Transcript Example

PowerPoint Project: Exam Review
By: Jack Dunn and Max Scheurell
Properties

Addition Property (of equality):



Multiplication Property (of equality)




Example: If a is a real number then a = a
Hint: a number equals itself
Symmetric Property (of equality)



Example: If a = b then ca = cb
Hint: If you multiply both sides by “c” the answer does not change
Reflexive Property (of equality)


Example: If x = y then 5 + x = y + 5
Hint: if you add a 5 to both sides the answer does not change
Example: If a and b are real numbers then a = b and b = a
Hint: a real number = a real number
Transitive Property (of equality)


Example: If a = b and b = c then a = c
Hint: can be assumed If a = b and b = c then a = c

Associative Property of Addition




Associative Property of Multiplication

Example: (ab)c = a(bc)

Hint: if the signs are the same they can be moved and the answer will not change
Commutative Property of Addition





Example: a(b + c) = ab + ac
Hint: you distribute the a to the b and the c and the answer doesn’t change
Property of Opposites or Inverse Property of Addition



Example: ab = ba
Hint: you commute to school one way and back another way
Distributive Property


Example: a + b = b + a
Hint: you commute to school one way and back another way
Commutative Property of Multiplication


Example: (a + b) + c = a + (b +c)
Hint: if the signs are the same they can be moved and the answer will not change
Example: a + (-a) = 0
Hint: if you add the opposite the answer will be zero
Property of Reciprocals or Inverse Property of Multiplication


Example: b × 1/b = 1
Hint: If you multiplied by a reciprocal to get 1

Identity Property of Addition



Identity Property of Multiplication





Example: if “a” is a real number and “b” is a real number then a + b is real number
Hint: a real number added to a real number is a real number
Closure Property of Multiplication



Example: a × 0 = 0
Hint: any number multiplied by zero = 0
Closure Property of Addition


Example: a × 1 = a
Hint: if you multiply by one “a” keeps its identity (does not change)
Multiplicative Property of Zero


Example: a + 0 = a
Hint: a keeps its identity the answer does not change because zero is added
Example: if “a” is a real number and “b” is a real number then ab is real number
Hint: a real number times a real number is a real number
Product of Powers Property


Example: 72 × 76 = 7(2 + 6) = 78
Hint: when finding the product of powers you add them

Power of a Product Property




Power of a Power Property

Example: (am)n = amn

Hint: in Power of a product property you multiply the exponents
Quotient of Powers Property




Example:
Hint: power of a quotient can be obtained by finding the powers of numerator and
denominator and dividing them
Zero Power Property



Example:
Hint: property states that to divide powers having the same base, subtract the
exponents.
Power of Quotient Property


Example: (ab)m = am × bm
Hint: the exponent outside the parenthesis applies to all that is in the parenthesis
Example: (3)0 = 1
Hint: a number to the zero exponent equals one
Negative Power Property


Example:
Hint: move the number to the denominator if in the numerator


Zero Product Property

Example: (1 + a) (6 + a) then 1 + a = 0 or 6 + a = 0

Hint: A product of factors is zero if and only if one or more of the factors is zero.
Product of Roots Property



Quotient of Roots Property



Example:
Hint: take the square of both the numerator and the denominator
Root of a Power Property



Example:√a ∙ √b = √ab
Hint: the variables are multiplied
Example:
Hint: none
=
Power of a Root Property


Example:
Hint: none
Solving First Power Inequalities in one
Variable (Practic Number lines

Solving inequalities with only one inequality sign (try problems before
looking at the answer

Problem: x < 4 (put on number line)

Hint: the problem reads “x” is less than four. Therefore the circle is open
and the arrow goes from the four toward the negative and carries on. If the
problem had this sign ≤ (less than or equal to) the circle would be colored in
rather than open
Continued

Conjunction

Problem: -2 < x and x≤ 4

Hint: The reason this is a conjunction is because it contains the word “and”
which tells you that both inequalities need to be contained in you number line. If
the problem read -2 < x ≤ 4 it is also implied that both inequalities need to be
contained in you number line. The problem reads “x” is greater than negative two
“and” “x” is less than or equal to four. You have to find an answer that works for
both. A special case is also a conjunction and in this special case the answer is
null set (Ø) because it is impossible to put both on the same number line
Example: -2 < x and x > 4
These would go in different directions on the number line and have no common
points


Continued

Disjunction

Example: x < -3 or x > -2

Hint: in a disjunction both inequalities can be labeled. There is a special case with
disjunctions in which all real number or the whole number line is contained. An
example would be:
x > -3 or x < -2
Contained would be all real numbers because it is a disjunction and the lines
would go toward one another making them go on forever in both directions


Linear Equations in two variables
Slopes


A slope is told by rise over run and in slope-intercept form (y = mx + b) is
“mx” and the y intercept is “b”
Rising Line: Has a positive rise over run or slope and all rising lines will
enter quadrant I and quadrant III eventually. A possible equation is
y = 2x + 3.


Is not the answer to the equation but is an example of a rising line
Falling Line: Has a negative slope or rise over run and all falling lines will
enter quadrant II and quadrant IIII eventually. A possible equation is
y = - 2x -9

Is not the answer to the equation but is an example of a falling line

Vertical line: Has an undefined slope and a rise over run of 1/0. The xcoordinate stays the same in a vertical line. A possible equation is x = 5 it is
as simple as that because there is no slope or y-intercept.

Horizontal line: Has a slope of 0 and a rise over run of 0/1. The y-coordinate
stays the same in a horizontal line. A possible equation is y = 7 because there
is no slope worth writing in the equation and the y-intercept is seven

Standard Form: for a linear equation in two variables, x and y, is usually
given as:





Ax + By = C
A, B, and C are integers, and A is non-negative, and, A, B, and C have no
common factors other than 1
Also called general form
You can convert it to slope-intercept form by isolating the “y” variable
Point-slope formula: is used to take a graph and find the equation of that
particular line.




Just plug x and y values into your point-slope formula above and you can find the
equation of a line simple as that
Plug in 5 for “m” and plug in points (5,5) and when you find your answer click
Answer: y = 5x -5
If you have an x and y coordinate and the slope you can find the slope-intercept
form
How to Graph a line


When you have your equation in slope-intercept form you can easily graph
the line
A refresher:




Slope-intercept form is y= mx + b
“b” is the y-intercept
“m” is the slope
In the problem y = -5x + 6 six is the y-intercept and can be put on the y axis at
positive six. -5x is the slope and x really has no meaning here -5 is the same as -5
over 1 so therefore you go down 5 pts. And over one making this line a falling line
7
This is the equation above
graphed
6
5
f x = -5x+6
4
3
2
1
-2
2

Another possible equation which would give you a rising line is


Y = 6x -5 The y intercept is -5 and the slope is 6 and over one
F(x) = 6x – 5 is the same problem f(x) is the same as y and is a function


A Function is only a relation when it has a distinct input and a distinct output. Not
all relations are functions though all functions are relations
See if you can graph this problem yourself

The problem will appear when you click
2
1
This is the equation above
graphed
A
-2
2
-1
-2
-3
-4
-5
4

Another possible equation which would give you a Horizontal line is:



Y or f(x) = 2
This problem results in a horizontal line because the y coordinate is always equal
to 2 and will go across the graph
See if you can graph this

The problem will appear when you click
4
This is the equation above
graphed
It is hard to see but the
horizontal line across the
graph is the line
3
fx = 2
2
1

Another possible equation which would give you a Vertical line is:



X=3
The x coordinate will always equal 3 and will go straight up the graph
See if you can graph this equation

The problem will appear when you click
f  y = 3
6
4
This is the equation above
graphed
2
-5
5
-2
-4
Linear Systems

Substitution Method




Replace a variable with an equal expression
To be able to do substitution method you need to be able to isolate a variable and
leave it with a coefficient of one
Take the equality of the variable and substitute it in for the variable in the second
equation and then solve
Example (see if you can complete the problem and find the right answer)










y = x + 7 and 2x + 3y = 11
Steps:
2x + 3 (x + 7) = 11
2x + 3x + 21 = 11
5x + 21 = 11
5x = -10
x = -2
Then you must plug the x variable into the beginning problem
y = -2 + 7
Final answer is: (-2,5)

Elimination Method



Combine the two equations to cancel out a variable
Find the LCM of either the x or y variable and its coefficient then multiply them
by the factor then add the variables together you will end up with either the x or
the y variable eliminated then solve the remaining for the variable and back
substitute
Example (try and solve by yourself first)









3x + 2y = 5 and 5x -3y = 15
9x + 6y = 15 and 10x – 3y = -34
19x = -19
x = -1
Then back substitute
-3 + 2y = 5
2y = 8
y=4
Final answer is: (-1,4)

Consistent system


Parallel lines don’t intersect and have the same slope
Parallel lines are a consistent system
4
2
-2
-4

Inconsistent System


Perpendicular lines which intercept and have opposite reciprocal
Perpendicular lines are an inconsistent system
fx = -3x+1
gx =

1
3
x
4
2
-5
5
-2

Dependent System


Two line with the same domain and range
All points on the line
4
2
-2
-4
Factoring and Rational
Expressions

GCF


For any numbers of terms
Examples of GCF try to do them before looking at the answer


Solution:



10x^3 - 15x^2
5x^2(2x - 3)
Both 10x^3 and 15x^2 are multiples of five and each contain 5x^2
Difference of Squares


Difference of squares factors into two conjugates
Examples of Difference of Squares


x^2 – 16
Solution:


(x+4)(x-4)
Unlike the sum of squares this can be factored and is factored into two conjugates

Sum and Difference of Cubes


End product will be a binomial and a trinomial
Example:


Solution:



X^3 - 27
You have to first cub the roots with the original sign in the middle. Then you
square those roots. The middle product is the opposite product of the cubed roots:
(x-3)(x^2+3x+9)
PST


A trinomial that is the exact square of a binomial
Example:


x^2 + 4x + 4
Solution:



(x+2)(x+2)
When factored the solution is two of the same binomials
If the 1st and third are squares and the middle number is twice the product the
answer is a PST

Reverse FOIL


Trial and Error
Example:


q^2+5q+6x
Solution:


(q+3)(q+2)
You foil by trial and error but if the signs are both addition signs then all of your
signs in the binomial are addition signs

Grouping 2 X 2


Find the GCF and then the leftover factors which will leave you w/two binomials
Example:


5x+10y-ax-2ay
Solution:




First regroup and find the GCF:
5(x+2y) - a(x+2y)
x+2y is the common factor and can be taken out the final answer is:
(x+2y)(5-a)

Grouping 3 X 1


If an expression consists of four terms and part of it is a perfect square, then the
expression can be factorized by grouping 'Three and One‘
Example:


a^2-6a+9-16y^2
Solution:




(a^2-6a+9) - 16y^2
(a-3)2 - (4y)2
Final answer:
(a-3+4y)(a-3-4y)
Quadratic Equations In One
Variable

By Factoring







Set the equation equal to zero so that all variables and numbers are on one side of
the equation
Then factor the problem
Example
y2 + 15y + 56 = 0
Now factor for your answer
(y + 8)(y + 7)
Square root of each side







Example:
9x^2 = 64
Solution:
9x^2 – 64 = 0
(3x + 8)(3x – 8) = 0
3x = -8 or 8
X = -8/3 or 8/3

Solve by completing the square x^2 + bx __?__








Find half the coefficient of x
Square the result of Step 1
Add the result of Step 2 to x^2 + bx + (b/2)^2
Example:
X^2 + 14x + ___?___
X^2 + 14x + 49 = (x + 7)^2
(14/2)^2 = 49
Quadratic Formula


Y = b  b  4ac
2a
Y = ax^2 + bx + c
2


A goes in for “a” in the formula B goes in for “b” in the formula etc…
Discriminant

Example: X^2 +4x + 1 = 0
4
4  42  4(1)(1)
2(1)
2
4  12 4  2 3

2
2
2  3
2
-5
-2
Functions

Domain




The domain of a function is the set of all possible x values which will make the
function "work" and will output real y-values
The range of a function is the complete set of all possible resulting values of the
dependent variable of a function, after we have substituted the values in the
domain.
Example: The domain of the line below is all real numbers. (X= All real
numbers)
If the line is vertical the domain would be what X is equal to. If X were 14 the
domain could only be 14 no other numbers because no point on that line would
have a coordinate of 14.

Range



The range is the set of all possible output values (usually y), which result from
using the function formula.
The Range of the Line below on the left is all real numbers. (Y = all Real
numbers)
The Range of the line below on the right is only 14 because all the points on that
line will have a Y coordinate of 14

How to find a linear function from two ordered pairs of data









Use the Formula: (y2 – y1)/(x2 – x1) to find the slope of the linear function
To find the Y intercept you set x= 0 in slope intercept form after you find the slope
Example if you have (-2, 5) and (4,8) as your ordered pairs
Step 1: find the slope (m = slope
 m = (8 – 5)/(4 – (-2))
 m = (8 – 5)/(4 + 2)
 m = 1/2
Step 2: substitute the slope or 1/2 for the variable m in y = mx + b
 y = 1/2x + b
Choose one of the points, say (4,8).
Now substitute 4 for x and 8 for y.
 y = 1/2x + b
 8 = 1/2(4) + b
Now solve for b
 8=2+b
 6=b
Then substitute 6 in for the variable b in the point slope formula (y = mx + b) like
you did with the slope for the variable m
Your final equation should be y = 1/2x + 6
Quadratic Functions

How to graph a quadratic function

Example 1: f(x) = x^2 – 2x - 2
Solution find the coordinates of selected points as shown in the table below. Plot the
points and connect them with a smooth curve

6
gy = 1
fx = x2 -2x-2
x
X^2 – 2x - 2
-2
(-2)^2 – 2(-2) – 2 = 6
-1
(-1)^2 – 2(-1) – 2 = 1
0
(0)^2 – 2(-1) – 2 =1
1
1^2 – 2(1) – 2 = -3
2
2^2 – 2(2) -2 = -2
3
3^2 – 2(3) -2 = 1
4
4^2 – 2(4) -2 = 6
4
2
5
-2
-4





This curve shown on the graph below (same graph as slide before) is called a parabola.
This parabola opens upward and has a minimum point at (1, -3). The y-coordinate of
this point s the least value of the function.
You can decide whether the parabola will open upward or open downward (close) is if
a in the quadratic function f(x) or y = (a)x^2 – bx – c is positive then the parabola will
open upward and if it is negative then the parabola will be closed or open downward
The dotted vertical line x= 1, containing the minimum point, is called the axis of
symmetry of the parabola. Meaning that the two halves are equally the same.
The equation for the axis of symmetry’s line is x = -b/2a
The maximum point is the highest point or ordered pair of a parabola and the y –
coordinate this point is the greatest value
6
gy = 1
fx = x2 -2x-2
4
2
5
-2
-4





The maximum point is the highest point or ordered pair of a parabola and the y –
coordinate this point is the greatest value
The minimum or maximum point of a parabola is called the vertex.
The average of the x- coordinates of any such pair of points is the x-coordinate of the
vertex
The average of these x-coordinates is -b/2a
The x-coordinate of the vertex of the parabola y = ax^2 + bx + c is -b/2a



Note: a cannot equal 0
Example 1: find the vertex of the graph of H: x  2x^2 + 4x -3
 Use the vertex and four other points to graph H
 Identify and graph the axis of symmetry
Solution
 Step 1: x – coordinate of vertex = -b/2a

= - 4/2(2)

= -4/4
= -1


Step 2: To find the y-coordinate of the vertex, substitute -1 for x





y = 2x^2 + 4x – 3
y = 2(-1)^2 + 4(-1) - 3
y=2–4–3
y = -5
The vertex is (-1, -5) because our solution for x = -1 and y = -5




Step 3: For values of x, select two numbers greater than -1 and two numbers less
that -1 to obtain paired points with the same y coordinate.
Step 4:Plot the points and connect hem with a smooth curve.
Step 5: The axis of symmetry is the line x = -1 (dashed line)
Red = vertex x and y coordinates
6
gy = -1
X
2x^2 + 4x – 3 = y
-3
2(-3)^2 + 4(-3) – 3 = 3
-2
2(-2)^2 + 4(-3) - 3 = 3
-1
2(-1) + 4 (-1) - 3 = 3
0
2(0)^2 +4(0) - 3 = 3
1
2(1)^2 + 4(1) - 3 = 3
4
fx = 2x2 +4x-3
2
-5
-2
-4
Vertex
Simplifying Expressions With
Exponents






simplify with exponents, don't feel like you have to work only from the for exponents. It is
often simpler to work directly from the definition and meaning of exponents. For instance:
Simplify x6 × x5 The rules tell me to add the exponents. But I when I started algebra, I had
trouble keeping the rules straight, so I just thought about what exponents mean. The " x6 "
means "six copies of x multiplied together", and the " x5 " means "five copies of x multiplied
together". So if I multiply those two expressions together, I will get eleven copies of x
multiplied together. That is:
x6 × x5 = (x6)(x5)
= (xxxxxx)(xxxxx) (6 times, and then 5 times)
= xxxxxxxxxxx
(11 times)
11
=x
Thus:
x6 × x5 = x11
Simplify the following expression:


The exponent rules tell me to subtract the exponents. But let's suppose that I've forgotten the
rules again. The " 68 " means I have eight copies of 6 on top; the " 65 " means I have five
copies of 6 underneath.

How many extra 6's do I have, and where are they? I have three extra 6's, and they're on top.

Simplify (–46x2y3z)0 This is simple enough: anything to the zero power is just 1.
(–46x2y3z)0 = 1
Simplify –(46x2y3z)0 The parentheses still simplifies to 1, but this time the
"minus" is out front, out from under the power, so the exponent doesn't touch it.
So the answer is:
–(46x2y3z)0 = –1

Simplify the following expression:





I can cancel off the common factor of 5 in the number part of the
fraction:


Now I need to look at each of the variables. How many extra of each do I
have, and where are they? I have two extra a's on top. I have one extra b
underneath. And I have the same number of c's top and bottom, so they
cancel off entirely. This gives me:
In the context of simplifying with exponents, negative exponents can create
extra steps in the simplification process. For instance:
 Simplify the following:


The negative exponents tell me to move the bases, so:



Then I cancel as usual, and get:
Simplifying Expressions with
Radicals




Square root problems are not difficult for a person when they are just one
number or a variable squared etc… but when we come across a problem that
cannot be squared we have a little more difficulty
These problems (an example is: 56 ) we have more difficulty but in truth
the problem is not that hard
56 can simply be reduced to 7  8 and that can be reduced to 4  2  7
since square can be taken of four you can take a two out as the square of four
and your final answer will be: 2 7 2
The answer does look complicated but with the simple steps you can find the
answer now you try one (click and examples will appear on the next page

Examples (try these before you look at the answer):
32
16  2
4 2
84
12  7
4  3 7
2 3 7
50
2  25
5 2
Word Problems

Mimi is four years older than Ronald. Five years ago she was twice as old as
he was. Find their ages now. (try to solve)



Step 1 the problem asks for Mimi’s age and Roald’s age now.
Step 2 Let m = Mimi’s age now, and let r = Ronald’s age now
Step 3 Use the facts of the problem to write two equations



M = 4 + r (now)
M – 5 = 2(r – 5)
(five years ago)
Step 4 Simplify the equations and solve





M=4+r
M – 5 = 2(r-5)
That goes to m – r = 4 and m – 2r = -5
Then it goes to m-r = 4 m - 9 = 4
M = 13 is your final answer


Bicyclists Brent and Jane started at noon from points 60 km apart and rode
toward each other, meeting at 1:30 pm Brent’s speed was 4km/h greater than
Jane’s speed. Find their speeds.
Solution



Step 1 the problem asks for Brent’s speed and Jane’s speed. (Draw a sketch to help
yourself out)
Step 2 Let r = Jane’s speed. Then r +4 = Brent’s speed
Step 3 The sketch helps you write the equation


1.5(r + 4) + 1.5r = 60
Step 4





1.5r + 6 + 1.5r = 60
3r + 6 = 60
3r = 54
R = 18 (Jane’s speed)
R + 4 = 22 (Brent’s speed)


Raoul says he has equal numbers of dimes and quarters and three times as
many nickels as dimes. The value of his nickels and dimes is .50$ more than
the value of his quarters. How many of each kind of coin does he have?
Solution



Step 1: the problem asks for the numbers of nickels, dimes, and quarters
Step 2: :et x = the number of dimes and x = the number of quarters Then 3x = the
number of nickels
Step 3: value of nickels + value of dimes = value of quarters + 50



15x + 10x = 25x + 50
Step 4: 25x = 25x + 50
Then 0 = 50 giving you a false statement and letting you know that the problem
doesn’t have a solution


Hector Herrera made a rectangular fish pnd surrounded by a brick walk 2m
wide. He had enough bricks for the are of the walk to be 76 sq. m. Find the
dimensions of the pond if it is twice as long as it is wide.
Solution:



Step 1: The problem asks for the dimensions of the pond. Make a sketch.
Step 2: Let x = the width of the pond. Then 2x = the length of the pond
Step 3: Area of walk = Area of pond and walk – area of pond





OR 76 = (2x + 4)(x + 4) – (2x)(x)
Step 4: 76 = 2x^2 + 12x + 16 – 2x^2
76 = 12x + 16
60 = 12x
5 = x and 2x = 10
Line of Best Fit or Regression Line

When do you use this?


How does your calculator help?




When you are sure you cannot graph the line using simple arithmetic
Your calculator can more quickly find and solve the function that you plug in to
it, but becomes not useful when you use your calculator for simple easy problems
that you can do on your own
Here is a problem for you to plug into your calculator and find out if you can
find the regression equation
Please get your calculator out
Do you remember how to find the regression equation?


Plug in: (4,6), (8,6), (4,4) and (6,8) (when you have found the answer click)
Answer: y = 0.37x + 3.9