flipper numbers. natural numbers: whole numbers: integers: rational numbers: fractions, decimals...

FlipperNumbers

Numbers

Natural Numbers:

Whole Numbers:

Integers:

Rational Numbers: Fractions, decimals that end, decimals that repeat

Irrational Numbers: Decimals that never end and never repeat

Real Numbers: The set of rational numbers and irrational numbers combined

{1, 2 ,3 , …

{0 , 1 ,2 , 3 , …

{…,−3 , −2 , −1 , 0 , 1 ,2,3 , …}

3=31

1.5=32

√4=2 .3=13

𝜋 √7 .65791698734 …√24

Variable and Terms

Variable and Terms Distributive Property

Variable: A letter or symbol that is used to represent one number or a set of number.Term: Consist of numbers and variables or a combination of numbers and variables. Like Terms: Must have the same variable(s) to the same exponent. Add or subtract coefficients when combining. Exponent does NOT change.

9 𝑥2+4 𝑥2+24 𝑥𝑦−7 𝑦2

Variable

Coefficient

Exponent

Term

Like Terms13 𝑥2

Distributive PropertyIf a, b and c are real numbers then:

Examples:

3 (𝑥+2 )=3𝑥+6 7 𝑦 (2 𝑥+2 𝑦 )=14 𝑥𝑦+14 𝑦 2

Be careful of negative signs outside the parenthesis.

3 𝑥−5 (𝑥− 8 )=3 𝑥−5 𝑥+40=− 2𝑥+40

Always SIMPLIFY

Order of Operations

Order of Operations Equations

Step 1 – Evaluate expressions inside grouping symbolsStep 2 – Evaluate PowersStep 3 – Multiply and divide from left to rightStep 4 – Add and subtract from left to right

PleaseExcuseMy DearAunt Sally

ParenthesisExponentsMultiplicationDivisionAdditionSubtraction

24 – (9 + 1)EX.24 – (32 + 1) == 24 – 10= 14

Equations

9 17x

8x 9 9

x = 427

–

( )(x ) =27

–72

–72

– 4

x = –14

One Step: Reciprocal:

Two Step:

2 5 15x 55

2x 102 2

5x

7x – 4x = 21

3x = 21

x = 7

= 33x

321

Combine Like Terms:

Equations (cont.)

7x + 2(x + 6) = 39

7x + 2x + 12 = 39

9x + 12 = 39

9x = 27

x = 3

7 – 8x = 4x – 17

7 – 8x + 8x = 4x – 17 + 8x

7 = 12x – 17

24 = 12x

2 = x

Distributive Property: Variables on Both Sides

2x + 10 = 2(x + 5)

2x + 10 = 2x + 10

Identity: ALL REALS 3x = 3(x + 4)

3x = 3x + 12

0 = 123x – 3x = 3x + 12 – 3x

No Solution:

Equations (cont.) Fractions or Decimals

Fractions or DecimalsTo solve an equation with fractions, multiply through by the LCD to eliminate the fractions.

To solve an equation with decimals, multiply through by either 10, 100, 1000, etc. to eliminate the decimals.

34𝑥−

12=

53

12( 34𝑥−

12=5

3 )

9 𝑥− 6=209 𝑥=26𝑥=

269

0.5 𝑥+0.02=−0.2100(0.5 𝑥+0.02=− 0.2)

50 𝑥+2=− 2050 𝑥=−22𝑥=− .44

Graph using a Table

Graph Using a Table Graph w/Intercepts

𝑥 𝑦

0 1 2 3

𝑦=2 𝑥−1

1. Draw Table2. Pick values3. Plug into equation to obtain y-

values4. Plot points on the coordinate

plane5. Draw line connecting the points

Graph w/InterceptsAx By C Use STANDARD FORM

C

A

C

Bx-intercept y-intercept

2 3 6x y

6

32

6

23

Slope Formula

Slope Formula Graph Slope-Intercept Form

mrise

run

y

x

y y

x x

2 1

2 1

( , )x y1 1

( , )x y2 2

x x2 1

y y2 1RISE

RUN Let (x1, y1) = (3, 5) and (x2, y2) = (6, –1).

–1 – 56 – 3

=

– 63= = –2

Horizontal lines have a slope of and vertical lines have an undefined

slope.

y x 2 3

Slope = 2

y-intercept = 3

2

1

Graph Slope-Intercept Form

Step 1: Put the equation in Step 2: Plot the y-intercept on the y-axis Step 3: From the y-intercept, two options to for slope

Positive slope (+) – Up and to the right Negative slope (-) – Down and to the right

Step 4: Draw a line through the two points

Graph Functions

Graph Functions Point-Slope Form

Function Notation –• f (x) is another name for y• Read “the value of f at x” or “f of x”

f(x) = 37x + 7

x f(x)

0 37(0) + 7 = 7

1 37(1) + 7 = 44

2 37(2) + 7 = 81

Point-Slope Formy y m x x 1 1( )

Slope

Point y + 2 = (x – 3).2

3

Graph in Point-Slope Form

Standard Form

Standard Form Slope-Intercept Form

Integers

Write an equation in standard form of a line whose slope is -3 and goes through the point (1, 1).

y – y1 = m(x – x1)

y – 1 = –3(x – 1)

𝑦−1=− 3𝑥+3

𝑦=− 3𝑥+43 𝑥+𝑦=4

𝐴𝑥+𝐵 𝑦=𝐶

Slope-Intercept FormWrite an equation of the line that passes through

(–2, 5) and (2, –1).

3m = y2 – y1x2 – x1

= –1 – 52 – (–2)

= –64

= – 2

y = mx + b

5 = – 32

(–2) + b

2 = b

Substitute – 32

for m and 2 for b.

y = – 32 x + 2

Parallel Lines

Parallel Lines Perpendicular Lines

Parallel Lines – Two lines the will never cross and have the same slope.

Write an equation of the line that passes through (–3, –5) and is parallel to the line y = 3x – 1.

y = mx + b

–5 = 3(–3) + b

4 = b

y = 3x + 4

Perpendicular Lines

1.Two lines are perpendicular if and only if the product of their slopes is -1.

2.You can also tell if two lines are perpendicular if their slopes are negative reciprocals of each other.

EX:

Opposite reciprocal would be

23

∙ −32=− 1

𝑚=2 𝑚=−12

Write an equation of the line that passes through (4, –5) and is perpendicular to the line y = 2x + 3.

–5 = – (4) + b1

2

y = mx + b

–3 = b

y = – x – 312

Inequalities

Inequalities Compound Inequalities

<

>

<

>

Less than or equal to

Greater than or equal to

Less than

Greater than

Closed Circle

Open Circle

When solving inequalities and you have to multiply or

divide by a negative number, reverse the inequality.

< 7x–6

x > –42

x–6

> –6–6 7

Compound Inequalities

x < –1 or x 4

OR Compound Inequality

= –3 x < 5 x –3 and x < 5

AND Compound Inequality

Absolute Value Equations

Absolute Value Eq. Absolute Value Ineq.

x – 3 = 8

x – 3 = 8 or x – 3 = –8

x = 11 or x = –5

Absolute Value: Distance a number is from zero. NEVER NEGATIVE

Absolute Value Inequalities

x – 5 7

x – 5 7 or x – 5 7

x –2 or x 12

Remember to flip the inequality for the negative!

Graph Systems

–x + y = –7

x + 4y = –8

1. Graph both lines on the same coordinate plane.

2. The ordered pair where the two lines cross is the solution to the system.

Special Cases: Lines are Parallel

Lines Coincide

no solutionmany

solutions

Graph Systems Substitution

Substitution1. Solve one of the equations for one of its variables.2. Substitute that expression into the other equation and solve.3. Once you find one variable, substitute into either original

equation to find the remaining variable.

y = 3x + 2

x + 2y = 11

7x + 4 = 11

7x = 7

x = 1

x + 2(3x + 2) = 11

x + 2y = 11 y = 3x + 2 y = 3(1) + 2 y = 3 + 2

y = 5

(1,5)

Elimination

Elimination Systems of Inequalities

• Multiply one or both equations to achieve same coefficient with the same variable.

• Add or Subtract the equations to eliminate one of the variables.• Solve the resulting equation for the other variable.• Substitute, in either original equation, to find the value of the

eliminated variable.

6x + 5y = 19

2x + 3y = 5

6x + 5y = 19

–6x – 9y = –15

–4y = 4y = –1

(4 ,−1)

Systems of Inequalities

y > –x – 2 y 3x + 6

• Graph both lines• Solid line and • Dashed line < and >

• Pick a test point• Shade if its true for both inequalities

Exponent Properties

Exponent Properties Exponent Properties

Product of PowersExample:

Power of a PowerExample:

Power of a ProductExample:

a a am n m n

5 5 54 2 6

( )a am n m n

( )3 37 3 21

( )a b a bm m m ( )3 2 3 24 4 4

Exponent Properties

Quotient of Powers

Example:

Power of a Quotient

Example:

a

aa a

m

nm n , 0

3

33 3 9

7

57 5 2

( ) ,a

b

a

bbm

m

m 0

( )4

3

4

3

16

92

2

2

Zero Exponents

Zero Exponents Negative Exponents

Let a be a nonzero number and let n be a positive integer. A nonzero number to the zero power is 1.

a a0 1 0 ,

Examples: (2 𝑥𝑦𝑧)0=1 (23423453 )0=1

Negative Exponents To make an exponent positive, write the

reciprocal.

aa

nn

1 1

aa

nn

2𝑥−2

𝑦2 = 2𝑥2 𝑦 2

𝑥3

𝑦− 4 𝑧2 =𝑥3 𝑦 4

𝑧 2

Exponential Growth

Exponential Growth Exponential Decay

EXPONENTIAL GROWTH MODEL

C is the initial amount. t is the time period.

(1 + r) is the growth factor, r is the growth rate.

y = C (1 + r)t

Exponential Growth Model

$250 at 2.5% interest for 6 years𝑦=250 (1+.025)6

𝑦=$ 289.92

Exponential DecayEXPONENTIAL DECAY MODEL

C is the initial amount.t is the time period.

(1 – r ) is the decay factor, r is the decay rate.

y = C (1 – r)t

$13,000 car depreciates at a rate of 6% per year. What is the value in 3 years?

𝑦=13000 (1− .06)3

𝑦=$ 10,797.59

Add/Sub Polynomials

Add/Sub Polynomials Mult. Polynomials

Add. (2x3 – 5x2 + x) + (2x2 + x3 – 1) 2x3 – 5x2 + x

+ x3 + 2x2 – 1

3x3 – 3x2 + x – 1

Subtract. (4x2 – 3x + 5) – (3x2 – x – 8)

(4x2 – 3x2) + (–3x + x) + (5 + 8)

x2 – 2x + 13

4x2 – 3x + 5 – 3x2 + x + 8

Polynomials

2 5 4 73 2x x x

Leading Coefficient

Degree (largest exponent)

Constant Term

Polynomials do not have variable exponents or negative exponents

Classify by Degree0. Constant1: Linear2: Quadratic3: Cubic4: Quartic5: 5th degree polynomial

Multiply a Monomial by a Polynomial

Mult. Mono by Poly Foil/Rectangle

Find the product 2x3(x3 + 3x2 – 2x + 5).

2x3(x3 + 3x2 – 2x + 5) Write product.

= 2x3(x3) + 2x3(3x2) – 2x3(2x) + 2x3(5) Distributive property

= 2x6 + 6x5 – 4x4 + 10x3 Product of powers property

FOIL/Rectangle Method

x x2 5 6

𝑥

2 𝑥

3 𝑥 6

𝑥2𝑥

2

3

( )( )x x 2 3

x2 2x3x 6

x x2 5 6

Special Products

Special Products Factor GCF

Examples: A)

B)

( )x 3 2

( )a b a ab b 2 2 22( )a b a ab b 2 2 22

( )2 5 2x

( )( )a b a b a b 2 2

(𝑥−11)(𝑥+11)• Example:• A) 𝑥2−121

Factor GCF

A. 4x3 – 44x2 + 96x = 4x(x2– 11x + 24)

= 4x(x– 3)(x – 8)

B. 50h4 – 2h2 = 2h2 (25h2 – 1)

= 2h2 (5h – 1)(5h + 1)

In order to factor out the greatest common factor, it has to be common to ALL terms. Look at coefficients first then variables.

Look for GCF first

Factor remaining polynomial

Look for GCF first

Look for special patterns(difference of two squares)

Factor Trinomials

Factor Trinomials Factor Special Products

x x x x2 _ _ ( _ )( _ )

x x x x2 _ _ ( _ )( _ )

x x x x2 _ _ ( _ )( _ )x x x x2 _ _ ( _ )( _ )

Two + #’s

Two - #’s

One + # One - #

One + # One - #

ADD Multiply

EX. x2 – 4x + 3 = (x – 3)( x – 1)

x2 + 3x + 2 = (x + 2)(x + 1)

Factor Special Products

2 2 ( )( )a b a b a b

a. y2 – 16 = (y + 4)(y – 4) b. 25m2 – 36 = (5m + 6)(5m – 6)

2 2 22 ( )a ab b a b 2 2 22 ( )a ab b a b

d. 9x2 – 12x + 4 = (3x – 2)2c. 4s2 + 4st + t2 = (2s + t)2

Factor by Grouping

Factor by Grouping Graph Quadratics

= x2(x + 3) + 5(x + 3)= (x + 3)(x2 + 5)

x3 + 3x2 + 5x + 15 = (x3 + 3x2) + (5x + 15)a.

y2 + y + yx + x = (y2 + y) + (yx + x)b.= y(y + 1) + x(y + 1)= (y + 1)(y + x)

Use factor by grouping if there is no common monomial to factor out.

Graph Quadratics 1. Find the x-coordinate of the vertex. 2. Make a table of values, using x values to the left

and to the right of the vertex. 3. Plot the points and connect them with a smooth

curve to form a parabola.

Graph y = 3x2 – 6x + 2.

x y

−𝑏

2𝑎

−−62 (3 )

=1

Solve Using Square Roots

Solve Using Square Roots Complete the Square

If , then has two solutions:d x d

x d

0 2

.

If then has one solution:d x d

x

0

0

2,

.

If then has no solution.d x d 0 2,

d > 0

d = 0

d < 0

a. 2x2 = 8x2 = 4

x = ± 4 = ± 2

b. m2 – 18 = – 18

m2 = 0

m = 0

b2 = – 7

c. b2 + 12 = 5

No Solution

Complete the Square (𝑏2 )2

x2 – 16x = –15

x2 – 16x + 64 = –15 + 64

(x – 8)2 = 49

x – 8 = ±7

x = 8 ± 7

𝑥=15 , 1

Find new c value

Factor the left side and simplify the right side

Take the square root of both sides

Add 8 to both sides

Simplify

Quadratic Formula

Quadratic Formula

xb b ac

a

2 4

2

2x2 – x – 7 = 0

x =

b2 – 4ac+ ––b2a

– (–1) –+ ( –1)2 – 4(2)(–7)

2(2)=

4=+ –1 57

The solutions are 2.14 and -1.64

Example:

* Solutions are where the graph crosses over the x-axis.

*Make sure quadratic is in standard form before identifying a, b and c.