Quadratics

By: Sharan Johel

Key features of Quadratic relations:

Maximum/minimum value: optimal value (k value). Labelled as: *y=k*


Optimal value (min/max value): The highest/lowest value the graph reaches. It is the highest/lowest point on the y-axis. This is the y –value of the vertex (labelled as y=)



Axis of symmetry: divides the parabola in half (h value). Labelled as: *x=h*


Y-intercept: When x=o and parabola crosses y-axis. Labelled as: *(0,y)*


X-intercepts: when y=o the parabola crosses the x-axis. Labelled as: *(x,0)*


Zeros: the zeroes are the same as x-intercepts or roots. Labelled as: *(x,0)*

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Table of contents:

There was lots of topics covered in the unit Quadratics. Here is a list of all topics which we have went through:

Introduction to quadratics:

1) Analyzing quadratics (second differences)


2) All about parabolas (intro) ex: all the components to it


3) Transformations


Types of equations:


- Standard form y= (ax^2+bx+c)

- Factored form y=a(x-r)(x-s)

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


4) Vertex form (the properties)


- graphing from vertex form

- finding an equation when only given the vertex



5) Expanding binomials and multiplying binomials


6) Factored form:

- graphing from factored form

- common factoring

-factoring simple trinomials

-complex trinomial factoring

- binomial common factoring

- factoring by grouping



Factoring special quadratics:

- perfect squares

-difference of squares


7) Standard form:


- completing the square (moving from standard form to vertex form)

- solving quadratics using the quadratic formula

- discriminant

- graphing from standard form


And lastly learned lots of types of word problems such as expand and simplify problems, factoring etc.

Parabolas

What is a parabola? A parabola is something we graph when given an equation or when trying to find one.


There are many components to a parabola.

- The axis of symmetry (A.O.S) is the midpoint of the two zeros. (h value)

- Vertex (h,k)

- Optimal value is found by subbing the A.O.S value into the equation. (k value)

- X-intercepts (the zeros)

- Y-intercepts - when graph crosses the y-axis. (add picture of diagram)

Everything about parabolas:

- Parabolas can open up or down


- The zero of a parabola is where the graph crosses the x-axis


- The y-intercept of a parabola is where the graph crosses the y-axis.


- The Zeroes can also be called the x-intercepts or the roots.


- The vertex is the point where the axis of symmetry and the parabola meet. (its the point where the parabola is at its maximum or minimum value.


- The axis of symmetry (A.O.S) divides the parabola into 2 equal halves.


- The optimal value is the value of the y coordinate of the vertex

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Forms of quadratic relations

First form is standard form: y=ax^2+bx+c


What do we know from this equation?

- y-intercept = c value

- direction of opening ( a value)


Example: y=x^2+7x+12 ----> For this equation the y-intercept =12 and a value is 1 which means the parabola opens up.



Second form is factored form/intercept form; y=a(x-r)(x-s)


What do we know from this equation?

- the zeros/x-intercepts (r and s)

- direction of opening (a value)


Example: y=-2(x-1)(x+3) -------> for this equation we know the x-intercepts are 1 and -3 which are opposite of what is inside the brackets, and a=-2 (parabola opens down)


Third form is vertex form; y=a(x-h)^2+k


What do we know from this equation?

- The vertex (H,K)

- max/min optimal value=k

- The axis of symmetry is x=h

- direction of opening (a value)


Example: y=3(x-2)^2+5 --------->for this equation the vertex is (2,5) NOTE: the x value is always opposite of what is inside the bracket.

- a=3 (parabola opens up)

- min=5

- A.O.S is x=2

Identifying quadratics

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When analyzing quadratics, you should always remember that it will ALWAYS have equal second differences but not first differences. Only linear relations have equal first differences.
3.1 Analyzing Quadratics

Identifying the step pattern:

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Graphing from vertex form and transformations

Equation: Y=a(x-h)(2)+k


a= - direction of opening(up or down)

- stretch or compression

- step pattern

h= - horizontal translation ( left or right, if h is negative its right, and if positive its left)


k= - vertical translation (up or down, when k is negative it moves down and if positive it moves up)


Vertex: (h,k)

Max/min (optimal value): y=k

Axis of Symmetry: x=h



Example: Describe the transformations that occurred to y=x(2) in order to become y=-2(x-3)(2)+6.


a= is -2 (vertical stretch by a factor of 2, opens down (reflection in the x-axis)


h= 3 - horizontal translation 3 units right.


k=6 - vertical translation 6 units up.

Example 1:

y=x^2+3


For this example we know that a=1, h=0, k=3

Vertex: (0,3)

The step pattern we would use is 1,3,5, we use the normal step pattern because there is no a value in the equation.

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Example 2: (graphing from vertex)

y=2(x+3)^2+6


Step 1: First plot the vertex at (0,0)


Step 2: Move the vertex 3 units to the left (h=3)


Step 3: Move the vertex 6 units up (k=6) *now we have found our vertex (-3,6)


Step 4: Graph all points, remember we have an a value of 2, so the step pattern is going to change up a bit. Taking the a value, we now multiply it by 2.


Normal step pattern: 1,3,5


New step pattern: 1x2, 3x2, 5x2


= 2, 6, 10 .. now you can plot it and you have a parabola that should open up!

3.2 Graphing from Vertex Form
3.3 More Graphing from Vertex Form

Finding the equation given the vertex (basics)

Example: Write an equation for the parabola with a vertex (3,5) opening upward, with no vertical stretch or compression.


The equation would be : y=(x-3)2+5 *always remember to put the opposite h value, for example the vertex for this is (3,5 but when subbing the vertex into the equation, don't forget to put the positive 3 into a negative 3.*

Finding equations in vertex form

Example: find the equation for the parabola with the vertex (3,-1) that passes through the point ( 1,7)


First identify the vertex which is (3,-1) and the point is (1,7)

Having that information given, you should sub it into the equation.


y=a(x-h)(2)+k

7=a(1-3)(2)-1

7=a(-2)(2)-1

7=a(4)-1

7=4a-1

7+1= 4a

4a/4 = 8/4

a=2


Therefore the vertex is (3,-1) and a=2

Equation is: y=2(x-3)(2)-1

3.4 Finding the Equation given Vertex
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Introduction to factored form

For factored form we will be going through all of the key concepts. We are going to identify the zeroes/x-intercepts (r and s) , the axis of symmetry (x=(r+s)/2) , optimal value (subbing in) , how to graph in factored form etc. The equation we will be using is y=a(x-r)(x-s).
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Graphing from factored form

A general quadratic function in factored form is: y=a(x-r)(x-s)


Note that r and s equal to the zeroes or x-intercepts.

- Axis of symmetry = r+s/2 and x=h.


a= direction of opening, stretch or compression, step pattern

Example 1

when y=-3(x+7)(x+9), find the vertex,axis of symmetry and optimal value.


First you would need to find the zeroes:


0=-3(x+7)(x+3)

x+7 x+3

x=-7 x=-3 *The zeroes are -7 and -3 because -7+7=0 and -3+3=0.*


Now that you found the zeroes, you need to find the axis of symmetry.

Formula for A.O.S (x=(r+s)/2)


x= -7+(-3)/2

Therefore x=-5


Now that you found the zeroes/x-intercepts, now sub it into the equation and find the y value. (remember that you need to sub it in to the FIRST equation you were given).


*the zeroes are always opposite of what they actually were, when you sub it in the equation*


y=-3(-5+7)(-5+3)

y=-3(2)(-2)

y=12


There we go! Now you know how to find the vertex, axis of symmetry, and optimal value. But... that's not it, we now need to make a graph using all the information we have.


(Our 2 zeroes are x=-7 and x= -3, our x value is -5 and y=12)

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3.5 Graphing from Factored Form
Below is a video I made explaining how to graph in factored form and hopefully it will help you expand your knowledge on it. :)
How to graph in factored form
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Common Factoring (adding brackets, opposite of expanding)

Factors are the numbers you multiply together to get another number.

When you find the factors of two or more numbers, and then find some factors are the same “common”, then they are the "common factors".

Example: 12 and 30
• The factors of 12 are: 1, 2, 3, 4, 6 and 12
• The factors of 30 are: 1, 2, 3, 5, 6, 10, 15 and 30
So the common factors of 12 and 30 are: 1, 2, 3 and 6

Example 1: 8x+6

Step #1: Find GCF (Greatest common factor, find what the greatest number in common between 8 and 6).


GCF=2


Step #2: *remember to write solution WITH brackets.*

8x/2 + 6/2 (dividing both the numbers with the GCF)


Therefore answer is: 2(4x+3) because 8x/2 = 4x and 6/2 = 3


Always put the GCF in front of the bracket.

Example 2: 14m+21n

Step #1: Find GCF.. For this example the GCF =7


Step #2: Divide both numbers with the GCF. ( 14m/7 + 21n/7 )


Therefore: 14m/7 is 2m and 21n/7 is 3n.


Answer: 7(2m+3n)



Below is a picture of factors for a number, and the GCF.

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3.7 Common Factoring

Factoring simple trinomials

Many polynomials such as x^2+7x+12 can be written as the product of two binomials of the form (x+r)(x+s)


EXPANDING:

x^2+7x+12 = (x+3)(x+4)


When factoring a polynomial of the form ax^2+bx+c (when a=1) we want to find; 2 numbers that have to ADD to give b and 2 numbers that MULTIPLY to give c.


Example: Factor x^2+6x+5


In order to factor this, we need to figure out ____x_____=5 and _____+_____=6


Now we know that 5x1= 6 and 5+1=6 so we have the numbers 5,1 so that would be our 2 digits we are going to use. Remember there are also many other numbers that can give us 6, like 3+3, but we know that isn't going to work because 3x3 does not equal to 5.


=(x+5)(x+1) *the order you put the numbers doesn't matter*





BELOW is a video explaining how to factor simple trinomials.

sharan

Example 2: n^2-13n+26

First, for this example we need to see _____x_____=26 and ______+_____=-13


Knowing right off the bat, you should be able to identify that this example is NOT POSSIBLE.

3.8 Factoring Simple Trinomials

More simple trinomial factoring

Factor: 3x^2+15x+18


First you need to figure out the GCF of 15 and 18 which is 3.

Now we need to make the equation into a simple trinomial.


=3(x^2+5x+6) -----> now that we have made this into a simple trinomial we need to factor.


* (3)(x^2) = 3^2 *

* (3)(5x) = 15x *

* (3)(6) = 18 * Therefore we have common factored correctly.



____x_____=6 (3x2=6)

____+_____=5 (3+2=5)


Therefore the answer is: 3(x+2)(x+3)

Communication question

Option 1: Determine 2 values of k, so that the trinomial can be factored.


x^2+kx+24


- Some possible values can be 6 and 4 because, since k represents the sum we know that 6+4=10 so k=10.


Another option: k=12

The factors would be 12 and 2 , and since k represents the sum, we know that 12+2=12 so k can be 12.

Factoring complex trinomials

Complex trinomials have a coefficient other then 1 in front of the x^2 term.


ax^2+bx+c (where a is NOT equal to 1)


Do you think you can factor the following?

3x^2+17x+10 ----> first you need to ask yourself if you can common factor, for this example you cannot.

Example 1: Binomial common factoring

Factor: 3x(z-3) +2(z-3)

---> *A binomial can be the common factor*

Answer: = (z-3)(3x+2)

Example 2: factor by grouping

Factor: df+ef+dg+eg

- There is NO common factor in all terms

- We can group terms together that have a common factor


---> f(d+e)+g(d+e)

Remember that the terms in both the brackets should ALWAYS be the same.


Therefore the answer would be =(d+e)(f+g)

Factoring complex trinomials: Decomposition method

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3.9 Complex Trinomial Factoring

Expanding Binomials

What is a binomial? A binomial is a polynomial with 2 terms: 5y^3+5


Example: (x+1)(x+2)


First you need to do (x)(x), then do (x)(2), after (1)(x) and finally (1)(2).


=x^2+2x+x+2

*collect like terms*

=x^2+3x+2

3.6 Expanding
Below is a video I made explaining how to expand binomials :)
Expanding binomials

Multiplying Binomials

When multiplying binomials always remember F.O.I.L (First, Outside, Inside,Last)


Example 1: (h+9)(h-5)

=h^2 - 5h + 9h - 45

=h^2 +4h - 45 ----> trinomial

Example 2:

= (n-4)(n+6)

=n^2 + 6n - 4n -24 *remember to always collect like terms*

=n^2 + 2n - 24

Special factoring (perfect squares, difference of squares)

Factoring Special Quadratics

What is the pattern?

1,4,9,16,25,36,49,64,81,100,121,144 ....


Pattern is: 1^2, 2^2, 3^2 , 4^2, 5^2 .... (perfect squares)


Example: Factor.

y^2 - 25 -----> (y)^2 - (5)^2


Remember: formula is a^2 - b^2 =(a+b)(a-b)

= (y+5)(y-5)

CHECK YOUR ANSWER: y^2 - 5y + 5y - 25

* the -5y and 5y cancel out*


Therefore answer is correct.

Example 2

9q^2 - 4r^2 -----> (3q)^2 - (2r)^2


=(3q+2r)(3q-2r)

Example 3

12k^2 - 27c^2


*Remember to always ask if you can common factor, for this example it is a YES*


=3(4k^2 - 9c^2) - difference of squares

=3(2k^2 - 3c^2) - simplify

=3(2k+3c)(2k-3c) - ANSWER

Perfect square trinomials

Example: factor the following

*expand and simplify*


(x+2)^2 -----> (x+2)(x+2)

= x^2 + 2x + 2x + 4

= x^2 + 4x +4


example 2: (x+3)^2 ---> (x+3)(x+3)

= x^2+3x+3x+9

=x^2+6x+9

* THESE ARE PERFECT SQUARE TRINOMIALS*

; a^2+2ab+b^2 = (a+b)^2

; a^2 - 2ab + b^2 = (a-b)^2


EXAMPLE: y^2 - 14y+49


__-7___x_-7__=49

__-7___+__-7__=-14


=(y-7)^2


EXAMPLE 2:

16m^2 + 24m + 9 --> 16 and 9 are perfect squares!


First, you need to find the square root of 16m^2 which is 4m, then you need to find the square root if 9 which is 3. Now to see if those two numbers are correct you need to do 2(4)(3) which should be equaled to 24.


NOW you have found your a value and b value.

*a=4m and b=3

Therefore answer is =(4m+3)^2

3.10 Special factoring

Solving quadratic equations by factoring

When solving an equation that requires factoring, ONE SIDE MUST ALWAYS = ZERO.

y=ax^2+bx+c ---> standard form (set y=o)


Example 1: x^2 - 3x - 28=0 (simple trinomial)


_-7___x__4__=-28

_-7___+_4__=-3 * to solve for x you must let each bracket =0*


=(x-7)(x+4) =0

x-7=0 x+4=0

x=7 x=-4

(These are called the ROOTS of the equation)



Example 2: m^2-8m=-15 ------> m^2-8m+15=0

(m-5) (m-3)=0

m-5=0 m-3=0

m=5 m=3

The video below basically summarizes the whole factoring topics.
3.11 Factoring

Standard form: Graphing

Standard form: y=ax^2+bx+c


To put this quadratic relation in factored form you need to FACTOR!


Reminder: Factored form is y=a(x-r)(x-s)

- the roots(x-intercepts) are the values of r and s.


Example 1: y=x^2+8x+15

STEP #1: Factor to get factored form!


So y=x^2+8x+15 will now be y=(x+5)(x+3) because... __5__x__3__=15 and __5__+__3___=8


STEP #2: Find the roots


(x+5)(x+3)

x+5=0 x+3=0

x=-5 x=-3 ----> the roots (x-intercepts)


STEP #3: Find the A.O.S (Axis of symmetry)


x= -5+(-3)/2

x=-4 ----> this is the x-value of the vertex (h value)


STEP #4: Find the optimal value

Sub x=-4 into the ORIGINAL equation ( y=x^2+8x+15


y=x^2+8x+15

y=(-4)^2+8(-4)+15

y=16-32+15

y=-1 ---> this is the y-value of the vertex (k value)


Therefore: the vertex is (-4,-1), the x-intercepts are (-5,0) and (-3,0)



EXAMPLE #2:

*Always see if you can common factor first, for this example you can*


y= -x^2 -2x+8

y= -1(x^2+2x-8) --> simple trinomial

__4___x__-2__= -8

__4___+__-2__=2

y= -(x+4)(x-2) - factored form

NOW set y=0 to find x-intercepts


-(x+4)(x-2)=0

x+4=0 x-2=0

x=-4 x=2 ---> roots


NOW find A.O.S

x=-4+2/2

x=-1 ---> x-value of the vertex


NOW find the optimal value

sub x=-1 into y=-x^2-2x+8

y=-(-1)^2-2(-1)+8

y=-1+2+8

y=9 ---> y-value


Therefore: the vertex is (-1,9), the y-int is (0,8) and the x-intercepts are (-4,0) and (2,0)

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3 15 Quadratic Formula
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Completing the square

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Discriminant

What is a discriminant? A discriminant can help determine the number of solutions a quadratic equation has.


REMEMBER: If b^2 -4ac > 0 then the quadratic has 2 real roots.


If b^2 -4ac = 0 then the quadratic has 1 real root.


If b^2 -4ac < 0 then the quadratic has no real roots.


Below is a video which thoroughly explains all the key concepts of it.

Discriminant of Quadratic Equations

WORD PROBLEMS

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3.12 Motion problems
3.13 Quadratics Economics

Some connections made between a topic:

Graphing and equations:


You can graph equations either in standard, factored or vertex form. From each of these equations you can easily find out the axis of symmetry(x value) , optimal value (y value) , and zeroes(x-intercepts). When finding all those things you will then begin to make your graph. When given an a value in the equation you can simply identify if the parabola is vertically compressed or stretched.

Reflection of the unit!

Quadratics for me was really easy at first but then started getting somewhat complicated. What I found easy was how to graph the parabola and figure out how to find the vertex. After we started learning lots of other new concepts, it just got to the point where I was really confused, and on top of that we started learning lots of new terminology, which was very new to me. I felt like I didn't do really well in the quadratics quiz #1 because I was really confused on the knowledge and thinking section. I felt like I needed to stop making small errors, sometimes I made those tiny mistakes here and there. I know I needed to get help in order to succeed during the test that was coming up. I think what brought my mark down was that it probably wasn't making sense to me or I just needed more practice. I was really sad with my mark, but I knew I shouldn't be sad for long because there was many more quizzes, tests, assignments etc, coming up.

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Quiz #2 reflection

I was now done with my first test for quadratics, and now it was time to write my Quiz #2. For this section of quadratics it was all about factoring. Simple trinomial factoring, complex trinomial factoring, perfect squares, difference of squares, common factoring, how to find the roots of an equation etc. I felt like I did really well on this quiz because I find factoring really simple. I sometimes forget which method to use because there’s just so much to remember. I was really scared to write this quiz because I knew if I did bad, then I needed to improve as soon as possible, and practice really hard for the upcoming test. Once I saw my mark, I was the happiest person you could ever imagine. I knew even though I did well, doesn't mean I shouldn't study for the test. So far I think I am doing somewhat well in the Quadratics unit, I just find that there’s so much to learn and just so little time. I think I really need to improve on my application skills, because I struggle a lot with word problems. I know I can improve on some sections and I will continue to try my hardest. All in all, I think if I don’t try I won’t succeed, and that I should come ask for more help when needed. I know I will continue working hard for the upcoming lessons.

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