5

1=p

he first term he second ter

We can find th

We know that

hus, the sum

s te series is thn infinite seri|r| ≥ 1, the

um Example: tterms will

ple: Find ∑∞

0=x

ecause this se

ill be similar ries

he formula is

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formula be

Mat

rm of a geomthe 13th term

ratio r is 21

is (1024)( 21

he sum of a g

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∑ p)2-(3

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he common r

n = 5 becaus

m is (-2)-1(-2-6(1-

e sum of a paies to be solvaseries will co

the series 2, simply keep

∑ x21

eries has a co

to the form

s sum = 1

first

rator differs f

∞, (21 )∞ ap

ecomes 1

th Power Guide

metric sequenm in the sequ

)13–1=41

geometric seq

of the first

–6 = 12

ratio r by div

se the index p)5

= 3

3+6(1-

attern of numable, |r| < 1ontinue to g

4, 8, 16… (getting bigge

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pproaches ze

| 32

nce, use the fouence that beg

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n terms is

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p goes from 32) = –2(33)

mbers with a

grow infinite

(r = 2) does er

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ng the sum o

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r-1term)(1 (first

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) = –66

an infinite nu

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not have a f

e formula we

of the first n

metric series f

e (1 – rn) in

term = (1st te24, 512, 256

)r-1 n

, wher

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fixed sum be

e use to find

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n the geomet

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e r is the

ms

pproach a

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7 The harmnaturally re

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Examp 3

81 Th Th

rat

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s progression

ven though tonvergent exan the harmon

mean

ric mean is t

ple: What is 27× = 81

11/2 = 9=81hus, the geomhe answer matio of 3

ts its name fromalf, a third, a fo

Math

2=2

11=

21

21

0

to find a numcan also diveres do not addseries7 is a co

n is 11 +

21 +

31

the terms becample, in whnic series, the

he product o

the geometri

9 metric mean akes sense be

m the way a strourth, etc. of th

Power Guide |

2

mber for the rge d up to a niceommon exam

+41 +

51 … o

come smallehich each term

terms keep a

of n terms rai

ic mean of 3

of 3 and 27 ecause 3, 9, a

ring vibrates. The length of the

33

sum, which m

e number mple of a div

or ∑∞

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r, they don’tm was half ofadding up to

ised to n1

and 27?

is 9 and 27 form

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vergent series

t scale downf the previouo infinity

a geometric

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series with a

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nverged

ms in the

common

uencies that

G

Graphing The follow

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0

10

20

30

40

50

60

wing are grap

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he above grap

The dots n

0

5

10

15

20

25

30

35

0

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Mat

phs of variou

e above graphdistances betwed the dots, tlinear) rate

ph models th

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2

4

A

th Power Guide

s sequences a

h represents ween each otthey would f

he series 1=n∑10

ve equal verti

4 6

n

Arithmetic S

6

n

Arithmetic Se

| 34

and series

an arithmetither form a straigh

n

ical distances

8

Sequence

8

eries

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10

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because the te

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Power Guide |

uence, each t

metric series

96094, and w

6

n

Geometric Se

6

n

eometric Sequ

35

term is twice

n1=n 2

4∑10

we can see th

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hat the graph

10

10

he previous o

approaches 4

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Power Guide |

uence, each t

metric series

96094, and w

6

n

Geometric Se

6

n

eometric Sequ

35

term is twice

n1=n 2

4∑10

we can see th

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Right T T

Sp

POWER

Geometrydimensionright trianexplore hoaddition t

Triangles The Pythagor The Pyth

a2 + b2 = c a and

is the We can a

whether a If a2 +

c iare

If a2 + A Pythago

Examp 3, 5, 7, 9, 8,

Any multi 6, 8, 1

53 ,

54

pecial triang The 45-45

Two a The tw The le

length Drawi

square

PREVIEW

y is the studynal). Of particngles) and quow to find tho several othe

rean theoremhagorean thec2 b are the twlength of the

also use the non-right tr b2 > c2, thenis the length e the lengths b2 < c2, thenorean triple iples of comm4, 5 12, 13 24, 25 40, 41 15, 17

iple of a Pyth10 is a multip

, 1 is also a m

gles: 45-45-95-90 right trangles are 45wo legs are eqength of the

h of each leg ing a diagone results in a

y of figures cular interest uadrilaterals. Ie area and vor topics.

Math

m eorem states

wo leg lengthse hypotenusePythagorean

riangle is acutn the triangleof the triang

s of the othern the triangleis a set of thr

mon Pythago

hagorean tripple of 3, 4, 5,

multiple of 3

90 and 30-6riangle is an i°, and the 3rd

qual in lengte hypotenuse

nal from co45-45-90 tri

(both two-are triangles (In this sectio

olume of such

Power Guide |

a special re

s of the righte n theorem tote or obtuse

e is acute gle’s longest lr two legs e is obtuse ree integers threan triples:

ple also satisfi, so it is also

3, 4, 5

0-90 isosceles righd angle is 90°h

e is always

orner to coriangle

and three-(specifically on, we will h figures, in

37

elationship t

t triangle; c

o determine

leg; a and b

hat satisfy th

fies the Pythaa Pythagorea

ht triangle °

2 times th

rner across

POWER NO

Accordquestiothis sec

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he Pythagorea

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he distance fo2

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be parallel oel lines are linlines m and

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orizontal anopes are 0 an

Mat

riangle is the triangle meas opposite th

e hypotenuse

other leg (thngth of the stude in an engles

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of the two co

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he slope is unthe slope is 0uation as “risormula to fin

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ce rived from thation of the y2, z2) in thr

212

2 )y-y(+or perpendicunes in the samn are parallelave the sames are lines thn are perpenperpendicula

If a line has a

nd vertical lid undefined,

th Power Guide

second speciasure 30°, 60°he 30° anglee is 2 times t

he leg opposishortest sideequilateral tr

the point eqwo end point

oordinates:

horizontal ch

two points (a

ndefined, and, and the twose over run”

nd the distanc

ween two poi

he Pythagoredistance form

ree-dimension2

12 )z-z(+ ular me plane thal, it is notatee slope at intersect to

ndicular, it is ar lines are ne

a slope of 53 ,

nes are perp, respectively

| 38

ial right trian°, and 90°

the length o

ite the 60° an

riangle resul

uidistant fros (a, b) and a c b d,

2 2+ +

hange

a, b) and (c, d

d the two poo points lie o ce between a

ints (x1, y1) a

ean theoremmula to findnal space

at never intered as m || n

o form 90° anotated as m

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x

ngle

of the

ngle)

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m both ends(c, d), the m

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oints lie on a on a horizont

any two poin

and (x2, y2)

d the distanc

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angles m n⊥ rocals of each

cular line has

o each othe

30°

x

3

s midpoint is

m = ΔΔ=

a-cb-d

vertical linetal line

nts

ce between tw

h other

a slope of −

er, even thou

60°

2x

x

found by

xΔyΔ

wo points

53

−

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P

8 The Princthe big ang

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and 8) Corres

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congru Altern

congru Conse

180°) Same-

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A trapezo The p The n An iso A righ

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gles are congrue

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nate interioruent

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measures of thormula for thid is a quadrarallel sides a

non-parallel siosceles trapezht trapezoid h

= b+b)(21( 21

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ent sides musus is a paralleoperties of a agonals are pe

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Math

that intersectt right, 1 andent gles (1 and 5

r angles (4

r angles (1

s (1 and 4, 2

angles (4 anquadrilateralur-sided polyhe interior anhe number ofilateral with are called basides are calle

zoid has conghas one right

)h)(

he lengths ofstem, the twohave differentuadrilateral wnd sides are cs are supplemect each otherns to halve an

of a base andstem, oppositelogram witharallelogramsongruent L is the lengtstem, opposist be perpendelogram withparallelogramerpendicular

above stuff reagle and a big an

Power Guide | 3

ts two paralled 3, 2 and 4

5, 2 and 6, 4

and 6, 3

and 7, 2

2 and 3, 5 a

d 5, 3 and 6)ls ygon ngles of a quaf angles in a pexactly one pses d legs

gruent legs, bbase angle

f the bases ano bases have tt slopes with two paircongruent mentary r n angle

d h is the perte sides have

h four right ans apply to rec

th and w is thite sides havedicular h four congrum apply to a to each othe

lly nicely in “Fngle are supplem

39

el lines 4, 5 and 7, 6

and 8, 3 and

and 5) ar

and 8) ar

and 8, 6 and

) are supplem

adrilateral adpolygon is (npair of parall

base angles, a

nd h is the hethe same slop

rs of parallel

rpendicular hthe same slongles ctangles

he width e the same sl

uent sides rhombus

er

Fred’s Theoremmentary. – Dean

6

d

re

re

d 7) are supp

mentary8

dd up to 360°n – 2)/180 el sides

and diagonals

eight pe (since they

sides

height ope and lengt

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8

lementary (a

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y are parallel

th

gth, and the

l angles are con

b2

h

1

34

5 6

7

b1

add up to

l)

slopes of

ngruent. All

2

2

3

1

Congru C

S

The d The d

Area =

d1

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uency and SCongruence Two figur

In oth The fo

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imilarity Two figur

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diagonals bisediagonals bise

= 21dd21

and d2 are thcoordinate syh other (becais a parallelogoperties of re

diagonals form= s2 is the length

diagonals are

Similarity

res are congruher words, coollowing figu

he triangle isea remain th

res are similaollowing figu

Mat

ect each otherect the corner

he lengths ofystem, the sloause they are gram that is bectangles andm 4 congruen

of one side perpendicula

uent if they hngruent figu

ures are all co

s rotated ande same

ar if they haveures are all sim

th Power Guide

r r angles to fo

f its two diagopes of the dperpendicul

both a rectand rhombuses nt isosceles ri

ar, bisect each

have the samures have sideongruent

d flipped seve

e the same shmilar

| 40

orm 4 congru

gonals diagonals arelar) ngle and a rhapply to squight triangles

h other, and

me shape and es and angles

eral different

hape

uent right tria

e negative rec

hombus uares s

have the sam

area of the same

t ways, but t

angles

ciprocals

me length

measures

the figure’s s

shape and

Plane an A

A

9 This form

Th

nd Solid FArea of triang There are

A = 21

b i Heron

a,

A = 21

a a Formulas

question These

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A = πr r i

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If

If

Area of regula Regular po

We cabase

mula is notoriou

he ellipses are

Figures gles, quadrilseveral form

bh2

is the length n’s formula9:

b, and c are

Csinab

and b are twofor finding t

formulas crties and typthe two-dim

r2 is the radius octor” of a circ

you have the

“Arc measu

you have the

ar polygons olygons havean divide the

usly difficult to

Math

e different siz

laterals, and mulas that allo

of the base a A = a)-s(s

the lengths o

o sides and Cthe area of qu

can all be fes of quadril

mensional set

of the circlecle is visually

e arc measure

ure” is the de

e arc measure

e sides of equese polygons

type into calcu

Power Guide |

zes, but they

circles ow us to find

and h is the hc)-b)(s-)(s

of the three s

C is the measuadrilaterals

found in thaterals”) of all points

analogous to

ement in deg

egree measur

ement in rad

ual length andinto isoscele

lators. Be carefu

41

y all have the

d the area of a

height; A is a

sides of the tr

ure of the anvary depend

he previous

equidistant

o a slice of pi

grees, A = πr2

re of the “cru

dians, A = πr2

d angles of eqes triangles, w

ful. – Steven

same shape

a triangle

area

riangle; s = a

ngle between ding on the ty

section (“C

from one cen

ie

2 360

measurearc

ust” of the sec2

radiansmeasurearc

qual size with each sid

2c+b+a

these two sidype of quadr

oordinate G

nter point

e

ctor slice e

de of the poly

des rilateral in

Geometry:

ygon as a

A

10 This is aexclusively

We catriangl

Examplength A

Th An

sid

Us

A

Thth

Area and volu A prism c

Surfacare rec

Volum A pyramid

This b SA = a

actually the defon regular pris

an then findles to find thple: Find the

h of 4.153 heptagon (so

he side lengthn apothem isde

Apothems In the drawLater, you For now, t

sing the f

= 15.4)(4(21

he heptagon e area of the

The area oume of prismonsists of tw

ce area = areactangles) me = (area of d is akin to abase rises up tarea of the ba

finition of a regms, so we will,

Mat

d the area of he area of thee area of a r

ometimes cal

h is 4, so thes the distanc

are always pwing above, twill be able the height (ap

formula for

306.8=)53

has 7 sides atriangle by 7f the heptago

ms, pyramidwo parallel anda of the 2 bas

f a base)(heigha prism, but ito a vertex (pase + area of

gular prism. Mtoo.

th Power Guide

f each isoscel whole polygegular hepta

lled a septago

base of the ice from the c

erpendicularthe apothem to use trigonpothem) is gi

r the area

and, therefor7 on is A = 8.3

ds, cylinders,d congruent ses + area of

ht) it has one baspoint of interthe lateral fa

Most basic math

| 42

les triangle agon agon with a

on) has 7 side

isosceles triancenter of a re

r to the sidesis the height

nometry to finiven as 4.153

of a tria

re, 7 isosceles

306 × 7 = 58, spheres, anbases and thlateral faces

se instead of rsection of th

aces (for our p

h (including De

and multiply

side length o

es

ngle is 4 egular polygo

t of the triannd the heigh3

angle, A =

s triangles, so

.142 nd cones he space betw(for our purp

f two he sides) purposes, tria

ecathlon math)

y it by the n

of 4 and an

on to the mi

gle ht of the trian

= bh21

, w

o we need to

ween the two poses, the lat

angles)

, however, focu

umber of

apothem

iddle of a

ngle

we have

o multiply

bases10 teral faces

uses almost

P

P

Volum

A cylinder SA = 2

r i Volum

A sphere i SA = 4

r i

Volum

A cone is

SA = π r i

Volum

Properties of Correspon There are

SSS siside le

SAS slengththen t

AA simare co

These theo If all

similaProperties of Angle mea

A circl π Ex

The m The m The m

interce The m

interce Tangents,

me = ( 31

)(are

r is essentially2πr2 + 2πrh is the radius ome = πr2h is the three-d4πr2 is the radius o

me = 34

πr3

a pyramid w

πr2 + πr 2 +ris the radius o

22 h+r is th

me = 31

πr2h

f similar figunding parts oa few ways timilarity theengths form aimilarity the

hs form a conhe two triangmilarity theongruent, thenorems can bethe correspor

f circles asures are an le has 360° oradians = 18xample: How

(1)( π

180

)

measure of a cmeasure of anmeasure of epted arcs (se

measure of aepted arcs secants, and

Math

ea of the base

y a circular p

of a base, and

dimensional s

of the sphere

with a circular2h+

of the base, a

he lateral heig

ures of similar figuo test triangl

eorem: if twoa constant rateorem: if twonstant ratio agles are similorem: if two n the trianglee extended toonding angle

important por 2π radians0°

w many degre

) = 3.57

central angle n inscribed anan angle in ee circle diag

an angle in t

d chords are t

Power Guide | 4

e)(height)

prism

d h is the hei

set of all poin

e

r base

and h is the h

ght, the dista

ures are proples for similaro triangles extio, then the o triangles exand the anglelar triangles exies are similaro other geomes in two fig

part of circle gs

ees is 1 radian

is equal to thngle is equal t

the interiograms on the the exterior o

the main thre

43

ight of the cy

nts equidistan

height

ance from the

ortional rity xist such thatwo triangle

xist such thaes included b

ist such that r

metric figures,gures are con

geometry

n?

he measure oto the half th

or of the cirlast page of tof the circle

ee types of lin

ylinder

nt from one

e edge of the

at all three pes are similar at two pairs obetween thos

two pairs of

, too ngruent, then

of the intercehe measure orcle is half this section) is half the

nes associate

center point

base to the t

pairs of corre

of corresponse sides are co

f correspondi

n the two fi

epted arc of the intercepthe sum of

difference of

d with circle

top point

esponding

nding side ongruent,

ing angles

igures are

pted arc f the two

f the two

s

A tang Ta Ta

A seca Se

A cho Th If

If are

Chordthat thchord’

Secantextern

Secantpart is

gent is a line angent lines “angents are pant is a line thecant lines gord is a line sehe longest chtwo chords a

Their intertwo chords ae the same did-Chord Powhe product o’s line segmet-Tangent P

nal part is equt-Secant Pows equal to the

Mat

that intersec“touch” circl

perpendicularhat intersects

o through circegment whoshord in a circare the same rsected arcs aare congruenistance fromwer Theoremof one chordnt lengths (s

Power Theoual to the squwer Theoreme product of t

th Power Guide

cts the circle les r to the radius the circle atcles se two endpocle is the diamdistance from

are also congrnt or if their

the center om: two interd’s line segmee circle diag

orem: the pruare of the lem: the produthe lengths o

| 44

at only one p

us drawn to tt two points

oints lie on thmeter m the center ruent intersected af the same cirsecting chorment lengthsgrams below)roduct of thngth of the t

uct of the lengof the other s

point

the point of t

he rim of the

of a circle, th

arcs are congircle rds form fous equals the ) he lengths otangent gths of one secant and its

tangency

e circle

hey are cong

gruent, the tw

ur line segmeproduct of

of the secan

secant and its external par

gruent

wo chords

ents such the other

t and its

ts external rt

⊥

⊥

∠ ∠

∠ ∠∠

Right T In

E

11 An easy w

POWER

Trigonomrelationshiare also caderived fro

Triangle Ren a right tria

sine of an

cosine of a

tangent of

Examples

sinA = cos

sinB = cos

tanA = cot

tanB = cot

secA = csc

secB = csc

csc is the r sinC =

sec is the r cosC =

cot is the r cotC =

way to rememb

PREVIEW

metry is the ips of planar alled the circuom the unit ci

elationshipangle ABC w

angle = hyp

op

an angle = h

f an angle =

sB = ca

sA = cb

tB = ba

tA = ab

cB = bc

cA = ac

reciprocal of = 1 cscC =reciprocal of = 0 secC ireciprocal of = 0 tanC i

ber these three p

study of anfigures. The

ular functionsircle.

Mat

ps where C is th

potenusepposite

hypotenuseadjacent

adjacentopposite 11

sin = 1 cos

is undefinedf tan is undefined

properties is wit

ngles and thtrigonometric

s because they

th Power Guide

he right angl

th the mnemon

he angular c functions y can all be

| 46

TR

le

nic “SOH-CAH

POWER NO

Accordquestiothis sec

Covers basic gu

RIGON

H-TOA.”

OTES

ding to the ons (20% of thction pages 33-35uide

NOMET

USAD ouhe test) will co

5 in the USA

TRY

utline, 7 ome from

AD math

Trigono T

T

Inverse B

12 My Algefunctions is

ometric FuTrig function The sign o

All thrpositiv

Sine is Tange Cosin

Each of thand cotancorrespon

Trig function We can u

of a trigon If the angl

Examp If the angl

Examp

If the angl Examp

If the angl

Examp

When usion the res

Examp

3π4

co

In

Th

TrigonomBasic informa The inver Basically,

Simila sin-1A is th

bra II teacher ts all functions,

unctions ns and quadrof the value oree main funve in Quadras positive in Qent is positivee is positive i

he three recipngent) is posiding “main”

ns and referese the refere

nometric funle θ is in Quple: 60 is inle θ is in Qu

ple: 4π3 is in

le θ is in Quple: 200 is ile θ is in Qu

ple: 3π5 is in

ng reference ult

ple: Find cos

3π is in Quad

os(21=)

3π

n Quadrant II

hus, cos( )3π4

metric Funcation

rse trig functiif sinA = B, t

ar relationshihe same as ar

taught me a tricsine, tangent, a

Math

rants of a function nctions (sine,ant I Quadrant IIe in Quadranin Quadrant procal functioitive in the sfunction

ence angles nce angle toction uadrant I, θn Quadrant Iuadrant II, 18

n Quadrant I

uadrant III, θin Quadrant uadrant IV, 3

n Quadrant IV

angles, follo

s( )3π4

drant III, so i

II, tangent is

21-=)

ctions

ions include then arcsinB ps apply for

rcsinA

ck to remembeand cosine. All

Power Guide | 4

depends on , cosine, tang

nt III IV12

ons (cosecantsame quadran

o determine t

is the referenI, so its refere80° – θ (or π

I, so its referθ – 180° or θIII, so its ref

360° – θ or

V, so its refer

ow the ASTC

its reference

s positive, and

arcsin, arcco= A the other inv

r this. If you gostudents take c

47

the quadrangent) are

t, secant, nts as its

the value

nce angle ence angle is π −θ ) is the

rence angle is

π−θ is the rference angle2π – θ is th

rence angle i

C rule mentio

angle is -3π4

d sine and co

os, arctan, arc

verse function

o in order fromclasses. – Dean

nt of the angl

60 reference an

s 4π=

4π3-π

reference ange is 200 – 1

he reference a

is 3π=

3π5-π2

oned above t

3π=π-

osine are neg

ccsc, arcsec, a

ns as well

m quadrants I to

S

T

e

ngle

gle 180 = 20

angle

3π

to put the co

gative

and arccot

o IV, the order

A

C

orrect sign

r of positive

E

T

Graphs P

All invEvaluating in Substituti

Examp

N

arc

The domains Trig funct To be able

ranges (see These

Period The perio

All trig Sine and c Tangent a The perio

coefficient

Examp

verse trig funnverse trig exon can be a p

ple: to evalua

ow, we’re jus

csin2π=

21

an

Notice tha

Otherw

s and ranges tions don’t pe to work wie table belowlimitations e

Function

Arcsin

Arccos

Arctan

Arccsc

Arcsec

Arccot

od of a functigonometric fcosine (and tand cotangenods of sine at of the angle

ple: the perio

Mat

nctions can bxpressions powerful too

ate cos(arcsin

st trying to so

nd cos2

=6π

at we use just

wise, cos =6π5

of inverse trpass the horizith the invers

w) ensure that th

Invers

n

(

(---

on is the intefunctions areheir reciproc

nt have periodand cosine (ae (here, x)

od of sin(kx)

th Power Guide

e notated eit

l in evaluatin

n( ))21 , let θ =

olve cosθ

23

the principa

23-= would

rig functionszontal line tesse functions a

he inverse fu

se Trig Func

Domain

[--- 1, 1]

[--- 1, 1]

−∞ ∞( , )

(--- ∞ ,---1]∪ [1, ∞

-- ∞ ,---1]∪ [1, ∞

( , )−∞ ∞

erval over whe periodic cal functions)ds of π and their rec

is kπ2

| 48

ther way

ng inverse tri

=arcsin21

al value of arc

d also be an a

s st, so their inas functions,

unctions pass

ctions

[

(

∞ ) π−[2

∞ ) π

[0,2

hich it repeat

) have period

ciprocal func

ig functions

csin

answer

nverses are no, we must lim

the vertical l

Range

[2

,2

-ππ

]

[0, π ]

2π,

2π- )

π∪,0) (0, ]2

π π π∪) ( , ]2 2

(0, π )

ts

ds of 2π

ctions) can b

ot functionsmit their dom

line test

be determine

mains and

ed by the

A

H

V

C

The perio(again, x)

Examp

Amplitude The amp

minimum Since

Sin and co Examp

The otherthe coeffic

Horizontal sh A constan

Examp N

ne If

Vertical shifts A constan

Examp N If

Combining a

Example:

The fi

f(x

O

This f

of 1 d

ds of tangen

ple: the perio

plitude of a m height of a w

amplitude mos have amplple: the amplr functions dcient can strehifts

nt term insideple: the horizote that the

egative the functions

nt term outsidple: the vertiote that this the function

all these prop

f(x) = 3cos(7

irst thing we

x) = 3cos[7(x

nly when x is

function has

own

Math

nt and cotang

od of tan(kx)

cyclical funwave

measures distalitudes that clitude of kco

don’t really hetch the graph

e the functionzontal shift o

shift is pos

n were sec(x +

de the functiocal shift of sishift is a pos

n were sin(x) perties

7x + 1-)2π7

need to do is

x + 1-)]2π

s by itself can

a period of

Power Guide | 4

gent can also

is kπ

nction is ha

ance, it is alwcan be determosx is |k| ave an “amph vertically

n can horizonof sec(x – k) iitive (to the

+ k), the shift

on can verticin(x) + k is ksitive shift – k, the shift

s factor out t

n we find the

7π2 , an ampl

49

o be determin

alf the distan

ways positivemined by the

plitude” becau

ntally shift ais k to the rige right) even

t would be n

cally shift theupward

t would be n

the coefficien

e period and

litude of 3, a

ned by the co

nce between

coefficient o

use their ran

function’s gght n though the

egative (to th

e function’s g

egative (down

nt attached to

horizontal sh

a shift of 2π

oefficient of

n the maxim

of the functio

nge is unboun

graph

e coefficient

he left)

graph

n)

o the x

hift

to the left, an

the angle

mum and

on

nded, but

( – k) is

nd a shift

Identiti P

R

Q

P

S

D

ies Purpose Oftentime You’ll havReciprocal id

sin x = csc

1

cos x = se

tan x = co

Quotient iden

tan x = cosi

cot x = sinco

Pythagorean sin2x + cos tan2x + 1 = 1 + cot2x =

um identitie sin(x + y) cos(x + y)

tan(x + y)

Difference id sin(x – y) cos(x – y)

tan (x – y)

vertical sh

es, problems ve to convert dentities

xc1 xcsc

xec1

xsec

xot1

xcot

ntities

xosxn =

xcscxsec

xnxos =

xsecxcsc

identities s2x = 1 = sec2x = csc2x es = (sinx)(cosy= (cosx)(cosy

= )x(tan-1t+xtan

dentities = (sinx)(cosy= (cosx)(cosy

) = x(tan+1-xtan

hift = A

Mat

with trig funfunctions us

= xsin

1

=

xcos1

x =

xtan1

y) + (cosx)(siny) + (sinx)(si

)y(tanytan

y) – (cosx)(siny) + (sinx)(si

)y)(tanxytan-

*

Gra

th Power Guide

nctions in thesing the iden

ny) ny)

ny) ny)

period =

*phase disp

aph of Bsin(C

| 50

em will not bntities below

2Cπ

placement =

Cx + D) + A

be solvable asto solve the p

amplitude =

amplitude =

DC

−

A

s presented problem

= B

= B

D

H

P

O

S

P

Trigono L

Double angle sin(2x) = 2 cos(2x) =

tan(2x) =

Half angle id

±=)2xsin(

±=)2xcos(

±=)2xtan(

Phase identit

sinx = cos

cosx = sin

Odd/even pro sin(–x) = – csc(–x) = – tan(–x) = cot(–x) = – cos(–x) = sec(–x) = s

um-to-prod

sin+xsin

sin-xsin

sin-xsin

cos-xcos

Product-to-su

ycosxsin

ycosxsin

ysinxsin

ometric EqLaw of Sines The law o

opposite s

e identities 2(sinx)(cosx)cos2x – sin2x

xtan-1xtan2

2

dentities

2xcos-1±

2xcos+1±

xcos+1xcos-1±

ties

)x-2πs(

)x-2πn(

operties –sinx –cscx –tanx –cotx cosx secx

duct identitie

2+xsin(2=yn

2+xcos(2=y

2+xcos(2=y

xsin(2-=ys

um identitie)y+xsin(=y

2)y+xsin(=y

2-)y-xcos(=

quations

of sines statesside is the sam

Math

= 1 – 2sin2x

es

2y-xcos()

2y+

2y-xsin()

2y+

2y-xcos()

2y+

2y-xsin()

2y+x

es

2)y-xsin(+

2)y-xsin(+

2)y+x(osc-

s that in a trime for all thr

h Power Guide |

x = 2cos2x – 1

)

)

)

)y

iangle, the raree angles

51

1

atio of the sinne of an anglle to the lenggth of the

L

A

a

Asin

As lonangle,

Law of Cosin The law o

Whereworks

Given(refer

c2 = a2

In

Algebraic equ Unless the

exist to a t To sol

If Fo

Check Ex

als To solve t

Chang Ex

Use su Ex

= b

Bsin = s

ng as we havewe can find

nes of cosines is aeas the Pythfor any trian

n two sides ato the above

2 + b2 – 2ab(cn a right trian

The cosinedisappears

uations invoere are restritrigonometrilve for all solx is a solutio

or tangent anThe period

k for other soxample: if a sso be a solutitrig equationsge all trigonoxample: cos2x

First use a Thus, (1 –

ubstitution ifxample: 2sin2

Let u = sinSubstitutio

The solutio

Mat

cCsin

e one angle-sithe rest of th

a general formhagorean thengle and the angle

triangle for cosC) ngle, c is the he of ninety in the Pythalving trig fuctions on doc equation utions, reme

on, then 360°nd cotangent,d of these funolutions solution to aion in Quadrs, isolate the

ometric exprex + sinx + 1 =Pythagorean

– sin2x) + sinxf necessary 2x + sinx – 1

nx on and factor

ons can be fo

A

th Power Guide

ide pair (A anhe variables

m of the Pytheorem only w

e between ththe following

hypotenuse, degrees is 0

agorean theorunctions omain and ra

ember that th° + nx, where, 180° + nx inctions is onl

a sine equatiorant II, since trigonometr

essions to the= 0 n identity to cx + 1 = 0

= 0

ring give us (

ound by solvi

C

c

b

| 52

nd a, B and b

hagorean theworks for ri

hem, we cang formula)

which mean, which is wrem

ange, an infi

he functions ae n is an inteis also a solutly 180°

on is found sine is positi

ric expressione same functi

convert all th

2u – 1)(u +

ing u = sinx =

Ba

b, or C and c

eorem ight triangle

n find the len

ns C is the rigwhy the last

inite number

are periodic eger, is also a tion

in Quadrantive in Quadr

n ion

he expression

1) = 0

= 21 and u =

B

c) and anoth

s, the law o

ngth of the t

ght angle term in the

r of possible

solution

t I, then therants I and II

ns to sine

= sinx = –1

er side or

of cosines

third side

e formula

solutions

re should I

Then solve x = 30°

Math

e for x ° and x = 270

Power Guide |

0°

53

TERMS

Arran

Com

Facto

Mult

Perce

Perm

Proba

TERMS

Abso

Arith

Arith

Arith

Asym

Com

Com

Com

Conv

Degr

Discr

S – GENERngement prin

bination

orial

tiplication pri

entage

mutation

ability

S – ALGEBlute value

hmetic sequen

hmetic series

hmetic mean

mptote

plex conjugat

plex number

posite functio

vergent

ee

riminant

RAL MATciple

inciple

BRA

nce

te

on

Mat

TH To find are indis

!r!n

An arran

matter; n

The prointegers l

To find several dnumber

Represen

An arran

n(n=Prn

The chaoutcomepossible

The nonnumber

A pattern

The sum

The aver

A line th

A pair of

Any numthe imag

A functio

Applies t

The high

In the qu

th Power Guide

the total numstinguishable,

ngement of a

-n)(!r(!n=Crn

oduct of a nless than n; th

the total nudifferent objecof choices for

nts100

n of the

ngement of a

r)!-!n

nce that a gies in which toutcomes

n-negative vais from 0 on

n of numbers

m of an arithm

rage of two or

hat a function

f complex num

mber in the foginary unit

on resulting f

to an infinite

hest exponent

uadratic form

| 54

mber of arrang, divide the t

a collection o

)!r-

non-negative his is expresse

umber of poscts (each withr each object

e whole

a collection

iven event wthe event occ

alue of a nuthe number l

s that have a c

metic sequence

r more numb

n approaches b

mbers in the

orm a ± bi wh

from using on

series which

t power of a p

mula, the part

POW

gements of ntotal number

of objects in

integer n wed as n!

ssibilities wheh several choi

of objects in

will happen; ecurs divided b

umber; in otline

common diffe

e

bers

but never reac

form a + bi a

here a and b a

ne function as

approaches a

polynomial; a

under the squ

WER LI

n objects wherof arrangem

which order

with all of th

en picking onices), multipl

n which orde

equal to the nby the total n

ther words,

ference

ches

and a – bi

are real numb

s the input of

a fixed sum (|r

also known as

uare root; b2

ISTS

re r objects ments by r!:

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