Algebra 2 – Things to Remember!
Exponents:
x0  1
x m • x n  x m n
m
x
 x mn
n
x
1
xm  m
x
n m
( x )  x n• m
n
x
x
   n
y
 y
n
( xy ) n  x n • y n
Factoring:
Look to see if there is a GCF (greatest
common factor) first. ab  ac  a (b  c)
x 2  a 2  ( x  a )( x  a)
( x  a) 2  x 2  2ax  a 2
( x  a) 2  x 2  2ax  a 2
Factor by Grouping:
proportionality, k. Find k, and then proceed.
with advertising and inversely
with candy cost.
by 4, use remainder, solve.
ka
y
c
Logarithms
y  log b x  x  b y
ln x  log e x natural log
Properties of Logs:
log b b  1
log b 1  0
e = 2.71828…
m
log b    log b m  log b n
n
log b (m r )  r log b m
Domain: log b x is x  0
(a  bi ) conjugate (a  bi )
(a  bi )(a  bi )  a 2  b 2
Change of base formula:
a  bi  a 2  b 2 absolute value=magnitude
log b a 
Exponentials e x  exp( x)
b x  b y  x  y (b  0 and b  1)
If the bases are the same, set the
exponents equal and solve.
Solving exponential equations:
1. Isolate exponential expression.
2. Take log or ln of both sides.
3. Solve for the variable.
Variation: always involves the constant of
Direct variation: y  kx
k
Inverse variation: y 
x
Varies jointly: y  kxj
Combo: Sales vary directly
Complex Numbers:
1  i
a  i a ; a  0
2
14
i  1
i  i 2  1 divide exponent
log x  log10 x common log
log a
log b
Quadratic Equations: ax 2  bx  c  0 (Set = 0.)
Solve by factoring, completing the square, quadratic formula.
b 2  4ac  0 two real unequal roots
2
b  b  4ac
x
b 2  4ac  0 repeated real roots
2a
b 2  4ac  0 two complex roots
Square root property: If x 2  m, then x   m
Completing the square:
x2  2 x  5  0
1. If other than one, divide by coefficient of x2
2. Move constant term to other side x 2  2 x  5
3. Take half of coefficient of x, square it, add to both sides
x2  2x  1  5  1
ln( x) and e x are inverse functions
ln e x  x
eln x  x
ln e  1
eln 4  4
2
e 2ln 3  eln 3  9
4. Factor perfect square on left side.
Absolute Value: a  0
Sum of roots: r1  r2  
 a; a  0
a 
 a; a  0
m  b  m  b or m  b
m  b  b  m  b
log b (mn)  log b m  log b n
( x  1) 2  6
5. Use square root property to solve and get two answers. x  1  6
b
c
Product of roots: r1 r2 
a
a
2
Inequalities: x  x  12  0 Change to =, factor, locate
critical points on number line, check each section.
(x + 4)(x - 3) = 0
x = -4; x = 3
m  b  m  b or m  b
ANSWER: -4 < x < 3 or [-4, 3] (in interval notation)
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Radicals: Remember to use fractional exponents.
a
n
xx
an  a
1
a
m
n
x  n xm 
n
ab  n a  n b
n
 x
n
m
a na

b nb
Simplify: look for perfect powers.
x12 y17  x12 y16 y  x 6 y 8 y
72 x 9 y 8 z 3  3 89 x 9 y 6 y 2 z 3  2 x3 y 2 z 3 9 y 2
Use conjugates to rationalize denominators:
5
2 3
10  5 3

 10  5 3

2 3 2 3 42 3 2 3  9
Equations: isolate the radical; square both sides
to eliminate radical; combine; solve.
2 x  5 x  3  0  (2 x  3) 2  (5 x ) 2
3
4 x 2  12 x  9  25 x  solve : x  9; x  1/ 4
CHECK ANSWERS. Answer only x = 9.
Functions: A function is a set of ordered pairs in which
each x-element has only ONE y-element associated with it.
Vertical Line Test: is this graph a function?
Domain: x-values used; Range: y-values used
Onto: all elements in B used.
1-to-1: no element in B used more than once.
Composition: ( f  g )( x)  f ( g ( x))
Inverse functions f & g: f ( g ( x))  g ( f ( x))  x
Horizontal line test: will inverse be a function?
Transformations:
 f ( x) over x-axis; f ( x) over y-axis
f ( x  a) horizontal shift; f ( x)  a vertical shift
f (ax) stretch horizontal; af ( x) stretch vertical
Working with Rationals ( Fractions):
Simplify:
remember to look for a factoring of -1:
3 x  1 1( 3 x  1 )

 1
1  3x
1  3x
Add: Get the common denominator.
Factor first if possible:
Multiply and Divide: Factor First
Rational Inequalities
x 2  2 x  15
 0 The critical values
x2
from factoring the numerator are -3, 5.
The denominator is zero at x = 2.
Place on number line, and test sections.
Sequences
Arithmetic: an  a1  (n  1)d
n(a1  an )
Sn 
2
n 1
Geometric: an  a1 r
a1 (1  r n )
1 r
Recursive: Example:
a1  4; an  2an 1
Sn 
Binomial Theorem:
n
n
(a  b) n    a n  k b k
k 0  k 
Solving Rational Equations:
Get rid of the denominators by mult. all terms by
common denominator.
22
3
2


2
2x  9x  5 2x 1 x  5
multiply all by 2 x 2  9 x  5 and get
22  3( x  5)  2(2 x  1)
22  3x  15  4 x  2
37  3x  4 x  2
35  7 x
5 x
Great! But the only problem is that
x = 5 does not CHECK!!!! There is no solution.
Extraneous root.
Motto: Always CHECK ANSWERS.
Equations of Circles: x 2  y 2  r 2 center origin
( x  h) 2  ( y  k ) 2  r 2 center at (h,k)
x 2  y 2  Cx  Dy  E  0 general form
Complex Fractions:
Remember that the fraction bar means divide:
Method 1: Get common denominator top and bottom
2 4 2  4x
1

2
x2 x  x2  2  4 x  4 x  2  2  4 x  x
 1
4 2
4x  2
x2
x2
4x  2
x2
 2
x x
x2
Method 2: Mult. all terms by common denominator for
all.
2 4
2
4
x2  2  x2 

2
x
x 
x
x  2  4 x  1
4 2
4
2
4x  2
 2
x2   x2  2
x x
x
x
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Trigonometry –
Things to Remember!
Radians and Degrees
Arc Length of a Circle =  r (in radians)
Change to degrees multiply by
Special Right Triangles
Quadrantal angles – 0, 90, 180, 270
Change to radians multiply by
180
180

CoFunctions: examples
sin   cos(90º  ) ; tan   cot(90º  )
30º-60º-90º triangle
side opposite 30º = ½ hypotenuse
side opposite 60º = ½ hypotenuse

3
45º-45º-90º triangle
Inverse notation:
arcsin(x) = sin-1(x)
arccos(x) = cos-1(x)
arctan(x) = tan-1(x)
Trig Functions
o
a
o
sin   ; cos  ; tan 
h
h
a
h
h
a
csc   ; sec  ; cot 
o
a
o
Reciprocal Functions
1
1
1
sin  
; cos 
; tan 
csc 
sec 
cot 
1
1
1
csc  
; sec 
; cot 
sin 
cos 
tan 
tan  
sin 
cos 
Trig Graphs
sin x
hypotenuse = leg 2
leg = ½ hypotenuse 2
Law of Sines: uses 2 sides and 2 angles
sin A sin B sin C


Has an ambiguous case.
a
b
c
cot  
cos 
sin 
cos x
sinusoidal curve = any curve expressed as
y = A sin(B(x – C)) + D
Law of Cosines: uses 3 sides and 1 angle
c 2  a 2  b 2  2ab cos C
amplitude (A) = ½ | max – min| (think height)
Area of triangle: A = ½ ab sin C
Area of parallelogram: A = ab sin C
period = horizontal length of 1 complete cycle
Pythagorean Identities:
sin 2   cos 2   1 tan 2   1  sec 2 
1  cot 2   csc2 
horizontal shift (C) – movement left/right
frequency (B) = number of cycles in 2  (period)
vertical shift (D) – movement up/down
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Statistics and Probability –
Things to Remember!
Normal Distribution and Standard Deviation
Statistics:
x1  x2  ...  xn 1 n
  xi
n
n i 1
median = middle number in ordered data
mode = value occurring most often
mean  x 
range = difference between largest and smallest
mean absolute deviation (MAD):
1 n
population MAD   xi  x
n i 1
variance:
population variance  ( x) 2 
n
1
2
 xi  x 

n i 1
standard deviation:
population standard deviation =
x
1 n
2
 xi  x 

n i 1
Empirical Probability
# of times event E occurs
P( E ) 
total # of observed occurrences
Binomial Probability
r
nr
“exactly” r times
n Cr • p • q
n
or   • p r • (1  p ) n  r
r
[TI Calculator: binompdf(n, p, r)]
When computing "at least" and "at most"
probabilities, it is necessary to consider, in
addition to the given probability,
• all probabilities larger than the given
probability ("at least")
[TI Calculator: 1 – binomcdf(n, p, r-1)]
Sx = sample standard deviation
 x = population standard deviation
Probability
Permutation: without replacement
and order matters
n!
n Pr 
(n  r )!
Combination: without replacement
and order does not matter
 n  n Pr
n!

n Cr    
 r  r ! r !(n  r )!
• all probabilities smaller than the given
probability ("at most")
[TI Calculator: binomcdf(n, p, r)]
Theoretical Probability
n( E )
# of outcomes in E
P( E ) 

n( S ) total # of outcomes in S
P(A and B) = P(A)•P(B)
for independent events
P(A and B) = P(A)•P(B| A)
for dependent events
P(A’ ) = 1 – P(A)
P(A or B) = P(A) + P(B) – P(A and B)
for not mutually exclusive
P(A or B) = P(A) + P(B)
for mutually exclusive
P ( B | A) 
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P( A and B)
(conditional)
P( A)