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