Algebra Honors (period 5 &6)

 Methods of Solutions 12-5

You have learned five different methods of solving quadratics.. picking the best one is the challenge
Here are some tips on when to use each method:
Quadratic Formula--> when to use: ax² +bx + c = 0 (It always works) Great when you have a calculator

Factoring--> when to use: ax² +bx + c = 0 you see the factors easily or you have ax² +bx = 0

Using the Property of SQRT's --> when to use: ax² + c = 0

Completing the square --> when to use: x² +bx + c = 0 especially when b is even
Graphing... this really does takes time but if done correctly... it really shows what are  the roots... and.... the x-intercepts, the solutions, the zeroes...

Which method would you use for the following?
x² +5x - 6 = 0
Answer: Factoring

x² -2x = 1
Answer: Completing the square

11x² = 44
Answer: SQRTing

3x² -5x = 4
Answer: Quadratic Formula

Pi Day, March 14, 2013

HAPPY PI DAY


HAPPY PI DAY! WHAT AN IRRATIONAL DAY!!!!
BE IRRATIONAL.. AND TRANSCENDENTAL
... here are the facts:
Today we celebrate Π (Pi), a very cool number. Π is a comparison between the measurements of the circumference to the diameter of a circle—any circle.
Pi is an IRRATIONAL number. That means it has no pattern and never terminates.
It CANNOT be written as a fraction with an integer in the numerator and denominator.

We use 3.14 and 22/7 as APPROXIMATIONS of pi.

These are not the exact values. The only symbol that tells the exact value is Π. Pi is a ratio, a comparison between two numbers.

You will be able to discover many interesting facts about pi—even finding it on your own.

Here's some links to click on if you're interested in the mystery of Pi:




History of pi
Find your birthday in pi
The first ten thousand digits of pi


Here is a rap I really like...


Lose Yourself in the Digits of pi


... and of course... here are the songs we sang

Happy Pi Day (to the tune of “Happy Birthday”)

Happy PI day to you
Happy PI day to you
Happy PI day everybody
Happy PI day to you



Oh, Number Pi (to the tune of “Oh, Christmas Tree”)

Oh, Number Pi
Oh, Number Pi
Your digits are unending.
Oh Number Pi
Oh Number Pi
No pattern are you sending
You’re three point one four one five nine
And even more if we had time,
Oh, number Pi
Oh, number Pi
For circle lengths unbending.

Oh, number Pi
Oh, number Pi
You are a number very sweet
Oh, number Pi
Oh, number Pi
Your uses are so very neat.
There’s 2 Pi r and Pi r squared
A half a circle and you’re there,
Oh, number Pi
Oh, number Pi
We know that Pi’s a tasty treat




Pi Day Song (to the tune of “Jingle Bells”)

Pi day songs
All day long
Oh what fun it is
To sing a jolly pi day song
In a fun math class
Like this

Verse;
Circles in the snow
Around and round we go
How far did we have to run?
Diameter times pi! (Refrain)

We wish you a Happy Pi Day (to the tune of “We wish you a Merry Xmas”)

We wish you a happy Pi Day
We wish you a happy Pi Day
We wish you a happy Pi Day
To you and to all

Pi numbers for you
For you and for all
Pi numbers in the month of March
So three point one four!!

Algebra Honors ( Periods 5 & 6)

 Graphs of Quadratic Equations: The Discriminant  12-4

Sometimes it is NOT necessary to draw the graph of an equation.
Knowing the relationships between an equation and its graph— you can obtain important information from the equation itself.

Today’s lesson illustrates the relationship between the number of roots of a quadratic equation and the number of x-intercepts of the related parabola.

In the parabola the x coordinate of a point where the curve intersects the x axis is called an x-intercept of the curve.

Looking at the parabola on page 572 you notice that this parabola has two x-intercepts, -1 and 3 because y = 0 for both of these values of x.

you can also see that
x²- 2x – 3 = 0 or (x +1)(x-3) = 0

has -1 and 3 as its roots, solutions, zeros… {-1,3}

The algebraic fact that a quadratic equation can have two, one or no real number roots corresponds to the fact that a parabola can have two, one or no x-intercepts…

and we can find that out by just working with the quadratic equations—rather than having to graph…

However… let’s graph a few
Example 1: x² + 4x + 1 = 0
find –b/2a
-4/2 = -2
f(-2) = (-2)² + 4(-2) + 1 = 4 -8 + 1 = -3
so (-2,-3) is the vertex

x = -2 is the axis of symmetry

Create an x y table

f(-1) = (-1)² +4(-1) + 1 = 1 -4 +1 = -2
f(0) = 02 + 4(0) + 1 = 1
x y
-1 -2
0 1

Now solve
Solving x² + 4x + 1 = 0

Now graph
Number of roots: two real roots Number of x-intercepts: two


Example 2 : x² + 2x + 1 = 0

find -b/2a -2/2 = -1
f(-1) = (-1)² + 2(-1) + 1 = 1 -2 + 1 = 0

(-1.0) is the vertex

and x = -1 is the axis of symmetry
create an xy table
f(0)=0² + 2(0) + 1 = 1
f(1) = (1)² +2(1) + 1 = 4
x y
0 1
1 4
Now graph


Now Solve
Solving x² + 2x + 1 = 0
We can solve easily but let’s use the quadratic formula

= -1






The solution set is {-1}

Number of roots: 1 real root Number of x- intercepts: one

Example 3: x² -4x + 7 = 0

find –b/2a=        +4/2 = 2
f(2) = (2)² -4(2) + 7 = 4 -8 + 7 = 3

vertex ( 2, 3)
and x= 2 is the axis of symmetry

create an xy table
f(1) = (1)² -4(1) + 7
f(0) = (0)² -4(0) + 7
x y
1 4
0 7
Now graph
Now solve
x² -4x + 7 = 0

There is NO REAL number root since does not represent a real number

Number of roots: no real number roots Number of x intercepts: None


In each of these examples the value of b²-4ac in the quadratic formula is the key to the number of real roots.

When b²-4ac is negative, there is no real number root of the equation ax² + bx + c = 0 because square roots of negative numbers do not exist in the set of real numbers.

Because the value of b²-4ac discriminates, or points out differences, among these 3 cases above, it is called the discriminant of the quadratic equation.

The discriminant is only part of the quadratic formula and simpliy helps one determine the number of roots.

If  b²-4ac <  0 "no real roots"
If  b²-4ac =  0 "one rational root"
If  b²-4ac  >   0 " two real roots"  we can break this down even more specific:

If  b²-4ac  is a perfect square then " two rational roots"
If  b²-4ac  is a non-square then  "two  irrational roots"

Algebra Honors (Period 6 & 7)

 The Quadratic Formula 12-3

Make sure to read the first post on the Quadratic Formula BEFORE or AFTER you watch these You Tube Videos....

How about creating your own video!! (For extra Credit!!)

Now if Emily can recite the Quadratic Formula after singing Twinkle, Twinkle Little Star... shouldn't you?

Math 6A (Periods 2 & 4)

Subtracting Integers 11-3

The life story about someone who was so negative-- you wanted to take a little negativity away but since you can't do that you add a little positiveness-- works in math as well!!
Instead of subtracting ... "ADD THE OPPOSITE!"
We proved it in class with our little red and yellow tiles... If you need to review, make your own out of red and yellow paper-- or whatever colors you want!!

In life-- to take away a little negative-- add some positive

To take away an integer... add its opposite.

Rule from our textbook
for all integers a and b

a - b = a + (the opposite of b) or
a - b = a + (-b)

Instead of subtracting.. "ADD THE OPPOSITE"
make sure you do the check, check.. you need to have two check marks.. one changing the subtraction to addition and the other changing the sign of the 2nd number to its opposite.

5 - ^- 2=
5 + ⁺ 2= 7

-2 - 5 =
- 2 + ^- 5 =
before I give you the answer look... we are looking at
-2 + ^-5
We are adding two negatives.. so we are back to the rules from Section 11-2...
when adding the same sign just add the number and use their sign so
- 2 + ^- 5 = -7

But what about -2 - ^-5 ?
adding the opposite, we get
-2 + ⁺ 5 .
Now, the signs are different so the rule from Section 11-2 is
ask yourself... "Who wins?" and "By how much?"

Okay, here the positive wins so I know the answer will be +
and by how much means.. to take the difference
5-2 = 3
so -2 + ⁺ 5 = +3
Do you need to put the + sign? No, but I like to in the beginning to show that I checked WHO WON!!

What about
-120 - ^-48?
add the opposite
-120 + ⁺ 48 follows the
Different signs rule... so
ask yourself
Who wins? answer: the negative.. so I know the answer will be negative..
And "By How Much?" take the difference 120-48 = 72
so
-120 + ⁺ 48 = -72
or

-120 - 48
-120 + (-48) = -168

NEVER EVER CHANGE THE FIRST NUMBER'S SIGN!!

WALK THE LINE, the number line-- that is!!
Remember... Attitude is such a little thing... but it makes a BIG difference!!
Always start with a positive attitude!!
When you walk the line, Which way are you always facing when you start???

Always attempt to get everything into addition so we can follow the rules of Section 11-2 Adding Integers.
1) SAME SIGN rule---> just add the numbers and use their sign
-4 + -5 = -9
2) DIFFERENT SIGNS rule
ask yourself those 2 important questions
a) Who wins? (answer is either negative or positive)
b) By how much? (Take the difference)

Page 376 answers to # 2-18 (evens)
2. -9
4. -5
6. 22
8. -9
10. -74
12. -32
14. -160
16. 498
18. -284

Algebra Honors (Periods 5 & 6)

The Quadratic Formula: 12-3

Method 5:
THE QUADRATIC FORMULA

Let's all sing it to "Pop Goes the Weasel!

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

x =
-b ± √ (b² - 4ac)
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!


You should know that for the quadratic formula, you don't need the "a" coefficient to be positive!

That's important if you use factoring, SQRTing each side, and completing the square.


But for the quadratic formula, either way, you'll get the same roots!


EXAMPLE: x² + 8x = 48 


You can move the 48 over or move the x² + 8x over and you'll get the same answers!
Let's look at:


"a" coefficient positive *****vs***** "a" coefficient negative



x² + 8x - 48 = 0 ****VS**** -x² - 8x + 48 = 0


First x² + 8x - 48 = 0
-8 ± √[64 - 4(1)(-48)]

2(1)



-8 ± √(64 + 192)

2



-8 ± √(256)
2



-8 ± 16
2



(-8 + 16)/2 and (-8 - 16)/2



8/2 and -24/2
4 and -12

Now let's compare -x² - 8x + 48 = 0


8 ± √[64 - 4(-1)(48)]

-2(1)

8 ± √(64 + 192)

-2

8 ± √(256)
-2


8 ± 16
-2

(8 + 16)/-2 and (8 - 16)/-2
24/-2 and -8/-2
-12 and 4

So all the signs are simply the opposite of each other and therefore the answers are the same

Math 6A (Periods 2 & 4)

Adding Integers 11-2

Rules: The sum of two positive integers is a positive integer.
The sum of two negative integers is a negative integer.

So- if the two numbers have the same sign, use their sign and just add the numbers.

-15 + -13 = - 28

-10 + -4 = -14

Rules: The sum of a positive integer and a negative integer is :

POSITIVE… IF the positive number has a greater absolute value

NEGATIVE… IF the negative number has a greater absolute value

ZERO… IF both numbers have the same absolute value

Think of a game between two teams- The POSITIVE TEAM and The NEGATIVE TEAM.

30 + -16 … ask yourself the all important question…
“WHO WINS?
in this case the positive and then ask
“BY HOW MUCH?”
take the difference 14

14 + - 52…
“WHO WINS?”
the negative… “BY HOW MUCH?”
38
so the answer is -38


(-2 + 3) + - 6 you can work this 2 ways

(-2 + 3) + - 6 = 1 + -6 = -5 or
using all the properties that work for whole numbers
Commutative and Associative properties of addition
can change expression to (-2 + -6) + 3 or -8 + 3 = -5 you still arrive at the same solution.

You want to use these properties when you are adding more than 2 integers.
First look for zero pairs—you can cross them out right away!!
3 + (-3) = 0
-9 + 9 = 0

Then you can use C(+) to move the integers around to make it easier to add them together rather than adding them in the original order. In addition, you can use A(+) to group your positive and negative numbers in ways that make it easier to add as well.

One surefire way is to add all the positives up… and then add all the negatives up.
At this point ask yourself that all important question… WHO WINS? …
use the winner’s sign..
and then ask yourself..
BY HOW MUCH?

example:

-4 + 27 +(-6) + 5 + (-4) + (6) + (-27) + 13

Taking a good scan of the numbers, do you see any zero pairs?
YES—so cross them out and you are left with
-4 + 5 + (-4) + 13
add your positives 5 + 13 = 18
add your negatives and use their sign – 4 + -4 = -8

Okay, Who wins? the positive
By how much? 10
so
-4 + 27 +(-6) + 5 + (-4) + (6) + (-27) + 13 = 10

Algebra Honors (Periods 5 & 6)

Completing the Square: 12-2
METHOD 4:


Now this is completely new to you!!!

When does the square root = ± square root method work well?

When the side with the variable is a PERFECT SQUARE! We saw that in the previous section.
perfect square = k ( when k ≥ 0)

So what if that side is not a perfect BINOMIAL SQUARED?

It may be possible to trasform it into one... by COMPLETING THE SQUARE
You can follow steps to make it into one!

Why is this good? 

Because then you can just square root each side to find the roots!


THIS METHOD ALWAYS WORKS!


EXAMPLE:
x² - 3x -18 = 0

(Head's up-- I wouldn't use this method here because I can see that it factors easily... into (x-6)(x+3) = 0 so the solution set is {-3,6}
However, knowing what I need to get for my solutions might be a good way to practice Completing the square..
so

x² - 3x = 18

Step 1) b/2

That is, take half of the b coefficient ,
or in this case (- 3/2)



Step 2) Square b/2

(b/2)²
(-3/2 ⋅ -3/2 = 9/4)


Step 3) Add (b/2)² to both sides of the equation

x² - 3x + 9/4 = 18 +9/4

Step 4) Factor to a binomial square

(x - 3/2)² = 18 + 9/4
 = 81/4


Step 5) Square root each side and solve

√(x - 3/2)² = √ 81/4

x - 3/2 = ± 9/2
x = 3/2 ± 9/2
x = 3/2 + 9/2 AND x = 3/2 - 9/2
x = 6 and x = -3
{-3, 6}
We got the same solutions !! Yay!!


x² - 10x = 0


Not a TRINOMIAL SQUARE so it would not factor to a BINOMIAL SQUARED.

But here's how you can make it one!


Step 1) b/2

That is, take half of the b coefficient ,
or in this case (- 10/2 = -5)


Step 2) Square b/2

(b/2)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

√(x - 5)² = √ 25

x - 5 = ± 5

Step 6) ADD 5 TO BOTH SIDES

x - 5 = ± 5
+ 5 = +5
x = 5 ± 5
Step 7) Simplify if possible

x = 5 + 5 and x = 5 - 5

x = 10 and x = 0



So the 2 roots (solutions/zeros/x intercepts) are 0 and 10.
YOU DON'T NEED TO GRAPH THE PARABOLA, BUT IF YOU DID, IT WOULD CROSS THE X AXIS AT 0 AND 10.
I don't know where the vertex is, but I don't need to because it's not the solution to the quadratic (although I certainly could find the vertex by using x = -b/2a)
Notice that I could also factor x² - 10x = 0 to get the solution more easily.
So don't complete the square if the quadratic factors easily!

IF THERE IS A "c", first move the c constant to the other side of the equation before completing the square:
x² - 10x - 11 = 0
x² - 10x - 11 + 11 = 0 + 11
x² - 10x = 11
NOW COMPLETE THE SQUARE AS ABOVE:
x² - 10x + 25 = +25 + 11
(x - 5)² = 36
√ (x - 5)² = √ 36
x - 5 = ± 6
x = 5 ± 6
x = 11 and x = -1

Again, this one factored easily so I wouldn't have even used completing the square. ALWAYS CHECK IF IT FACTORS FIRST!
Now an example that DOES NOT FACTOR: x² - 10x - 18 = 0
x² - 10x - 18 = 0
x² - 10x - 11 + 18 = 0 + 18
x² - 10x = 18
NOW COMPLETE THE SQUARE AS ABOVE:
x² - 10x + 25 = +25 + 18
(x - 5)² = 43
√ (x - 5)² = √ 43
x - 5 = ± √ 43
x = 5 ± √ 43
x = 5 + √ 43 and x = 5 - √ 43
When there is an IRRATIONAL square root, always SIMPLIFY if possible!

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 find the completing the square term and add it to both sides:
[(-3/2)(-3/2)]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
√[(x - 3/4)² ] = ±√ [17/16]

x - 3/4 = ±[√17] /4

x = 3/4 ±[√ 17] /4

x = (3 ± [√17]) /4
or written better x =
3 ± √ 17
4

Math 6A (Periods 2 & 4)

Negative Numbers 11-1

On a horizontal number line we use negative numbers for the coordinates of points to the left of zero. We denote the number called ‘negative four’ by the symbol -4. The symbol -4 is normally read ‘ negative 4’ but we can also say ‘ the opposite of 4.’

The graphs of 4 and -4 are the same distance from 0—>but in opposite directions. Thus they are opposites. -4 is the opposite of 4.

The opposite of 0 is 0

Absolute Value is a distance concept. Absolute value is the graph of the distance of a number from 0 on a number line. The absolute value of a number can NEVER be negative!!

Counting (also known as Natural) numbers: 1, 2, 3, 4, ….
Whole numbers 0, 1, 2, 3, 4….
Integers are natural numbers and their opposites AND zero
…-4, -3, -2, -1, 0, 1, 2, 3, 4….

The opposite of 0 is 0.

The integer 0 is neither positive nor negative.

The farther we go to the right on a number line--- the bigger the number. We can compare two integers by looking at their position on a number line.

if x < 0 what do we know? x is negative number if x > 0, what do we know? x is a positive number

We have been practicing representing integers by their graphs, that is, by points on a number line.

Make sure that your number line includes arrows at both ends and a line indicating where zero falls on your number line.
The graph of a number MUST have a closed dot right on the number line at that specific number.
Please see our textbook page 366 for an accurate example.

Algebra Honors ( Periods 5 & 6)

Quadratic Equations with Perfect Squares 12-1

In Chapter 5 you learned how to solve certain quadratic equations by factoring and in Chapter 11 you learned how to solve quadratics in the for
x² = k
as in x² = 49
x = ±7

This lesson extend to any quadratic equation involving a perfect square

if we have x² = k
if:
k > 0 then x² = k has two real-numbered roots with x = ±√k
if k = 0 then x² = k has one real numbered root x = 0
if k<0 nbsp="" sup="" then="" x="">2
  = k has no real numbered roots
m² = 49
√(m²) = ±√49
m = ±7
{-7, 7}


5r²= 45

5r² /5= 45/5
r²= 9
√r²= ±√9
r = ±3
{-3,3}

(x + 6)² = 64
√(x+6)² = ±√64
x+6 = ±8

be careful here
you now have
x = -6±8 which means

x = -6-8 = -14 AND x = -6+8 = 2
{-14, 2}

9r² = 121

9r²/9 = 121/9
r² = 121/9
√r²= ±√121/9
r = ±11/3

Make sure to check your solutions to insure that they both work

(x-3)² = 100
√ (x-3)² = ±√100
(x-3) = ±10
x = 3 ±10
x = 3 +10 = 13 and x = 3-10 = -7
{-7, 13}

5(x-4)² = 40
5(x-4)²/5 = 40/5
(x-4)² = 8
√ (x-4)² = ±√8
Now you need to simplify the Radical

√ (x-4)² = ±2√2
(x-4) = ±2√2
x = 4 ±2√2

{4 - 2√2, 4 + 2√2}



7(x - 8)² = -28
7(x-8)²/7 = -28/7
(x-8)² = -4
WAIT!!! that can't happen no real numbered solutions...

An equation that has a negative on one side and a perfect square as the other has NO REAL NUMBER Solutions

y² + 6y + 9 = 49
(y+3)² = 49
√(y+3)² = ±√49
(y+3) = ±7
y = -3 ±7
y = -3-7 = -10 and y = -3 +7 = 4
{-10. 4}

Perfect Squares like (x -4) ² and ( y + 3) ²

The square of any real number is always a non negative real number.
2(3x-5)² + 15 = 7 becomes
2(3x -5)² = -8
or (3x -5)² = -4... NO REAL number solution!!

Algebra Honors ( Periods 5 & 6)

Introduction to Quadratic Equations

We learned from Chapter 8 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 12 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 where 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 5)
but...some quadratics are not factorable

3) Square root each side (+ or - square root on the answer side) We did this in Chapter 11
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! :)

REVIEW:

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 5 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!!

METHOD 3:

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

METHOD 4:
The Quadratic Formula!!
We will get to this one...

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 Honors (periods 5 & 6)

Quadratic Functions 8-8

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)

Quadratics have a squared term, so they have 2 possible solutions (roots)
You already saw this when you factored the trinomial and used zero products property.

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.


Let's look at f(x) = x² - 2x - 2

When we plotted a few points, we discovered that definitely was NOT a straight line.
We used f(0), f(1), f(2), f(3), f(4), and then f(-1), f(-2)
we connect with a smooth curved line  a U shape with arrows at either end--> because the parabola continues without end. Do not put the equation of the parabola on it, however.
Notice this parabola opens upward and has a minimum point or lowest point  at (1, -3)
the y coordinate at this point is the least value of the function.
Domain: {x Ι x = R} This is read as "the Domain is x such that x equals all real numbers."  and the
 Range: { y Ι  y  ≥ -3} This is read as "the Range is y such that y is greater than or equal to -3."

the vertical line x = 1 contains this minimum point and is called the axis of symmetry.   If you fold the graph along the axis of symmetry the two halves coincide. You could always just plot points-- but how would you know which points to pick?

Let's look at f(x) = -x²  + 2x + 2
Just by looking at this function I can tell that it opens downward. That (0,2) is where one side of the parabola crosses the y-intercept.

f(x) = ax²  + bx + c ( a ≠ 0 )   is a quadratic function.  Why can't a =0?
If the domain of f is the set of all real numbers, then the graph is a parabola
if a in the function f(x) =ax²  + bx + c is positive, the graph opens upward. It's a happy face!
If a in the function f (x) = ax²  + bx + c is negative, the graph opens downward. It's a sad face!

The minimum or maximum point is called the vertex of the parabola.
Notice that all the other points ( except the vertex) occur in pairs with the same y-coordinate-- and the same distance away from the axis of symmetry. That makes sense, doesn't it?

The x coordinate of the vertex of a parabola f(x) = ax²  + bx + c
is  -b/2a
YES... you do need to memorize this!
The axis of symmetry is the line  x = -b/2a

H:x  --> 2x²  + 4x -3
Find the vertex
Remember -b/2a is the x coordinate of the vertex
so
-b/2a = -4/2(2) = -1
The x coordinate of the vertex is -1. Plug in this x value to find the corresponding y value for the vertez
f(-1) = 2(-1)²  + 4(-1) -3 = 2 -4 -3 = -5
so the vertex is
(-1, -5)
and
x = -1     is the line of symmetry or  the AXIS OF SYMMETRY

What happens as the "a" coefficient gets really big or really small (fraction/decimal)? We'll look at that together on my graphing calculator.

But think, what happened when the "m" (slope) coefficient got big?

The slope got steeper.

So now think that both sides of the U get steeper at the same time.

What's happening to the shape of the U???

Now think, what happened when the "m" (slope) coefficient got tiny?
The slope was a bunny slope.
So now think that both sides of the U are bunny slopes at the same time.
What's happening to the shape of the U???

Putting in standard form:
Standard form is:
y = ax² + bx + c

OR

f(x) = ax² + bx + c

You can't read the sign of a, b, or c until it's in standard form (just like y = mx + b!)

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:

First MAKE SURE THE EQUATION IS IN STANDARD FORM!

y must be isolated on one side and then you can read the a and b coefficients.

y = ax² + bx + c


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


Plug that into the equation and then find the y value of the vertex



Next, draw the AXIS OF SYMMETRY :

x = -b/2a


a line through the vertex parallel to the y axis


Point 2) Pick an x value IMMEDIATELY to the right or left of the AXIS OF SYMMETRY and find its
y by plugging into the equation.



Point 3) Graph its mirror image on the other side of the AXIS OF SYMMETRY by counting from the axis of symmetry


Points 4 and 5) Repeat point 2 and 3 directions with another point ONE STEP FARTHER from the AXIS OF SYMMETRY.



JOIN YOUR 5 POINTS IN A SMOOTH "U" SHAPE ( not a V shape!)

AND 
EXTEND LINES WITH ARROWS ON END



Parabolas are functions whose domains 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 frowny face


The vertex is called the maximum.


The x value of the vertex is -b/2a 


a = -3 and b = 0 (it's missing!)


The x value of the vertex = -b/2a = -0/2(-3) = 0


Plug the x value of 0 back into the function to find the y value of the
vertex:


y = -3(02) = 0
So the vertex is (0, 0)

The domain is all real numbers.


The range is y is less than or equal to zero (It's a frowny face)



To graph this function:


1) Graph vertex (0, 0)


2) Draw the AXIS OF SYMMETRY –
a dotted line at x = 0 (actually this is the y axis!)


3) Pick x value immediately to the right of axis of symmetry, x = 1


Plug it in the equation to find the y value: y = -3(1) = -3


Plot (1, -3)


4) Count the same 1 step from axis of symmetry on the other side of the axis
and place another point to the LEFT of axis at the same y value
(-1, -3)


5) Pick another x value to the right 2 steps away from the axis of symmetry, x = 2

Plug it in the equation to find y:
y = -3(22) = -12
Plot (2, -12)


6) Count 2 steps from axis of symmetry on the other side of it and
place another point to the LEFT of axis at the same y value
(-2, -12)



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

Algebra Honors (Periods 5 & 6)

Direct & Inverse Variations  8-9 and 8-10

Direct Variations  8-9

 f(x) = mx + b

It is a linear function. the f(x) is dependent on the x value.

A direct variation is a function defined by an equation in the form

y = kx, where k is a non zero constant. When graphing, k is  the slope

You can say that y varies directly as x

that is, as x goes up in value so does y or

as x decreases à so does y

When the domain is the set of all real numbers, the graph of a direct variation is a straight line with slope k that passes through the origin.

Given that m varies directly as n, and that m= 42 when n = 2, the constant of variation can be found by writing m = kn and substituting in those values. You find that

42 = 2k , so k = 21.
You can write the function as m = 21n.
Now when asked to find the value of m when n = 3 you just plug and chug…
m = 21(3)  so m = 63

Could we have found this a different way? YES..

Suppose ( x₁, y₁) and  ( x₂, y₂) are two ordered pairs of a direct variation defined by

y = kx and you know that neither  x₁ nor  x₂  are zero.

You know that

y₁ = kx₁    and that y₂= kx₂ 

Solving both for k you discover that

y₁/x₁ = k and so does y₂/x₂ = k

Since each ratio equals k, the ratios are equal and you can set them in an equation

y₁/x₁ = y₂/x₂  and you read this… “y₁ is to x₁ as y₂ is to x₂.”

This is a proportion!!

For this reason k is sometimes called the constant of proportionality  and
y is said to be directly proportional to x

When you use a proportion to solve a problem, you will want to recall that the product of the extremes equals the product of the means.

We reviewed a few equations and found the following to be direct variations:

y = 3x

p = 9s

d = 3.3t

even y/x = -5

But the following were determined NOT to be direct variations:

y = 3x²

xy = 4

Eample:

y varies directly as x

y = 6 and x = 72 Find the constant of variation

y = kx

6 = k(72)

6 = 72k

k = 1/12

Inverse Variations 8-10

An inverse variation is a function defined by an equation in the form:

xy = k where k is a NON ZERO constant

or  y = k/x  where x ≠ 0

You say that y varies inversely as x or that y is inversely proportional to x.

The constant k is the constant of variation.

The graph of an inverse function is NOT a straight line!

xy = k is NOT linear!

Your graph can be a hyperbola

When k is positive the branches of the graph are in Quadrants I and III

When k is negative the branches of the graph are in Quadrants II and IV

Similarly to direct variation, you can compare two ordered pairs of the same inverse variation. Since the coordinates must satisfy the equation xy = k you know that

x₁y₁=k and x₂y₂= k

or

x₁y₁= x₂y₂

Reviewing :

Direct Variation is y = kx    or y₁/x₁ = y₂/x₂

Inverse Variation  is xy = k or x₁y₁= x₂y₂

These equations show that for direct variation the quotients of the coordinates are constant and for inverse variation the products of the coordinates are constant.

Is it Direct or Inverse Variation?

y/x = k  (careful this becomes   y = kx    so its direct)

y= k/x           inverse

p = k/z          inverse

xy = 25         inverse

d= 40t          direct

m/n= 5/8      direct

x/y = 1/k      direct

kxy = 5        inverse

WORD PROBLEMS FOR DIRECT & INVERSE VARIATIONS
( 2^nd day of lesson)

Several examples were given in class:

Truck rental Company charges $35 a day plus 21 cents per mile. Normally your questions in the past were "What is the cost for a rental of a truck for ...:

1 day and 340 miles?" or

"2 days and 450 miles?"

You would just plug in and figure out the exact cost...

Then we discussed the rental of a chain saw (from the textbook)... remember it is for cutting down trees... like those which were knocked down by those tremendous winds we had last year… remember those winds…

We are given that the rental is $5.90 a hour and you must pay $6.50 for 1 can of gas. Again, in the past your questions would be something like...

"How much would it cost for 7.5 hours?"

But... you could write a linear function to represent the cost for all different rental hours.

Let h represent the hours

s(h) = 5.9h + 6.5

Now, no matter how many hours you rent the chain saw, you can figure out the cost.

Phone bills in the past charged 15 cents per message + a base charge. Let's say you were given the July bill of $18 which included 62 messages. What was the base charge?

First find out the cost of the messages (.15)(62) = $9.30 and subtract that from $18.

Or just realize it would be 18 - (.15)(62) = $ 8.70

Then you could write a linear function p(x) = .15x + 8.7

August had 76 messages... what was the bill becomes easy to solve-- just use the linear equation.

p(76) = .15(76) + 8.7 = 20.10. August bill was $20.10

Turn to Page 394

#20

distance on a map varies directly to actual distance

m= distance on the map

d= actual distance

m = kd

Given that 1 in on the map ---> 10 miles

1 = k(10)

1 = 10k

k = 0.1

so formula is m = 0.1d

writing as a proportion you would have

1/10 = m₂ / d₂

# 22 Volume directly proportional to temp T in Kelvin

5 Liters 300 degrees

V = kT

5= 300k

k = 1/60

so formula is V = (1/60)T

and as a formula before you simplify

5/300 = V₂ /T₂

or 1/60 = V₂ /T₂

Algebra Honors ( periods 5 & 6)

Linear and Quadratic Functions 8-8

This lesson will be completed in two parts:

Linear Functions:

The function g defined by g(x) = 2x - 3  is called a linear function

Notice the word line in linear

If its domain is the set of all real numbers  then the straight line that is the graph of

y = g(x) = 2x - 3  is the graph of g . 

The slope of the graph is 2 and the y-intercept is -3.

A function f defined by f(x) = mx + b is a linear function

If the domain of f is the set of all real numbers, then its graph is the straight line with slope m and y-intercept b.

Relations:

A relation is any set of ordered pairs.

The set of the first coordinates of the ordered pairs is the domain of the relation

The set of the second coordinates of the ordered pair is the range.

A function is a relation in which different ordered pairs have different FIRST coordinates.