Quadratics:

Unit 1,Unit 2, Unit 3

Vertex Form

Learning Goals For Unit

1: I will be able to graph vertex form equations using the mapping notation.

2: I am able to determine if a relation is Linear, Quadratic, or neither using finite differences.

3. How the values of a,h, and k affect the shape of the parabola.

Summary of Unit

Finite differences:

First we learned about using finite differences to find out if an relation is linear, quadratic, or neither.

How?

Parabola

Parabola is a U shaped line that consists of multiple parts. These parts include the X- Intercept(s), the Y-Intercept, The axis of symmetry, The vertex, and the direction of opening

Mapping Notation

Mapping Notation is a strategy used to accurately graph any quadratic relation by finding the x and y values of each and every point.


Remember to substitute the variables into the mapping notation expression.

(x+h), (ay-k)


Also you must change h if it is a positive it must become negative and if it is negative it must become positive.


For example:

y=2(x-3)^2-4


In order to find the points of the parabola, you must substitute the variables into the expression with each of the corresponding numbers in the vertex form equation.

(x+h)=(x+3), (ay+k)=(2y-4)

Using this Information We Can Find The Coordinates

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Create y=x^2 T chart
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Using the y=x^2 chart, plug in and solve for the values of x and y with (x+h) and (ay+k)
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Calculate and plot points.

Transformations

The vertex form equation can tell us a lot about how the parabola will look like.

Looking at y= a(x-h)^2+k, we can analyze a few things.

a:

If the variable 'a' is a negative the parabola is opening downwards, however if the variable "a' is positive, the parabola is opening upwards.

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(equation: y=x^2)
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(equation: y=-x^2)
The variable 'a' vertically stretches/compresses the graph.


  • If a is -1 < a < 1, then the graph is compressed
  • if a is -1 > a > 1, then the graph is stretched
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(equation: y=0.2x^2)
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(equation: y=2x^1)

h:

h is the x-coordinate of the vertex. It horizontally translates/shifts the graph left or right.

  • If the value of h is negative the x-coordinate shifts right
  • If the value of h is positive the x-coordinate shifts left

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[equation: y=2(x-3)^2]
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[equation: y=2(x+3)^2]

k:

  • If K is negative, then the graph of y=x^2 vertically translates the graph down.
  • If K is positive, then the graph of y=x^2 vertically translates the graph up
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(equation: y=x^2+2)
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(equation: y=x^2-2)

Graph

The vertex form equation of the following parabola is

y=2(x-1)^2+4

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Therefore, h is the x-coordinate, k is the y-coordinate, and a determines if it will be opening down or up as well as determining how vertically compressed and stretched it is.

Word Problem

At a baseball game, a fan throws a baseball from the stadium back onto the field. The height in meters of a ball x seconds after being thrown is modeled by the function y=-4.9(x-2)^2+45. What is the maximum height of the ball? When did the maximum height occur? What is the height of the ball after 1 second? What is the initial height of the ball? How long is the ball in the air?
Looking at the relation, we know that the maximum height (k) is 45 meters y=-4.9(x-2)^2+45.
Also, from analyzing the relation we can tell that the maximum height occurred after 2 seconds y=-4.9(x-2)^2+45.
In order to find the height of the ball after 1 second, you must set the value of x to the number of seconds. In this case it is 1, therefore the relation is y=-4.9(1-2)^2+45.

Now we solve.

y=-4.9(1-2)^2+45

y=-4.9(-1)^2+45

y=-4.9(1)+45

y=-4.9+45

y=40.1


Therefore, the height of the ball after 1 second is 40.1 meters

To find the initial height of the ball, you must set the value of x to 0.

Now we solve.

y=-4.9(0-2)^2+45

y=-4.9(-2)^2+45

y=-4.9(4)+45


y=-19.6+45


y=25.4


Therefore, the initial height of the ball is 25.4 meters

Finding the time of the ball in the air is similar to finding the initial height, but instead of setting x to 0, you set y to 0.

Now we solve.

0=-4.9(x-2)^2+45

-45=-4.9(x-2)^2

-45/-4.9= -4.9(x-2)^2/-4.9 We divide by -4.9 to get rid of it

√9.18= √(x-2)^2 We square root it to get rid of the exponent

3.03=x-2

3.03+2=x

5.03= x


Therefore, the ball was in the air for 5.03 seconds

Vertex form equations

Factored Form

Learning Goals For Unit






1. I will learn how multiply polynomials to get a simplified factor formed equation.

2. I will learn how to factor simple and complex trinominals.

3. I will learn how to factor by grouping.

Summary of Unit

Understanding the Factored Form Equation

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

The value of a is gives the shape and direction of opening for the parabola.

The value of x and s gives you the x-intercepts

Section 5.1: Multiplying polynomials

1. Find the product of two binomials by multiplying each term in one binomial by each term in the other binomial.

2. Simplify by collecting like terms

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

For Example:

(X+2)(X+3)

=2x^2+3x+4x+6

=2x^2+7x+6

Section 5.2: Special products Squaring Binomials

1. Square the first term

2. Twice the product of the terms/ add two middle terms after expansion.

3. Square the last term.

(a + b)^2

= a^2 + 2ab + b^2

or

(a - b)^2

= a^2 - 2ab + b^2


For Example:

(x+4)^2

=x^2+8x+16

Product of a Sum and a Difference

When you multiply the sum and difference of two terms, the two middle terms are opposite, so they add to zero.

(a - b)(a + b)

= a ^2– b^2

For Example:

(x-4)(x+4)

=x^2+16

Section 3: Common Factoring

Monomial Common Factoring:

1. Find GCF of coefficients and variables.

2. Divide each term by GCF.

Example:

5c-10d

5c/5-10d/5

5(c+2d)


Binomial Common Factor:

1. If there are two binomials that are exactly same, consider that as binomial common factor.

Example:

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

(x+2)(x-2)


Factor by grouping:

1. Factor groups of two terms with a common factor to produce a binomial common factor.

Example:

9x^2 + 15x + 3x + 5

3x

3x(3x+5)1(3x+5)

(3x+1)(3x+5)

Section 4: Simple Trinomial Factoring

Simple Trinomial: ax^2+ bx + c

A simple trinomial is a quadratic where a = 1

1. Find the Product and Sum


  • Two numbers that multiply to get C but add to get b.
For Example:


x^2+10x+25

5+5=10

5x5=25

Therefore the product and sum in 5 and 5.

Answer:

(x+5)(x+5)


2.Look at the signs of b and c to determine the factored form.

x^2 + bx + c

(x-r)(x-s)


  • If b and c are positive, both r and s are positive.

For Example: x^2+8x+16 = (x+4)(x+4)



  • If b is negative and c is positive, both r and s are negative.

For Example: x^2-16x+64 = (x-8)(x-8)

  • If c is negative, one of r or s is negative.

For Example: x^2+10x-24 = (x+12)(x-2)

  • If b and c both are negative, one of r or s is negative.

For Example: x^2-5x-23 = (x-8)(x+3)

Section 5: Complex Trinomial Factoring

Factoring ax^2+bx+c When a is not equal to 1.

1. Multiply a by c.

2. Now decompose b so that it adds up to the value of b but multiplies to the product of ac.

3. Now factor by grouping to get answer.


Example: 3x^2+14x+8

3x8=24

3x^2+12x+2x+8


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

3x


2x+8 =2(x+2)

2


3x(x+2)2(x+2)


(3x+2)(x+2)

How to Solve Complex Trinomial Factoring questions By Khan Academy on Youtube

Factoring trinomials with a non-1 leading coefficient by grouping

Section 6: Differences of Squares

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

How do you tell if differences of squares: Two terms are being subtracted and both of them are perfect squares.


Consider the Following: x^2-49

1. Square root both values

a=(x)^2 b=7

2.Sub a and b into a factor form. y=a(a-b)(a+b)

y=(x-7)(x+7)

When that becomes expanded it becomes x^2-49.

Section 6 Cont: Perfect Squares

In a perfect square trinomial, the first and last terms are perfect squares, and the middle term is twice the product of the square roots of the first and last terms.


Consider the Following:

Verify that x^2+6x+9 is a perfect square trinominal.

1. Square root a and c


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

2. Check the middle term. It should be twice the products of the square roots of the first and last terms.

2(x)(3)= 12x



Now we know that x^2+6x+9 is a perfect square trinominal.


Next, the square rooted a and c go into the factor form equation to solve,

Therefore the answer is (x+3)^2

Section 6.2 and 6.3: Solving Quadratic Equations in Factored Form

Finding X-Intercepts


With a factored form equation, make y equal to 0.

Consider the following:

y=(x-4)(x-2)

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


The factored terms inside the bracket should equal to 0.


x-4=0

x-2=0


Solve for x.


x-4=0

x=4


x-2=0

x=2.


Therefore the x-intercepts of y=(x-4)(x-2) are (4,0),(2,0)


Finding the Vertex of a Factor Form Equation

Consider the Following:

y=x^2-16x+60

y=(x-10)(x-5)


1. Find the x-intercepts

0=(x-10)(x-6)

0=x-10

10=x

0=x-6

6=x

Therefore the x intercepts are (5,0),(9,0).


2. Find the Axis of Symmetry

x1+x2 = AOS

2

6+10 = AOS

2

16 = AOS

2

8 = AOS


3. Now sub AOS into x

y=(8)^2-16(8)+60

y=64-128+60

y=-4

Therefore the Y-int is (0,-4)

The vertex is (AOS,Y-int)

Therefore the Vertex is (8,-4)

Word Problem

The height of a ball thrown from the top of a ladder can be approximated by the formula h=-2t^2+4t+48. where t is the time in seconds and h is the height in meters.


a) Write the formula in factored form

h=-2t^2+4t+48

-2

h=t^2-2t-24

(t+4)(t-6)


b) When will the ball hit the ground?

0=(t+4)(t-6)

0=t+4 0=t-6

-4=t 6=t

Therefore the x-int are (-4,0),(6,0)

Therefore the ball will hit the ground at (6,0)


c) What is the maximum height the ball will reach in meters?

6-4 = AOS

2

2 = AOS

2

1=AOS

h=-2(1)^2+4(1)+48

=-2+4+48

=50

Therefore the maximum height is 50m

Standard Form

Learning Goals:

1.I will be able to find the X intercepts using the Quadratic Formula

2.I am able to complete the square.

3.I can solve revenue, rectangle and triangle problems.

Summary

Completing the Square:

Squaring a Binomial = Perfect Square Trinomial

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


Example:

(x+3)^2 = x^2+3x+9


How to make it a perfect square, if it's not given:

Consider the following:y=x^2+ 8x+5

1. Put '5' outside of the bracket

y=(x^2+ 8x)+5


2. Divide 'b' by 2 then square it. In this case it's 8.

(8/2)^2 = 16

Then the equation becomes y=(x^2+8x+16-16)+5


3.Move the negative 16 outside the bracket.

y=(x^2+8x+16)-16+5


4.Write as squaring a binomial

y=(x+4)^2-11


Therefore the vertex is (-4,-11)

The Quadratic Formula:

Consider the Following:

All quadratic equations of the form ax^2+bx+c=0 can be solved using the quadratic equation.


This is the quadratic equation:

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The quadratic formula is used to find the x-intercepts of a parabola.

The values for the variables a,b, and c are taken from standard form quadratic equation.


For Example:

Use the quadratic formula to solve for the X-Intercepts

3x^2+5x+2=0


We know that a=3, b=5, and c=2

1. Substitute in the values.

2. Add the vaules together in the square root.

3. Square root the result.

4. Add and subtract the outside number with the product of the square root.

5. Then divide the number you get from subtracting and adding with 2a.

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Now from solving the quadratic formula, you can now plot the the x-intercepts.


Here is a video by mahalodotcom on Youtube on how to solve the quadratic formula.

How to Solve Quadratic Equations

The Discriminant:

Inside the the quadratic formula there is the discriminant.


The equation for the discriminant is:

b^2-4ac


If D>0 there are two x-ints

If D<0 there are no x-ints

If D=0 there is one x-int

Word Problem: Completing the square

Ms. Dhaliwal runs a snowboard rental business that charges $12 per snowboard and averages 36 rentals a day. She discovers that for each %0.50 decrease in price, her business rents out two additional snowboards per day. At what price can Ms. Dhaliwal maximize her profit.


R=(price)(quantity)

let 'x' represent # of decreases


p=(12-0.5x)


q=(36+2x)


R=(12-0.5x)(36+2x)

=432+24x-18x-x^2

=-x^2+6x+432

-1

=-(x^2-6x)+432 (6/2)^2=9

=-(x^2-6x+9-9)+432

=-(x^2-6x+9)+432-9(-1)

=-(x^2-6x+9)+432+9

=-(x-3)^2+441


x=3


p=(12-0.5(3))

p=12-1.5

p=10.5


Therefore, Ms. Dhaliwal should change her price to $10.5 because she can maximize her profit to $441

Reflection:

The quadratics unit was a fairly simple and easy unit to understand in my honest opinion. It certainly boosted my average by a significant amount but also thought me the importance of math in our society. I learned that math is in everything that we do, that we can predict where a ball will land and how fast it is going using the vertex form and standard form equations. Since I have completed the unit I have leaned several things about Vertex, Factored, and Standard form equations such as how in the vertex form equation the variables can change the shape and direction of the parabola and how to turn a factored form equation into a Standard form and a Standard form into a Vertex form by completing the square. The word problems were also a good learning experience since it made you read out the question and really made you think about how to solve it. For example the revenue problems made you make your own factored form equation with the information given, then turn it into a vertex form to find the value of x to put into the equation you made to find the maximum revenue.The unit was thought very well by my teacher Ms. Dhaliwal, her using the IPad really helped me to understand how to do the problems and she also helped us when we didn't understand something. She also had a good habit of doing the same type of problems over and over again which helped me do the problems better. She also posted all of our lessons on Edmodo which helped me a lot when it came time to review for the tests. All in all this unit was understandable due to the way Ms. Dhaliwal thought the class which boosted my mark up and was also a great learning experience for me.

Connections:

During the Quadratic unit, I was able to make many connections that can be used throughout the unit that we used from past units. An example of a connection that I made was how you could change a Factored form equation into a standard form then complete the square to turn the standard form equation into a vertex form equation. This allowed us to plot the vertex of a parabola on a graph using the value of x and the maximum and minimum value which is an alternative to finding it through the AOS method. This shows how other parts of the unit are interconnected to help solve these types of problems.

Assessments: Quadratic Standard Form Test

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