intro pg.1-3 standard form to vertex form – jesus pg. 4-11 solving quadratics – markita pg.12-15

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Page 1: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15
Page 2: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Greetings, we have designed this booklet to share our knowledge of Quadratics, to EAHS students who may be experiencing trouble with this topic. Quadratics is a unit covered during Algebra II class. We will cover everything from factoring to complex numbers and more, we hope this booklet will help you understand this topic and make your life easier.

Page 3: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15 Factoring Quadratics – Antonio pg.16-18 Graphing Quadratics – Joceline pg.19-22 Finding Min and Max – Anallely pg. 23-28 Complex Numbers – Joceline pg. 29-36 Completeting the Square – Anallely pg. 37-40 Real World Applications – Antonio pg.

Page 4: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

The term: Quadratic Equation: Definition of Vertex Form to Standard Form (vice versa) Mathematical definition of the term. Vertex Form: The vertex form is defined as: y = a(x-h) ^ 2+k, where the vertex of the graph

is (h, k) and the axis of symmetry is x = h. The h-value of the vertex is found by using the formula h = -b/2a . Now, we

know the x value of the vertex, we can apply the x value in the given function to determine the y-value.

Standard Form: The Standard form is expressed as y = ax^2 + bx + c The role of “a” : If a > 0, the parabola opens upwards : If a < 0, the parabola opens downwardsThe axis of symmetry (a.o.s): The axis of symmetry is the line x = -b 2a Meaning of the term in own words. The first thing I do when dealing with Vertex form is labeling my a and b terms.

Find your axis of symmetry which equals your h. Plug you’re a.o.s back to the original equation that answer will become your k value. When you have your vertex (h, k), just plug in your values into the vertex form equation. Pg.1

Page 5: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Example of the term and something that looks like an example but is not.

EXAMPLE 1 EXAMPLE 2y = 3x^2 + 12x + 5 y = 4(x +

2)^2 - 6A= 3 B = 12 4(x + 2)(x +

2) - 6 First identify a.o.s 4(x^2 + 4x +

4) - 6 -(12)/2(3) = -2 4x^2 + 16x

+ 16 - 6So… -2 is also your h value y = 4x^2 +

16x +10 Next plug in -2 back to the equation3(-2) ^2 + 12(-2) + 5 = -7So… -7 is your k value Your vertex is (-2, -7)Finally use your vertex form and substitute h & k valuesY = 3 (x + 2) ^2 - 7Difference: This are both correct examples, the only true issue and

difference is that example 1 is going from standard form to vertex form, while example 2 is going from vertex form to standard form. Also you use different steps to find the answer.

Pg. 2

Page 6: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

 The vertex form of a parabola's equation is:

y= a(x-h)2+k (h, k) is the vertex If a is positive then the parabola opens upwards like a regular "U". If a is negative, then the graph opens downwards like an upside

down "U". If |a| < 1, the graph of the parabola widens. This just means that the

"U" shape of parabola stretches out sideways. If |a| > 1, the graph of the graph becomes narrower(The effect is the opposite of |a| >1).     

      

(Illustration 1)

Pg. 3

Page 7: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

How to graph in Vertex form

Y = (x - 1)^2 + 2 has a vertex of (1,2)

( Illustration 2 ) Pg. 4

Page 8: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

How to graph in Standard form Y = -x^2 + 4x + 1

Step 1: Find x- coordinate of vertex via –b/ 2aX- coordinate = -(4)/2(-1)= -4/-4 = 2

Step 2: Find the y- coordinate by solving for yY = -(2)^2 + 4(2) + 1 = -4 + 8 + 1 = 5

Vertex is at (2,5)

Pg. 5

Page 9: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

How to Write a Skill Description Skill TitleVertex form to Standard form (vice versa)  Skill DescriptionGoing from standard form to vertex form is an important part of graphing quadratic

equations, because it gives you the vertex of the parabola in the process.You may ask when I use Standard form or Vertex Form…. It all depends on what the

teacher wants you to find. ExamplesStandard Form to Vertex Form

Y = -3x^2 + 12x + 5A = -3 B =12 C = 5 A.O.S = -12/2(-3) = 2H = 2K = -3(2)^2 + 12(2) + 5 = 17Y = a(x-h) ^ 2+kY = -3 (X - 2)^2 + 17 Vertex Form to Standard Form Y = (x - 4)^2 + 29 (x - 4)(x - 4) + 29X^2 – 8x + 16 + 29X^2 – 8x + 45

Pg. 6

Page 10: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Application with word problem I really don’t think vertex form and standard form can be used in a word

problem, well its possible if the word problem asks you to use either equation to find the answer. But in case they just follow the steps and you’ll be ok. I have seen word problems that depend on you finding the a.o.s and the vertex and this are part of the standard and vertex form. Actually, if a word problem asks for you to find the revenue of a product and find the number of packages, its essential for the a.o.s and vertex to be found.

Description of the processTo go from standard form to vertex form follow this stepsLook at your standard form equation that is given : What I normally do is label my a, b ,c values : It helps me get everything in order plus it helps me not get confuse Now find your axis of symmetry as we know the (a.o.s)= h value: Its important to find you’re a.o.s because h is your x value and your part of

your vertex : It may sound confusing but it goes like this a.o.s = h = x : the x values its just a way to remember what your h value represents Ok… now you have you’re a.o.s, plug that answer back into the standard form

equation (original equation): This is how you find your ‘’ k ‘’ value :Vertex = (h,k)When you have your vertex (h,k) , plug in your two values into the vertex

form equation : y = a(x - h)^2 + k

Pg. 7

Page 11: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

CAUTION: Remember to plug in your “a” value before the (x - h)^2 part of the equation, also remember to put the square part where it should go. Also, remember to locate in your “h” and “k” values correctly. If “h” is a negative number make sure it ends up being and addition symbol inside the parentheses (x + h)^2

To go from vertex form to standard form follow this stepsThe first thing you want to do is expand the a(x +h) by this I

mean a(x + h) (x + h)… you may not always have an “a “value that’s ok. (Refer to Illustration 2 for better explanation of this step)

: You do this step because it will help you get a step closer to the standard form equation

Your new equation should look like this for now a(x + h)(x + h)+ k or (x + h)(x + h) + k

: Now you could either use the FOIL method or the “box method” to factor you expanded form

: Note I’m using addition for this example but you could have subtraction depending on the problem

If you have an “a” value then you distribute a(x^2 +bx + c) if not then you just combine like terms

: Finally your answer should look like this y = ax^2 + bx + c

Pg. 8

Page 12: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Methods of solving Quadratic Equations There are four methods with which to solve a

Quadratic equation. 1. The Quadratic formula can be used in any

situation but can get a little tricky because of all the operations involved.

2.You can also complete the Square whenever you have perfect squares and your Equation is in Standard form.

3. You can square root when you have two square roots .

4. you can factor whenever it's simple to do so.

Page 13: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

A Quadratic equation is an equation that is written as ax² + bx + cThe solution of a quadratic equation is the value of x when you set

the equation equal to zero.0= ax² + bx + cThe quadratic formula below solves the equation for zeroQuadratic Formula ExampleX² + 4x – 5a= 1 b= 4c= -5 Using the quadratic formula you will just substitute a, b and c into

the formula.-4± (square root of) 4² - 4(1) (-5)/ 2 (1)-4± (square root of) 16 – 20 / 2-4± (square root of) -4/ 2

Page 14: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Solving QuadraticsExample: Using Zero- Product

PropertySolve(x +5)(2x – 4) = 01. Using Zero- Product Prop, set

each factor = 0 & solve for x.x + 5 = 0 2x – 4 = 0 - 5 = -5 +4

= + 4______________ _____________ X = -5 2x =4 x =

2

2. Check -5 and 2(-5 + 5)(2(-5)-4) =0 (2 + 5)

( 2(2) -4) (0)(-10-4) =0 (7)(4-

4) = 0 (0)(-14) = 0 (7)

(0) = 0 0 = 0 0

=0

Page 15: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Solving through square rooting Ex: (x^2-25) Note both the x^2 and the 25 have √(x^2-25) = square roots. Square root the

equation (x-5) (x+5) find both the pos. And the neg.

answers x-5=0 x+5=0 use 0 product property and set

both +5 +5 -5 -5 answers=0. Solve both equations. X=5 x=-5 you find that x=±5

Page 16: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Factoring A quadratic equation is normally expessed as ax^2+bx+c.

When trying to solve the equation by factoring there are several things to keep in mind.

If c is positive that means both your factors will be the same sign. If c is positive than b determines what sign both factors will be. If b is pos. So are both factors. If b is neg. So are both your factors. This will always be true.

If c is neg. Than you will have a neg.factor and a pos. factor. This will alwatys be true.

Page 17: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

If a=1 The most impostatn thing to determine is if a=1. If a=1

then your two factors will multiply to c and add to b. Ex: x^2+8x+12. Your factors will be 6 and 2 (resulting in the equations (x+6) (x+2). why? Because 6X2=12 and 6+2=8. once you've checked that your factors do follow the “c” rules you simply solve the equations.

X+6=0 X+2=0 -6 -6 -2 -2x=-6 x=-2

Page 18: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

If a≠1 If a≠1 then things get more complicated. You have to use what's called

the box method to factor your equation.Ex:4x^2+9x+5 Start off bydrawing a rectangle with four rectangles inside it. In the top left rctangle write your ax^2 in this case 4x^2. In your bottom right box write in the c value. In this case 5.

Now multiply aXc or in this example 5X4=20 List the variables of 20. (1,20) (2,10) (4,5). determine which set adds

to 9. 1+20≠9 2+10≠9 5+4=9 Your factors are 5 and 4. fill them in the remaining boxes. (it doesn't

matter which number goes in which of the remaining boxes.) Now pull the greates common factors from one row and one column.

You should end up with one equation along the row and another along the column.

Your equations are (x+1) and (4x+5) now solve both of them to find X. Result x=-1 and x=-5/4

Page 19: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

When graphing quadratics you should start by identifyng each of your terms.

It's important to know what each of your terms mean. That way you have a basic idea of what your graph should look like and can descibe it without necessarily

needing a graph. Since each term controls certain aspects of the graph such

as whether it's a max. Or a min. And whether the vertax moves from (0,0).

Page 20: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

When the “a” is positive your parabola opens up. (which means your parabola is

a min)When it is Negative your parabola will

open down. (which means your parabola is a max)

If the absolute value of a<1 the parabola will be wider.

If the absolute value of a >1 the graph will be narrower.

Page 21: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

It shows Horizontal displacement; affects position of axis of symmetry.

Basically it shows if and where the vertex moves as far as the x-intercept is

concerned.If the “b” is positive, the parabola moves to the left that many units. If the “b” is negative the parabola moves to the right that many units.

Page 22: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

It shows vertical displacement.Basically it shows if and where the vertex moves in relation to the y-

intercept.It the “c” is positive the graph moves

that many units up.If the “c” is negative the parabola moves

that many units down.

Page 23: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

The term: Finding Min/MaxThe mathematical definition of the term.Minimum: lowest value of the graph [once graphed]

or the “a” term will determine thisMaximum: highest value of the graph [once

graphed] or the ‘a’ term will determine this The meaning of the term in own words. As the graph shows two ways that a parabola

can go when its maximum the graph is shown going down and minimum when the point is at the bottom and it’s the lowest point.

 

Page 24: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Example of the term and something that looks like an example but is not.

Minimum

Page 25: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Difference: As you can tell the first two graphs

make sense with their point but as for the

3rd you can compare with the other two to know that its wrong and get the right answer.

 

Page 26: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Another way that you can think about this term. “a” termAnother way is to use formulas to figure out which is which then to check

you answer, you would then graph.   “a” term   

   

 

   

Because the “a” term is negative then the graph has a maximum which is 20

maximum

Page 27: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Skill Description Min/Max is when you either find the highest or lowest value.

This comes in handy when trying to find a value and determining which option is better.

ExamplesA rocket is fired straight up in the air. Its equation is

where h as for feet and t as for seconds. Find the maximum or minimum height that the rocket reaches and how long it will take to land.

 

In 3 seconds it will reach its peak [maximum] at 144 feet and it will take 6 seconds to land.

Page 28: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Application with word problem If the world problem is asking you to find the

revenue or the highest or lowest point and sometimes when you are asked to compare products.

 Description of the process • Determine whether it will be minimum or

maximum depending on the “a” value• Find your “x” coordinate by using formula• Find your “y” coordinate by plugging in a.o.s into

original equation• Your “y” coordinate is your min or max

Page 29: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Complex numbers are numbers that havea real and an imaginary part. They are sometimes written in this from a+bi where a and b represent real numbers (such as 2,3,4 etc) and i represents the imaginary number. Also remember that i= √ -1. Another important rule to remember is that an i can not be in the denominator.

Page 30: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Rule #1 pay attention to the operation! Subtraction- subtract like terms Addition – add like terms Multiplication – use the box method or

FOIL

Page 31: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

The chart below is the i power chart. This chart will be usseful in pretty much any sittuation where i is used to a power. It is also usseful when simplifing complex numbers, as you will see in later examples

i^0 0

i^1 i

i^2 -1

i^3 -i

i^4 1

Page 32: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

i^4=1Ex: 2+4i^4 2+4(1) 2+4=5 Ex:(3+i) +(4+i) Combine like terms. Meaning (3+4) + (i+i) add real numbers together and

= 7+2i Imaginary numbers together

Page 33: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Ex: (5+i) – (3-i) Subtract like terms. (real-

(5-3) – (i-(-i)) real, imaginary - imaginary)

(5-3) – (i+i) remember a term- an neg. = 2-2i term is the same as a

term+a pos. Term Ex: (7+i) – (2+i) Subtract like terms. (real- (7-2) – (i-i) real, imaginary-imaginary) 5-(-i) Remember a term – a neg. 5+1 term = a term + a pos.

term

Page 34: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

i²= -1 Ex: √ -625/196

Answer: ±25i/ 14 Why ±? Because when you find the square root

you find both ± Why 25? Because 25 is the square root of 625 Why i? because when you find a square root of a

neg. number you pull out the -1 which =i Why 14? Because 14 is the square root of 196

Page 35: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

i^2=-136/12i An i is in the denominator36/12i * i/i the way to fix this is to

multiply the fraction by i/i36i/12i^2 simplify the i^2-36i/12 simplify completely-6i/2

Page 36: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

i^2=-1 i^2=-1

Page 37: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

The term: Completing the Square The mathematical definition of the

term.When a=1 and b is an even #The meaning of the term in own

words.Completing the square is used when you

are trying to figure out the third number and when you cannot factor easily.

 

Page 38: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Example of the term and something that looks like an example but is not.

Difference: Well for these particular problem there is no

difference you can see the different solutions for the examples.

X²+4x=21

B=4

b/2 = 4/2 = 2

(2)² = 4

(x+2)² =21

(x+2)² = 21+4

(x+2)² = 25

√(x+2)² =√25

X+2 = ±5

X=-2±5

X=-2-5= -7

X=-2+5 = 3

x2 – 4x – 8 = 0.

x2 – 4x – 8 = 0 x2 – 4x = 8

x2 – 4x + 4 = 8 +x2 – 4x + 4 = 12

(x – 2)2 = 12

(x – 2)2 = 12

Page 39: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Skill Description If you cant factor the problem easily and

if a=1 and b is evenExamples

(x+2)² =21 (x+2)² = 21+4

(x+2)² = 25√(x+2)² =√25

X+2 = ±5X=-2±5

Page 40: Intro pg.1-3 Standard form to Vertex form – Jesus pg. 4-11 Solving Quadratics – Markita pg.12-15

Application with word problemsOnce you set your equation you would look for the

signs on whether or no you can solve by completing the square.

Description of the process • Move the # w/o the x to the other side if all of the

equation is all on one side• Find “b”• Divide “b” by 2• Square root of (b/2)• Square whatever is in the parentheses• Put the square root and the # on the other side• Take the square root of both sides to solve for x