60°

30°

Chapter 1y = af[b(x-h)]+k

a – vertical dilation (x, ay)if 0<a<1 → vertical compressionif a>1 → vertical expansion

b – horizontal dilation (xb

, y)

if 0<b<1 → horizontal expansionif b>1 → horizontal compression

h – horizontal translation (x+h, y)if h>0 → rightif h<0 → left

k – vertical translation (x, y+k)if k>0 → upif k<0 → down

Reflections-If a is negative, reflect over x-axis (y values become opposite) (y=-f(x)) (-y=f(x)) (x, -y)-If b is negative, reflect over y-axis (x values become opposite) (y=f(-x)) (-x, y)

Inversey=f(x) → x=f(y) [f-1(x), reflect over line y=x (swap x & y values) (y, x)to solve interchange x & y in the fn & solve for y

Reciprocals

y=f(x) → y=1f (x)

(where y=0 on original graph →

vertical asymptote(s)) (x, 1y

) (if y is pos, reciprocal

also remains pos; if neg, stays neg) (x-axis: horizontal asymptote) (y=1, y=-1 stay the same

Absolute Valuesy=|f(x)| (to graph reflect all parts of original graph that are below the x-axis to above the x-axis; all parts above the x-axis stay the same)

y=f(|x|) (to graph first graph y=f(x) then reflect only the positive x values over the y-axis, also keep the points with a positive x value)

Chapter 4Conversions

n°(π180°

) = mrad

mrad(180°π

)= n°

Sector Anglea=rӨ (Ө in rad)

Angle DefinitionsStandard position – angle has a vertex on the origin & initial arm on positive x-axis; rotation counter clockwise → angle is positive; rotation clockwise → angle is negative

Coterminal angle - ±360° or ±2πradPrincipal angle – the least of the positive coterminal angles

Special Right ∆s

*factor out coefficient before finding phase shift*y=asin[b(Ө-c)]+d (b=2π/per; per=2π/b

a-amp, b-per, c-ps, d-vd*put calc in RAD mode for q’s like ‘cos2’ or ‘sec2’

Chapter 5ex. find the exact solutions

a) sinӨ = 12

Ө = π6

Ө = π - -π6

=5π6

b) cosӨ = 12

Ө = π - -π3

= 2π3

Ө = π + -π3

= 4 π3

ex. use graphs to solve a) sinӨ = 0 Ө = 0, π

b) cosӨ = 1 Ө = 0

c) tanӨ = 0 Ө = 0, π

General SolutionAdd the per to the smallest positive solutionex. find the general solution of sinӨ = 0 Ө = 0, π GS: Ө = nπ where n is an integer

*for 0≤Ө<2π cosӨ has 2 solutions → cos4Ө has 4

times as many solutions. For 1 per sol’s are π8

&

3π8

; to get the sol’s for 0≤Ө<2π continuously add

the period until you reach 2π

ex. solve for x a) cscx=3.2 sinx=1/3.2 x1=0.3178 x2=π-0.3178=2.824

Solving 1st Degree Equationsex. Solve (0≤Ө<360°) cosӨ = -0.468 ref angle = 62.095 Ө = 180° - 62.095° = 117.905° Ө = 180° + 62.095° = 242.095°

ex. Solve (0≤Ө<360°) cscӨ = -1.346

sinӨ = 1

−1.346 ref angle = 47.98° Ө = 180° + 47.98° = 227.983° Ө = 360° - 47.98° = 312.017°

Solving 2nd Degree Equationsfactor or use quadratic formula ex. solve for 0≤Ө<2π cosӨ

a) cos2Ө = 13

√cos2Ө = √ 13

cosӨ = ± 1√3

cosӨ = 1√3

Ө = cos-1( 1√3

)

Ө = 0.955 Ө = 2π – 0.955 = 5.328

b) 4tan2Ө – tanӨ – 3 = 0 (tanӨ – 1)(4tanӨ + 3)

tanӨ = 1

Ө = π4

Ө = 5π4

c) csc2Ө + 5cscӨ + 6 = 0 (cscӨ + 3)(cscӨ + 2)

cscӨ = -3

sinӨ = - 13

ref angle = 0.34

Ө = π + 0.34 = 3.481 Ө = π + 0.34 = 5.943

d) secӨcscӨ – 3secӨ = 0 secӨ(cscӨ – 3) = 0

cosӨ = - 10

no sol. b/c cos Ө is defined everywhere

Chapter 2

y=bx b>1

y=bx-h+kasymptote: y=ldomain: x∈Rrange: y>k if y=+bx

y<k if y=-bx

A=P(1+r/n)nt

A=AoXt/T

X: growth or decay factorIncrease by 10% → 1.10Decrease by 10% → 0.90

xm * xn = xm+n

xm / xn = xm-n

(xm)n = xmn

(x/y)m = xm/ym

x-n = 1/xn

x0 = 1xm/n = (x1/n)m = n√(xm) = (n√x)m

inverse of y=2x is x=2y ... log2x=ylogbx=y ↔ by=x b>0, b≠1

y=bx

b>1

logbxb>1x>0

y= logb(x-h)+kasymptote: x=hdomain: x>h if y=logx

x<h of y=log(-x)range: y∈R

b) csc3.2=x 1/sin3.2=x x=-17.1308

cosӨ = - 1√3

Ө = cos-1( - 1√3

)

ref angle = 0.955Ө = π - 0.955 = 2.186Ө = π + 0.955 = 4.097

tanӨ = −34

Ө = tan-1( - 34

)

ref angle = -0.644Ө = π – 0.644 = 2.498

cscӨ = - 2

sinӨ = - 12

ref angle = π6

Ө = π + π6

= 7π6

sinӨ = 13

Ө = 0.34Ө = π – 0.34 = 2.802

y=bx 0<b<1

y=bx

0<b<1

logbx0<b<1x>1

Inverse x=by

logbbx=xblogbx=xlogbb=1

logb(xy) = logbx + logby

logb(x/y) = logbx - logby

logb(xn) = nlogbx

logb(n√x) = logb(x1/n) = (1/n)logbx

ex. b/w ’56 & ’76 the annual pH of pption dropped from 5.6 to 4.3. how many times as acidic as the pption in ’56 was the pption in ’76?

105.6

104.3=101.3=20׿

logba = logca/ logcb a>0, b>0, c>0; b≠1, c≠1

ex. log512 = log 12log 5

= 1.54

ex. simplify log92=x log32 = 2xex. solve for x a) 2logx = log32 + log2 logx2 = log(32 * 2) logx2 = log(64) x2 = 64 x = ±8

reject x = -8∴ x = 8

b) log2(-x) + log2(2-x) = 3 log2(x2 – 2x) = 3 23 = x2 – 2x x2 – 2x – 8 = 0 (x + 4)(x + 2) x = 4, -2

reject x = 4∴ x = -2

Continuous Growth/DecayA=Aoekt

Natural Logarithm (ln)y=ex ↔ logey=x ↔ lny=xlne = 1

ex. solve for x a) 45 = e0.075x

ln45 = lne0.075x

ln45 = 0.075x(lne)

x = ln 450.075

= 50.755

b) 7 = e1-x

ln7 = (1 – x)lne x = 1 – ln7

Chapter 6Geometric Sequence Formulatn = arn-1

Finding Termsex. find the indicated term 8, 16, 32, 64, ... t12

t12 = 8(2)12-1

t12 = 16384

Finding Number of Termsex. find the # of terms in the sequence 8, 16, 32, ... 131072 131072 = 8(2)n-1

16384 = 2n-1

log16384 = (n-1)log2 log16384 = nlog2 – log2 log16384 + log2 = nlog2

n = log 16384+ log2

log 2 n = 15

Finding the Sequenceex. find the geometric sequence with t3 = 45 &t7 = 3645 from t1 to t7

45, __, __, __, 3645 a = 45 r = ? n=5 t5 = 3645 3645 = 45r5-1

81 = r4

r = 3, -3r = 3: 5, 15, 45, 135, 405, 1215, 3645r = -3: 5, -15, 45, -135, 405, -1215, 3645

Geometric Meanex. Given the geometric sequence: a, b, c

→ (geometric ratio) ba= cb

→ b2 = ac

thus, geometric mean b/w a & c is √ac, -√ac

ex. Between 8 & 128, insert 1 geometric mean 8, x, 128

x8=128x

x2 = 1024 x = 32, -32

8, 32, 1288. -32, 128

ex. $8500 invested at 734

% per year compounded

monthly for 15 months

tn = nth term r = common ratioa = 1st term n = # terms

1

1

1 2 3 4 5 6 7 8

A=P(1+ rn )nt

A=8500(1+ 0.077512n )1512x12

= $9361.73

ApplicationsUse tn = an-1 to fill in all given info & find unknownex. If an apt. was purchased last year for $21000 and is now worth $226800. find the value 10 years from the original price. (Assume value continues to depreciate at same rate each year). 210000, 226000, ...

r = 226000210000

= 1.08

t11 = 210000(1.08)11-1

Geometric Series

Sn=a(1−rn)1−r

or Sn=a−rl1−r

l = last termSn = sum of 1st n termsa = first termr = common ration = # of terms

ex. find sum of 1st 8 terms of 14 - 7 + 3.5 - 1.75 + ...

S8

❑=14 (1−(−1

2)8n

)

1−¿¿

¿ 59564

= 9.27

ex. find t1 & t3 if t2 = 24 & S3 = 126 for a sequence

t1, t2, t3 → t1, 24, t3 → 24r

, 24, 24r

24r

+ 24 + 24r = 126

24 + 24r + 24r2 = 126 4r2 – 17r + 4 = 0 (r - 4)(4r - 1) = 0

r = 4, 14

6, 24, 9696, 24, 6

To find the last term of a series given the sum useSn-1 +tn = Sn or tn = Sn – Sn-1

ex. If Sn = 2n2 +3n find tn

tn = 2n2 +3n – [2(n-1)2 + 3(n-1)] = 2n2 +3n – [2(n2 – 2n + 1) + 3n-3] = 2n2 +3n – [2n2 – 4n + 2 + 3n-3]

= 4n + 1

Applicationsex. Suppose you drop a ball from a window 45m above the ground. It bounces to 65% of its previous height w/ each bounce. Find the total distance that the ball travels up to the 8th bounce.

S8=a (1−r n)1−r

=45(1−0.658)1−0.65

=124.47m

up & down: = 2 x 124.47 - 45 = 204.4 m

Sigma Notation ∑Greek letter used to abbreviate a series/sum

∑k=i

j

f (x )

k – index of summationi – lower limit of summationj – upper limit of summationf(x) - function

ex. Find the sum ∑k=2

5

7 k−3

k=2: 7(2)-3=11 k=3: 7(3)-3 =18 k=4: 7(4)-3=25 k=5: 7(5)-3=32

∑k=2

5

7 k−3 = 11+18+25+32 = 86

# terms in seq. = upper limit – lower limit + 1

ex. Find the sum

∑k=2

8

2k−1

k=1: 21-1=1 k=2: 22-1=2 k=3: 23-1=4

Sn=a (1−rn )1−r

S8=1(1−28)1−2

=255

Writing Sigma Notation ∑ex. Given 4+8+16+...+128 write ∑ notation1. need to know # of terms tn = 128 a = 4 r = 2 tn = arn-1

128 = 4(2)n-1

32 = 2n-1

(since n=1 corresponds to original amount at year 0 would make n = 11)

1) substitute lower limit into expression & solve2) add 1 to lower limit & repeat step 13) repeat steps 1 & 2 until k = upper limit4) sum all solutions

60

55.8

n = 62. find general term tn = 4(2)n-1 → tn = 22(2)n-1 → tn = 2n+1

∑k=1

6

2k+1

Geometric Series2+8+32+128+... 512+128+32+8+2

S∞ = ᴓ S∞ = 20483

sum of infinite series: S= a1−r

a = 1st termr = common ratio-1< r <1

ex. Find the sum of the series ∑k=3

∞

( 43)1−k

k=3: 916

k=4:2764

a = 34

r = 916

S=

916

1−34

S = 94

ex. Solve for x ∑k=1

∞

xk=25

k=1: x k=2: x2

25= x1−x

2-2x = 5x 2 = 7x

x = 72

ex. A weather balloon rises 60m the 1st min. & each min. After that it rises 7% less than the previous height. What is the max. Height reached by the balloon?

S= a1−r

a = 60 r = 0.93

S= 601−0.93

=857.14