Algebra Period 3 (Monday)

Simplify, Multiply, and Divide RATIONAL EXPRESSIONS 10-1, 10-2, and 10-3

Rational Expressions = Expressions in fraction format (division) with a variable in the denominator

You have already been simplifying, multiplying and dividing these throughout this year!

SIMPLIFY: 10-1

You will need to FACTOR (Chapter 6) both the numerator and denominator and "cross out" common factors in both (their quotient is 1!)
EXAMPLE: Simplify
y² + 3y + 2 =
y² - 1


(y + 2)(y + 1) =
(y - 1)(y + 1)

y + 2
y - 1

MULTIPLY: 10-2

FACTOR if possible, cross cancel if possible, multiply numerators, then denominators, simplify
EXAMPLE:
(y + 4)³[y² + 4y + 4] =
[(y + 2) ³(y² + 8y + 16)]

(y + 4) ³][ (y + 2) ² =
(y + 2) ³(y + 4) ²

y + 4
y + 2



DIVIDE: 10-3


Same as the previous example, only this time you will need to

“FLIP the SECOND” fraction,

FACTOR then
MULTIPLY!!!!!
EXAMPLE:
x + 1 ÷ x + 1 =
x² - 1 x² - 2x + 1


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

( x + 1)( x - 1)(x - 1) =
(x + 1)( x - 1)(x + 1)


x - 1
x + 1







Add and subtract rational expressions with LIKE DENOMINATORS: 10-4
Add and subtract with UNLIKE DENOMINATORS: 10-5
When adding with LIKE DENOMINATORS,
simply add the numerators,
simplify

When subtracting with LIKE DENOMINATORS,
CHANGE THE SIGNS
OF EACH TERM IN THE NUMERATOR AFTER THE SUBTRACTION SIGN,
THEN ADD (double check!!!)

When adding or subtracting with UNLIKE DENOMINATORS,
find the common denominator (the least common multiple of all denominators),
then use equivalent fractions to restate each numerator using the new common denominator.
EXAMPLE:
2 + x
x² - 16 x – 4

First, factor the denominators if possible



2x + x
(x + 4)(x - 4) x – 4

Find the LCM = (x + 4)(x - 4);

2x + (x + 4)(x)
(x + 4)(x - 4) (x + 4)(x - 4)


Restate both fractions with the LCM

2x + x² + 4x
(x + 4)(x - 4)


Now add the numerators

x² + 6x =
(x + 4)(x - 4)

Simplify and re-factor numerator, if possible
x(x + 6)
(x + 4)(x - 4)

Algebra Period 3 (Friday)

How about one more video..
let me know if you find one you think everyone in our class would like


Solving Rational Expressions: 13-5

YOU NEED TO GET RID OF ALL DENOMINATORS!
After you do that, you may end up with a Quadratic that you can solve using any of your methods.

Easy one first

(a+1)/2 = 1/a
You could find the LCM of the denominators which would be 2a but what about using cross products—as you did with proportions—last year.
then the above becomes
a(a + 1) = 2(1)
0r
a² + a = 2
a² + a - 2= 0 using ZERO products property
Factor
(a + 2)(a-1) = 0
so a = -2 and a= 1


When you MUST Find the LCM of the denominators:
MULTIPLY EACH & EVERY TERM ON BOTH SIDES BY THE LCM!
You should end up with NO DENOMINATORS!
(Otherwise, you don't have the right LCM!)

EXAMPLE:
4/x – 4/(x+2) = 1
In this case the LCM is discovered just by multiplying both of the existing denominators. It is as if they are relatively prime!!
The LCM is x(x + 2)
Multiply each term on BOTH sides by x(x + 2)
[x(x +2)4]/x – [4 x(x +2)]/(x+2) = 1 [x(x +2)]

Simplify the denominators with the numerators:
(x + 2) 4 - x(4) = x(x + 2)
SIMPLIFY more:
4x + 8 - 4x = x² + 2x

USE THE ZERO PRODUCTS PROPERTY:
x² + 2x - 8 = 0
FACTOR OR QUADRATIC FORMULA:
(x + 4)(x - 2) = 0

x = -4 and x = 2


Solving Radical Equations 13-6
THIS IS A REVIEW OF CHAPTER 11

You square both sides to get rid of the radical sign
For these problems, you'll get a quadratic on one side after you square
BE SURE TO CHECK BOTH ANSWERS TO MAKE SURE THEY BOTH CHECK!

x – 5 = SQRT (x+ 7)
squaring both sides gives us
(x -5)² = [SQRT(x + 7)]²

x² - 10x + 25 = x + 7
x²- 11x + 18 = 0
(x – 9)(x -2) = 0
or x = 9 and x = 2

BUT when you check you discover
x -5 = SQRT ( x + 7)
9 – 5 =?= SQRT (9 + 7)
4 = 4
BUT

x -5 = SQRT ( x + 7)
2 – 5 =?= SQRT (2 + 7)
-3 DOES NOT EQUAL 3

so only solution is x = 9

Solve
[SQRT ( 27 - 3x)] + 3 = x
move the 3 to the other side using transformations

[SQRT ( 27 - 3x)] = x -3
NOW squaring both sides

[SQRT ( 27 - 3x)]² = (x -3)²
27 – 3x = x² - 6x + 9
move everything to the right side – to use the ZERO PRODUCTS PROPERTY

0 = x² - 3x – 18
0 = ( x – 6) (x +3)
so x = 6 or x = -3

WE MUST CHECK again:
Start with the ORIGINAL equation
[SQRT ( 27 - 3x)] + 3 = x

Plug in for x = 6
[SQRT ( 27 – 3(6))] + 3 =? = 6
[SQRT ( 27 - 18)] + 3 =? = 6
[SQRT ( 9)] + 3 =?= 6
3 + 3 = 6 YES

But we discover

[SQRT ( 27 - 3x)] + 3 = x
Plug in for x = -3
[SQRT ( 27 – 3(-3))] + 3 =?= -3
[SQRT ( 27 +9)] + 3 =?= -3
[SQRT ( 36)] + 3 =?= -3
6 + 3 DOES NOT EQUAL -3
so only x = 6 is the solution

Algebra Period 3 (Thursday)

The Quadratic Formula

Wait until you see and hear these videos--
This is the one from class





this is to the Flintstones



pretty funny stuff




So what do you think? Create one of your own...

THE QUADRATIC FORMULA:
-b + or - SQRT b² - 4ac
2a

-b plus or minus the square root of b squared minus 4ac all over 2a

Notice how the first part is the x value of the vertex -b/2a
The plus or minus square root of b squared minus 4ac represents
how far away the two x intercepts (or roots) are from the vertex!!!!

Very few real world quadratics can be solved by factoring or square rooting each side.
And completing the square always works, but it long and cumbersome!

All quadratics can be solved by using the QUADRATIC FORMULA.

(you will find out that some quadratics have NO REAL solutions, which means that there are no x intercepts - the parabola does not cross the x axis! Think about what kinds of parabolas would do this....ones that are smiles that have a vertex above the x or ones that are frowns that have a vertex below the x axis. You will find out in Algebra II that these parabolas have IMAGINARY roots)

So now you know 5 ways that you know to find the roots:
1. graph
2. factor if possible
3. square root each side
4. complete the square - that's what the quadratic formula is based on!
5. plug and chug in the Quadratic Formula -
This method always works if there's a REAL solution!



DON'T FORGET TO PUT THE QUADRATIC IN STANDARD FORM BEFORE PLUGGING THE VALUES INTO THE QUADRATIC FORMULA! ax² + bx + c = 0



DISCRIMINANTS - a part of the Quadratic Formula that helps you to understand the graph of the parabola even before you graph it!
the discriminant is b² - 4ac

(the radicand in the Quadratic Formula, but without the SQRT)

Depending on the value of the radicand, you will know
HOW MANY REAL ROOTS IT HAS
1) Some quadratics have 2 real roots (x intercepts or solutions) - Graph crosses x axis twice
2) Some have 1 real root (x intercept or solution) - Vertex is sitting on the x axis
3) Some have NO real roots (no x intercepts or solutions) - vertex either is above the x axis and is a smiley face (a coefficient is positive) or
the vertex is below the x axis and is a frown face (a coefficient is negative)

In both of these cases, the parabola will NEVER CROSS (intercept) the x axis!
b² -4ac > 0 if it's positive, 2 roots
b² -4ac = 0 if it's zero - 1 root
b² -4ac < 0 if it's negative - no real roots

Math 6 H Periods 1, 6 & 7

Review of Solving Combined Operations 8-5
Follow these general procedures to solving equations:

Step 1: Simplify each side of the equation

Step 2: If there are still indicated additions or subtrations, use the inverse operations to undo them

Step 3: If there are indicated multiplication or divisions involving the variable, use the the inverse operations to undo them

YOU MUST ALWAYS PERFORM THE SAME OPERATION ON BOTH SIDES OF THE EQUATION.

DO ONTO ONE SIDE WHAT YOU WOULD DO TO THE OTHER... hmmm... where have you heard that before??

Balance, balance , balance... it's all a balancing act!!


Solve the equation 3/2(n) + 7 = 22
subtract 7 FROM BOTH SIDES
(3/2)n = 15
Multiply BOTH SIDES by 2/3, the reciprocal of 3/2

(2/3)(3/2)n = 15(2/3)
simplify before you multiply out
n = 10

40 - (5/3)n = 15
There are two ways to solve this one
add (5/3)n to both sides
40 = (5/3)n + 15
then subtract 15 from BOTH SIDES
25 = 5/3(n)
Multiply both sides by 3/5, the reciprocal of 5/3
(3/5)25 = 5/3(n)(3/5)
Simplify before multiplying
15 = n

OR
40 - (5/3)n = 15
subtract 30 from both sides
- (5/3)n = 15 -40
need to ask yourself who wins? the negative... and by how much 25
- (5/3)n = -25

multiply by -3/5, which is the reciprocal of -5/3
(-3/5)-(5/3)n = -25(-3/5)
n = 15
we arrived at the same solution

Algebra Period 3 (Tues/Wed)

Solving Quadratics by Completing the Square 13-3
Okay, up until now the methods of solving quadratics should have been fairly familiar to you—but this method is a NEW strategy!!

When does the "square root = +/- square root" method work well?
When the side with the variable is a PERFECT SQUARE!
So what if that side is not a perfect BINOMIAL SQUARED?
You can follow steps to make it into one!

This method is great because then you can just square root each side to find the roots!
THIS METHOD ALWAYS WORKS!

EXAMPLE: x² - 10x = 0
Not a TRINOMIAL SQUARE so it would not factor to a BINOMIAL SQUARED. (However, we know that we could just factor out an x to solve—I want you to see how completing the square works—even with easy ones)
Here's how you can make this into a trinomial square

Step 1: b/2
Take half of the b coefficient in this case (- 10/2 = -5)

Step 2: Square b/2
(-5 x -5 = 25)

Step 3: Add (b/2)² to both sides of the equation
(x² - 10x + 25 = +25)

Step 4: Factor to a binomial square
(x - 5)² = 25

Step 5: Square root each side and solve
SQRT (x - 5)² = SQRT 25
x - 5 = + and - 5

ADD 5 TO BOTH SIDES
x = 5 + and - 5
x = 5 + 5 and x = 5 - 5
x = 10 and x = 0
Now the check for this one—just because we are learning this strategy--is to factor and make sure you get the same results.
that is
x² - 10x = 0 we know becomes
x(x-10) = 0
x = 0 and x = 10

x²-4x -7 = 0
add 7 to both sides
x²-4x =7
take –b/2 or -4/2 = -2 now square that (-2),sup>2 = 4
add 4 to both sides
x²-4x + 4 = 7 +4
x²- 4x + 4 = 11
(x -2)² = 11
take the square root of both sides
SQRT(x -2)² = + or – SQRT 11
x - 2 = + or - SQRT 11

IF THERE IS AN "a" COEFFICIENT, YOU MUST DIVIDE EACH TERM BY IT BEFORE YOU CAN COMPLETE THE SQUARE:

Example: 2x² - 3x - 1 = 0
Move the 1 to the other side of the equation:
2x² - 3x = 1
Divide each term by the "a" coefficient:
x² - 3/2 x = 1/2
Now follow the step to find the completing the square term and add it to both sides:
-b/2 = (-3/2)(1/2) remember (instead of dividing by 2, when you have a fraction, multiply by 1/2)= – 3/4
NOTE: REMEMBER this term you will use it again!!
square that [(-3/2)(1/2)]² = (-3/4)²
= 9/16
x² - 3/2 x + 9/16 = 1/2 + 9/16
(x - 3/4)² = 8/16 + 9/16
(x - 3/4)² = 17/16
SQRT[(x - 3/4)2 ] = + or - SQRT [17/16]
x - 3/4 = + or -[SQRT 17] /4
x = 3/4 + or -[SQRT 17] /4
x = 3 + or - [SQRT 17] /4

When our textbook ask you to complete the square
as in x² - 6x
all it is asking is that you follow the steps
-b/2 is -6/2 = -3
then square that (-3)² = 9
so the answer is
x² - 6x + 9

Algebra Period 3 (Monday)

Introduction to Quadratic Equations 13-1

We learned from Chapter 12 that a quadratic function is a function that can be defined by an equation of the form
ax² + bx + c = y where a is not equal to 0. This is a parabola when the domain is the set of REAL numbers.
When y = 0 in the quadratic function ax² + bx + c = y we have an equation of the form ax² + bx + c = 0. An equation that can be written in this form is called a quadratic equation.

STANDARD FORM is ax² + bx + c = 0
4x² + 7x = 5 write in standard form and determine a, b, and c
4x² + 7x – 5 = 0
a= 4
b = 7
c = -5


CHAPTER 13 gives you several different ways to SOLVE QUADRATICS
Solving a quadratic means to find the x intercepts of a parabola.
There are different ways of asking the exact same question:
Find the.....
x intercepts = the roots = the solutions = the zeros of a quadratic

We'll answer this question one of the following ways:
1) Read them from the graph (read the x intercepts) That’s were y = 0 or where the parabola crosses the x-axis!!
but...graphing takes time and sometimes the intercepts are not integers

2) Set y or f(x) = 0 and then factor (we did this in Chapter 6)
but...some quadratics are not factorable

3) Square root each side (+ or - square root on the answer side)
but...sometimes the variable side is not a perfect square (it's irrational)

4) If not a perfect square on the variable side, complete the square, then solve using #3 method
Now this method ALWAYS works, but...it takes a lot of time and can get complicated

5) Quadratic Formula (works for EVERY quadratic)
Really easy if you just memorize the formula and how to use it! :)

Section 13-1
Reading the x intercepts from a graph or factoring and solving using the zero products property.

METHOD 1:
Where the graph crosses the x axis is/are the x intercepts. (Remember, y = 0 here!)
The x intercepts are the two solutions or roots of the quadratic.

METHOD 2:
When we factored in Chapter 6 and set each piece equal to zero, we were finding the x value when y was zero.
That means we were finding these two roots!

y² – 5y = 6 = 6y – 18
first put this in standard form
y² – 11y + 24 = 0
(y -8) (y-3) = 0
y = 8 or y = 3

Substitute to verify that 8 and 3 are solutions!!
More Solving Quadratic Equations 13-2

You did this in Chapter 11 for Pythagorean Theorem!
If there is no x term, it's easiest to just square root both sides to solve!

DIFFERENT FROM PYTHAGOREAN: NOT LOOKING FOR JUST THE PRINCIPAL SQUARE ROOT ANYMORE. NEED THE + OR - SYMBOL!!

This is also different from what we did when we had radical equations. Before we squared both sides to solve. It looks like these, but only after we squared both sides. Before we had to carefully check each answer—we had changed the equations by squaring. However, always check your solutions for any mistakes!!

3x² = 18
divide both sides by 3 and get: x² = 6
square root each side and get x = + SQRT 6 or - SQRT 6


(x - 5)² = 9
SQRT each side and get: x - 5 = + or - 3
+ 5 to both sides: x = 5 + 3 or x = 5 - 3
So, the 2 roots are x = 8 or x = 2

(x + 2)² = 7
SQRT each side and get: x + 2 = + or - SQRT of 7
-2 to both sides: x = -2+ SQRT 7 or x = -2 - SQRT 7



FORMULAS THAT ARE QUADRATICS:
Many formulas have a variable that is squared: compounded interest, height of a projectile (ball)
The formula for a projectile is h = -5t² + v₀t
We can find when a projectile is a ground level ( h= 0) by solving for
0 =-5t² + v₀t \If the projectile begins its flight at height c, its approximate height at time t is h = -5t²+v₀t + c We can find when it hits the ground by solving 0 = -5t² + v₀t + c



For example: a slow-pitch softball player hits a pitch when the ball is 2 m above the ground. The ball pops up with an initial velocity of 9m/s If the ball is allowed to drop to the ground, how long will it be in the air?
When the ball hits the ground h = 0 so
0 = -5t² + 9t + 2 or
-5t² + 9t + 2 = 0
( 5t + 1)(t -2) = 0
5t = 1 and t = 2
t = -1/5 can’t be a solution since the answer should be positive t must be 2 seconds
Check out Purple Math for great help on quadratics

Algebra Period 3 ( Review)

Quadratic Equations Review

Maybe another teacher reviewing how to graph quadratics might be just what you need:





So did that help?

Math 6 Honors Periods 1, 6 & 7 (Thursday)

Combined Operations 8-5

In order to solve an equation of the form

ax + b = c or ax –b = c or b – ax = c

where a, b, c are given numbers and x is the variable, we must use more than one transformation

Solve the equation 3n - 5 = 10 + 6

Simplify the numerical expression

3n - 5 = 10 + 6
3n – 5 = 16

add 5 to both sides

3n – 5 + 5 = 16 + 5
or
3n – 5 = 16
+ 5 = +5

3n = 21

divide both sides by 3 (or multiply each side by the reciprocal of 3)

3n/3 = 21/3

n = 7

General procedures for solving equations

Simplify each side of the equation
If there are still indicated additions or subtractions, use the inverse operation to undo them
If there are indicated multiplications or division involving the variable, use the inverse operations to undo them

The books says you must always perform the same operation on both sides of the equation. I say, “do to one side what you have done to the other side.”

Solve the equation
(3/2)n + 7 = 22

subtract 7 from both sides

(3/2)n + 7 - 7 = 22 - 7

(3/2)n =15


multiply both sides by 2/3, the reciprocal of 3/2

(2/3)(3/2)n = 15(2/3)

n = 10

Solve the equation
40 – (5/3)n = 15

add (5/3)n to both sides

40 – (5/3)n + (5/3)n = 15 + (5/3)n


40 = 15 + (5/3)n
subtract 15 from both sides

40 – 15 = 15-15 + (5/3)n

25 = (5/3)n multiply both sides by 3/5

(3/5)(25) = (5/3)n (3/5)

15 = n

Math 6 H Periods 1, 6 & 7 (Wednesday)

Equations: Decimals and Fractions 8-4


You can use transformations to solve equations which involve decimals or fractions

Solve 0.42 x = 1.05
Divide both sides by 0.42
.42x/.42 = 1.05/.42

now, do side bar and actually divide carefully and you will arrive at
x = 2.5

Solve: n/.15 = 92
multiply both sides by .15 to undo the division

(n/.15)(.15) = 92 (.15)
Again, do a sidebar for your calculations and you will arrive at

n = 13.80

How would we solve the following: (2/3)x = 6?

Let’s look at 2x = 6. What do we do?
We divide both sides by 2—or multiply both sides by the reciprocal of 2—which is ½
Remember the product of a number and its reciprocal is 1

Reminder: the ultimate objective is applying transformations to an equation is to obtain an equivalent equation in the form x = c
(where c is a constant.) Also remember that the understood (invivisble) coefficient of x in the equation x = c is 1. [Can you picture the poster in the front of the room?]
So to solve (2/3)x = 6 you would divide both sides by 2/3 but that is the same as multiplying by the reciprocal of 2/3, which is 3/2.

If an equation has the form

(a/b)(x) = c,
where both a and c are nonzero,
multiply both sides by b/a, the reciprocal of a/b
Solve (1/3)y = 18

(3/1)(1/3)y = 18(3/1)

y =18(3)
y = 54

Solve the equation: (6/7)n = 8

(7/6)(6/7)n = 8(7/6)

n = 8(7/6)

simplify first , then multiply
n = 28/3
n = 9 1/3
Let’s check

(6/7)n = 8
well, we said that n = 9 1/3 so substitute back, but change to 28/3 first
(6/7)(28/3) ?=? 8
[read ?=? as ‘does that equal?’]

Now really do a side bar with the left side of the equation to see what
(6/7)(28/3) really equals. Simplify before you multiply
2(4) = 8 so
8 = 8



Try:
1. (1/7)a = 13

2. b/8 = 16


3. 3.6d = 0.9


4. (3/8)f = 129

Math 6 H Periods 1, 6 & 7 (Tuesday)

Equations: Multiplication and Division 8-3

If an equation involves multiplication or division, the following transformations are used to solve the equation:

Transformation by multiplication: Multiply both sides of the equation by the same nonzero number.

Transformations by division: Divide both sides of the equation by the same nonzero number.

Remember: Do undo on one side what you would do undo the other!!

Our goal is to get the variable alone and to find an equivalent equation of the form
“n = a number”
Our goal is to arrive at the “world’s easiest equation”

Solve 3n = 24
Use the fact that multiplication and division are inverse operations
3n/3 = 24/3

n = 8

Or you could have use the reciprocal of 3-- which is 1/3 and multiplied both sides by 1/3
(1/3)(3n) = 24(1/3)
n = 8 and still arrived at the SAME solution

Solve 5x/5 = 53/5

x = 10 3/5

Solve n/4 = 7

(4)(n/4) = 7(4)

n = 28
How could you know for sure your answer is correct?

Substitute your solution into the ORIGINAL equation
Try:
1. 3r = 57

2. 714 = 7t

3. Solve A = bh for h

4. Solve P = 4s for s

5. Solve C = 2πr for r

Algebra Period 3 (Monday)

Direct Variation, Indirect Variation and Joint Variation: 12-5 TO 12-7


DIRECT VARIATION
As x increases, y also increases (graphs as a LINE)
or as x decreases, y also decreases
x and y go in the SAME DIRECTION
For example: y = 2x
the "2" is called the constant of variation and is "k" in the formula:
y = kx
To find k, you need one (x, y) coordinate to solve.
It's like solving for "m" in y = mx + b, but there is no b!
Example: y varies directly with x. When y = 8, x = 2. Find y when x = 3.
FORMULA FOR DIRECT VARIATION: y = kx
PLUG IN THE (x, y):
8 = k(2)
SOLVE FOR k:
k = 4
WRITE THE EQUATION:
y = 4x
PLUG IN VARIABLE GIVEN:
y = 4(3)
SOLVE FOR MISSING VARIABLE:
y = 12

INDIRECT VARIATION
As x increases, y decreases (graphs as a RATIONAL function...I'll show you it in class)
x and y go in OPPOSITE DIRECTIONS
For example: y = 2/x
the "2" is still called the constant of variation and is "k" in the formula:
y = k/x
To find k, you need one (x, y) coordinate to solve.
Example: y varies indirectly with x. When y = 8, x = 2. Find y when x = 3.
FORMULA FOR INDIRECT VARIATION:
y = k/x
PLUG IN THE (x, y):
8 = k/2
SOLVE FOR k:
k = 16
WRITE THE EQUATION:
y = 16/x
PLUG IN VARIABLE GIVEN:
y = 16/3
SOLVE FOR MISSING VARIABLE:
y = 5 1/3

JOINT VARIATION
TWO VARIABLES VARY WITH y AT THE SAME TIME!
As x AND z increase, y also increases (graphs as a LINEAR function)
The PRODUCT xz and y go in the SAME DIRECTION
For example: y = 2xz
the "2" is still called the constant of variation and is "k" in the formula:
y = kxz
To find k, you need one (x, y, z) coordinate to solve.
Example: y varies jointly with x and z. When y = 12, x = 2, and z = 3. Find y when x = 3 and z = 5
FORMULA FOR JOINT VARIATION:
y = kxz
PLUG IN THE (x, y, z):
12 = k(2)(3)
SOLVE FOR k:
k = 2
WRITE THE EQUATION:
y = 2xz
PLUG IN VARIABLE GIVEN:
y = 2(3)(5)
SOLVE FOR MISSING VARIABLE:
y = 30

Math 6 H Periods 1, 6 & 7 (Monday)

Equations: Addition and Subtraction 8-2



If the replacement set for an equation is the set of whole numbers, it is not practical to use substitution to solve the equation. Instead we transform or change the given equation into a simpler, equivalent equation. When we transform the given equation, our goal is to arrive at an equivalent equation of the form
variable = number

for example

n = 5

the number 5 is then, the solution of the original equation. The following transformations can be used to solve equation

Transformation by addition: add the same number to both sides
Transformation by subtraction: subtract the same number from both sides

solve x – 2 = 8

our goal is to find an equivalent equation of the form

x = a number

The left side of the given equation is x – 2. Recall that addition and subtraction are inverse operations. If we add 2 to both sides the left sides simplifies to x

x-2 = 8
x – 2 + 2 = 8 + 2 (We usually show the +2 right below each side of the equation)
x = 10

The solution is 10

Solve x + 6 = 17

Subtract 6 from both sides of the equation to get an equivalent equation of the form “ x = a number”

x + 6 = 17
x + 6 – 6 = 17 – 6 (Again, we usually show the -6 right below each side of the equation)
x = 11
the solution is 11

In equations involving a number of steps, it is a good idea to check your answer. This can be done easily by substituting the answer in the original equation.
What about the following

34 – x = 27

add x to both sides

34 – x + x = 27 + x
34 = 27 + x
subtract 27 from both sides

34 – 27 = 27 – 27 + x
7 = x

You may be able to solve some of the equations in the homework without pencil and paper. Nevertheless, it is important to show all the steps in your work and to make sure you can tell which transformation you are using in each step.

Which transformation was used to transform the first equation into the second

x – 3 = 5
x – 3 + 3 = 5 + 3


Remember when we stated the properties as well-- back in our 1st quarter!!

Algebra Period 3 ( Review)

Review of How to Graph a Quadratic

Find the vertex
first find the x value of this ordered pair. (x = -b/2a)
then (using that value) plug in to the equation to find the y coordinate

Find the axis (line) of symmetry – just use the x value of the vertex that is
x = -b/2a
draw the line of symmetry using a dotted or dashed line

Find 2 more points in an x y table (go either left OR right of the vertex and use 2 points close to the line of symmetry for the x, then plug into the equation to find the y values)

Find the 2 shadow (mirror) points by counting over from the axis at the exact same y value

Draw a U (not a V) through the points. Extend it past and put arrows at the end

Remember to label the x and y axes and put arrows on the axes.

Function vs. Relations: Functions are special relations where there is a unique x for each y

Therefore, there will never be 2 x’s that will repeat and if you use a vertical line on the graph, it will only hit one point on the graph.

Domain vs. Range:
DOMAIN:
The domain for any quadratic function is ALL REAL NUMBERS

RANGE:
The range depends on where the vertex is and whether the quadratic is a smile or a frown.

Generally, it will be of the form: y such that y is either greater than or equal to (≥ ) or less than or equal to (≤) the y value of the vertex.

f(x) is just a different way of saying y. What is better about it? Without seeing the work before the solution, you can actually tell the x value as well as the y value in the solution.

For example, if the solution is f(-3) = 12 you know that the point is (-3, 12)

Compare that to the solution y = 12. For that solution, you would not know the x value unless you look back in the problem.
Check out Purplemath.com Great website for help

Algebra Period 3 (Wednesday)

Quadratic Functions: 12- 4
A QUADRATIC FUNCTION is not y = mx + b
(which is a LINEAR function),
but instead is
y = ax² + bx + c
OR
f(x) = ax² + bx + c
where a, b, and c are all real numbers and
a cannot be equal to zero because
it must have a variable that is squared ( degree of 2)
[If a = 0, then we would end up with y = bx + c which is really y= mx + b]

Quadratics have a squared term, so they have TWO possible solutions also called roots. You already saw this in Chapter 6 when you factored the trinomial and used zero products property!! ( CHAPTER 6-- again)

If the domain is all real numbers, then you will have a PARABOLA which looks like
a smile when the a coefficient is positive or
looks like a frown when the a coefficient is negative.

Graphing quadratics:
You can graph quadratics exactly the way you graphed lines
by plugging in your choice of an x value and using the equation to find your y value.

Because it's a U shape, you should graph 5 points as follows:
STEP 1: determine Point 1: the vertex -
the minimum value of the smile or
the maximum value of the frown

The x value of the VERTEX = -b/2a
We get the values for a and b from the actual equation
f(x) = ax² + bx + c
just plug in the b and the a value from your equation into -b/2a and you have the x-value of the vertex.
Now to find the y value -- take that x value and PLUG it into the equation


STEP 2: Next, draw the AXIS OF SYMMETRY : x = -b/2a
a line through the vertex parallel to the y axis . Draw this line as a dashed line. REMEMBER: It will be a dashed line parallel to the y-axis


STEP 3: Point 2- Pick an x value to the right or left of the axis and find its y by plugging into the equation.

STEP 4: Point 3- Graph its mirror image on the other side of the axis of symmetry by counting from axis of symmetry

STEP 5: Points 4 and 5- Repeat point 2 and 3 directions with another point even farther from the vertex

JOIN YOUR 5 POINTS IN A "U" SHAPE AND EXTEND LINES WITH ARROWS ON END

Parabolas that are functions have domains that are ALL REAL NUMBERS
Their ranges depend on where the vertex is and also if the a coefficient is positive or negative

EXAMPLE: f(x) = -3x² (or y = -3x²)
the a coefficient is negative so it is a frown face
the x value of the vertex (maximum) is -b/2a or 0/2(-3) = 0
the y value of the vertex is 0
So the vertex is (0, 0)


To graph this function:
1) graph vertex (0, 0)
2) Draw dotted line x = 0 (actually this is the y axis!)
3) Pick x value to the right of axis of symmetry, say x = 1
Plug it in the equation: y = -3(1) = -3
Plot (1, -3)
4) count steps from axis of symmetry and place another point to the LEFT of axis in same place
5) pick another x value to the right of the axis of symmetry, say x = 2
plug it in the equation y = -3(2) = -6
plot ( 2, -6). Count the steps from the axis of symmetry and place another point to the LEFT of the axis in the same place.

The domain is all real numbers.
The range is y is less than or equal to zero