Quadratic Relationships

Design to teaches students about the Quadratic Relationships

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Contents

1. Introduction
  • What's a Parabola?
  • What can you figure out by using Parabola?
  • Types of Quadratic Equation

2. Terminology

3. Diagram of the terms

4. Vertex Form

  • Transformations
  • How to create equations in vertex form?
  • How to graph equations in vertex form?
  • Tips
  • Word Problem
5. Factored Form
  • How to solve equations in factored form?
  • Graphing equations in factored form
  • How to determine an equation from a graph?
  • Factoring Trinomials
  • Graphing Trinomials
  • Word Problem
6. Standard Form
  • Quadratic formula
  • How to solve equations by converting standard form to factored form?
  • How to graph quadratic equations in standard form?

7. Converting equations to

  • Standard form
  • Factored Form
  • Vertex form

8. Perfect square and difference of square

9. Quiz

10. My Reflection

  • Assessment
  • Reflection

11. Other websites to check out!!

Introduction

What's a Parabola?

A Parabola is an u-shaped curve that is created by using vertex, the y-intercepts, x-intercepts and other coordinates on a graph. The parabola also has a mid line called the Axis of symmetry (AOS) and the optimal value (minimum or maximum point of the parabola) that help to identify the vertex. To sketch the parabolas, first we have to figure out these key points. This is done by using the equations.

What can you figure out by using Parabola? Why is it important?

Parabolas are not only used in Math class to solve problems and equations. But, they are using in real life scenarios too. Parabolas can help an Olympic diver to figure out that angle he/she needs to jump from, to reach maximum height and how long it will take to reach the swimming pool. Or help a gamer identify the prefect angel and height it will take the angry bird to hit its target. Parabolas are an essential tool for our day to day life.
Parabolas in real life

Types of Quadratic Equations?

There are 3 types of quadratic equations:
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How to identify a quadratic relationship?

In order to calculate the first differences, you need to subtract the second y value from the first y value. They do this to each the pairs of y values. All the x-values must be in order. If the first difference is constant then its a linear relationship. The first differences also tell you about the rate of change. If the second differences are constant then its a Quadratic relationship. If the 1st and 2nd difference are not constant, then it will have neither relationships.
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Terminology

1. Vertex

Vertex in the top most or the lowest point in a parabola. It is made up of two values, x value and y value. X value is the axis of symmetry and y value is the optimal value.

2. Axis of Symmetry

The axis of symmetry is the x value of the vertex. To identify the axis of symmetry, you have to divide the parabola into two parts. You have to add up the two x-intercept and then divide it by two to get the axis of symmetry. Also you can use the formula -b/2a. This is very beneficial while doing standard form equations.


For example: The x-intercept are (6,0) and (2,0). 6+2=8. 8 divided by 2 equals 4. The axis of symmetry is 4.

3. Optimal Value

It is the maximum or minimum value in a parabola. It is also the y value of the vertex. To identify the optimal value you can simply just sub in the x value of the vertex in the equation.


Example:

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4. X-Intercept

The x- intercept occurs when the parabola touches the x axis. A parabola can have one, two or none x-intercept. Few other names for the x-intercept are zeros, roots and solutions.

5. Y-Intercept

When the parabola touches the y axis it called a y-intercept. The y- intercept is really easy to find. All you need to do is sub in x as 0 (zero).


Example:

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Diagram of the Terms

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Vertex Form

For parabolas, the equation that is generally used is y = a(xh)^2+ k.

Transformations

How does "a" effect the orientation and shape?


  • If a > -1, the parabola opens up
  • If a < 1, the parabola opens down
  • If -1 < a < 1, the parabola is vertically stretched
  • If a >1 or a < -1, the parabola is vertically compressed


How does "k" effect vertex?

  • If k > 0 then the vertex moves up by k units
  • If k < 0 then the vertex moves down by k units


How does "h" effect the movement of the vertex?

  • If h > 0 then the vertex will move to the left side
  • If h < 0 then the vertex will move to the right side

How to create equations in vertex form?

Steps to solve the question:


  1. Plug in the given information in the formula [ y = a(x - h)^2+K]
  2. The x value in the vertex is h value in the formula.
  3. The y value in the vertex is k value in the formula.
  4. To complete the formula you have to solve for a value.
  5. Plug in the intercepts given and replace them with y and x.
  6. Check your work
  7. At the end when you figure out the value of a, rewrite the formula that includes a, h and k values.



*Use the steps to solve the problem below.


Information given:


  • Vertex: (8,2)
  • Goes through the point: (6,-14)
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How to graph equations in vertex form?

Quick Way of Graphing a Quadratic Function in Vertex Form

Tips

  • Remember to always check your work whether if it makes sense or not.
  • Be careful of determining the positive and negative signs.
  • Remember the formula

Word Problem

A badminton player hits the birdie which follows a parabolic path. It is represented as a equation h = -1(t - 3)^2 +11. Where h is the height in meters, and t is time in the air in seconds.


When does the birdie reach its maximum height?


A) 8 seconds

B) 3 seconds

C) -3 seconds

D) -4 seconds


The correct answer is B (3 second). The vertex is the highest point in this graph and 3 seconds will be the t value for the vertex. This is because h value ( where 3 is) is the x coordinates of the vertex.



What is the maximum height reached of the birdie?


A) 11 meters

B) 10 meters

C) 4 meters

D) -4 meters


The correct answer A (11 meter). A easy trick to solve this type of question is that when ever it says maximum or minimum, they are referring to the optimal value and the optimal value is in the vertex. We know that K=Y, in this case the y value is replace with h but it is the same thing. So therefore we look in the formula and see what the k value and its 11 meters.


What was the height of the birdie when its released?


A) 0 meter

B) 5 meters

C) 1 meter

D) 2 meters


The correct answer is D (2 meters). In order to identify the answer first you have to make the t value to 0 (zero). This is because when the birdie is released, there has not been any travel time for the birdie. Then your solve the equation to get the value of h.

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Factored Form

Definition: An algebraic form in which none of the expression can be simpler by putting a common factor because it factored to the max.

Equation: The equation used for factored form is y=a(x-r)(x-s)

How to solve equations in factored form?

X-Intercept/Zeros/Roots


For the factored form, we have sub in 0 as y to find the x-intercepts. When the parabola touches the x axis, it will have 0 y coordinates so therefore you can sub it in as 0 (zero). Once that done then you take both the pairs individually [(x-r) and (x-s)] and make them equal to 0 (zero). Then you solve for the value of x.


*Lets try it

y=3(x-5)(x-10)

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Axis of symmetry


Like I mentioned before, axis of symmetry is the middle x line of a parabola. Also, its the x value in the vertex. To find the axis of symmetry you have add up the x-intercepts and divide by two. [ x = (r+s)/2]


x = (5+10)/2

x = 15/2

x = 7.5


Therefore our axis of symmetry for this equation is 7.5.

Optimal Value


In order to find the optimal value, you must first find the axis of symmetry. Axis of symmetry and optimal value are interlinked. The optimal value is the y value in a vertex. We already know the x value ( axis of symmetry) in the vertex. Now all you need to do is sub in the x value (axis of symmetry) to find the optimal value.


y = 3(x-5)(x-10)

y = 3(7.5-5)(7.5-10)

y = 3(2.5)(-2.5)

y = 3(-6.25)

y = - 18.75


Therefore our optimal value is -18.75

Vertex


The vertex is made up of optimal value and axis of symmetry. The optimal value for this equation is -18.75 and the axis of symmetry is 7.5. Therefore the vertex is (7.5, -18.75)

Graphing equations in Factored form

To graph equations in factored form, you need first get all the values such as x-intercepts, optimal value and axis of symmetry. Once that is done, you have mark the coordinates and draw a parabola on a graph. Remember to put arrows at the end of the parabola. This helps indicate that the parabola is still continuing and its not just ending right there.
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For information on how to do graphing factored form, please watch this video.
3.5 Graphing from Factored Form

How to determine an equation from a graph?

Factored Form (Determining an equation from a graph)

Factoring Trinomials

Common factoring


The first thing you should always do while factoring equations in standard form to factored form is common factoring. You should look to find a similar component that each term can divide into. The common factor is not always a numerical number only, sometimes it can be a variable ( for example x or x^2) with a coefficient. Some equations might not have any common factors then you should leave as it is and do the problem.


Example of Common factor:

1. In order to wirte the common factor, you must first find the number that can be divided by all the numbers.


4x^2+ 8x

4x can be divided by all the numbers


2. Once you found the number by dividing it, then you rewrite the factored equation with the number that you divided being first followed by brackets and the left over equation that can not be divided any further.


4x(x^2+2x)


Example of a non-common factor:

x^2+21x-6

Simple Trinomials
Simple trinomials
Complex Trinomials
Complex trinamiols

Graphing Trinomials

X-Intercept

In order to graph a trinomial equation, u have to first find the x-intercepts. First of all, you have factor the equation ( look above to see how to factor trinomials). Then just like vertex form, you have to make the y value equal to 0 (zero) to find the roots.


Vertex

Also, in factored form, you can find the vertex by solving for axis of symmetry and optimal value. The axis of symmetry is solved by adding the two x-intercept and dividing by two. To find the optimal value, you have plug the axis of symmetry in the factored equation. This will give you the vertex.


Y-Intercept (optional)

If you want to go one step further, then you can solve for the y-intercept. In order to find the y-intercept, you have to let x=0 (zero).

Word Problem

Example:

Graph y = 2x^2 - 24x + 20 using the x-intercept and vertex. If needed, also you the y-intercept


Solution:

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Standard Form

For standard form, the quadratic formula is y= ax^2 + bx + c


How solve

Two ways to solve:


  • Quadratic formula
  • Changing standard form into factored form

Quadratic Formula

What is the Quadratic formula?

The coolest thing about the Quadratic formula is that it will always work. Sometimes, when using factored form, you might not be able to get the answer because it does not work with it. But, in the Quadratic formula, the equation will always works does not matter is the equation can be factored or not.


  • There are 3 coefficients in the Quadratic formula. They are a, b and c, and in the equation they are replaced with numerical numbers to help solve the equation.
  • Most often, Quadratic formula is used when the standard form is unable to be converted into factored form.
  • The equation is Quadratic formula must equal to 0 (zero)
  • Discriminant is the part in the Quadratic formula after the square root sign. It is b^2-4ac.
  • If discriminant is more than 0 then there are 2 x intercepts. If its less then 0 then there are 0 (zero) x-intercept. And if the discriminant is 0 then there is 1 x-intercept.



Fun tool to learn the Quadratic Formula

Quadratic Formula Pop Goes the Weasel
How use the Quadratic Formula?
Using the Quadratic Formula

How solve equations by converting standard form to Factored form?

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How to graph quadratic equations in standard form?

How to Graph a Quadratic Equation

Word Problem

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Converting Equations

Vertex form

Converting from Factored form to vertex form


  • Find axis of symmetry by adding the two x-intercepts and dividing by two.
  • Sub in the axis of symmetry (h value) to get optimal value (k value).
  • Sub in one of the point to get the a value.
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Converting from Standard form to vertex form.

  • Completing the squares
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Standard Form

Converting from vertex form to standard form


  • Expand and simplify


Example:

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

=2(x+3)(x+3)-2

=(2x+6)(x+3)-2

=2x^2+6x+6x+18-2

=2x^2+12x+16


Converting from factored form to standard form


  • Expand and simplify


Example:

y=-3(x+1)(x+5)


=-3(x^2+5x+1x+5)

=-3(x^2+6x+5)

= -3x^2-18x-15

Factored form

Converting from vertex form to factored form


  • Make y equal 0 (zero) and find the x-intercepts.
  • Remember when you square roots a number, it will be positive and negative.
  • Once you found the x-intercepts then place them into a factored form equation with keeping the "a" value as the same and removing the "k" value.


Example:

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Converting from standard form to factored form

  • Factor the equation using common factoring, grouping, difference of squares, perfect square, simple and complex trinomial. ( Go above to see more examples on how to use it)


Example:

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Prefect square and Difference of squares

Difference of squares
Factoring a difference of squares
Perfect Square
Factoring Perfect Square Trinomials - Ex1
Factoring Perfect Square Trinomials - Ex 2

Quiz

Please go on the link below to test yourself on everything you learned on this website. Also when doing equations and problems please remember to read the question properly and understand what its asking for before answering. Remember to check your work and be careful of using appropriate sign ( positive and negative signs)


http://www.cliffsnotes.com/math/algebra/algebra-i/quadratic-equations/quiz-solving-quadratic-equations

My Reflection

Assessment

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Things I did good on:


  • I clearly understood the different types of quadratic equations and I knew how to use it properly.
  • I did well on the word problem questions because I knew what they were asking for.
  • I had a good understanding of factoring equations from standard form. I had no mistakes in that.



Things I can improve on:


  • I did not check my work properly so therefore I lost a lot of marks by putting a positive sign where there should be a negative sign. Next time I should check all my work before handing in the test.
  • I messed up on identifying the difference between difference of square and perfect square. I should revise the lesson about difference of square and perfect square. This will prevent me from making mistakes.
  • I need to completely simplify when asked in a question. I lost few marks because of this.

Reflection

I really liked this unit because I got learn a lot of mathematical equations that I did not know before. Also this unit will be really essential for me in the following years because I will be learning how to use this equations but with hard problems. At the start of this unit, I thought quadratics will be really easy because we were doing grade 9 review stuff. But, as the unit progressed I was proved wrong. The quadratics unit is actually really difficult. There are a lot of equation and methods that I need to memorize. But, the different activities we did in class helped me understand the unit deeply. For example the fireball activity at the start of the unit taught me the basic terminology such as axis of symmetry and optimal value, that I needed to succeed in this unit. There was a lot of homework assigned on a regular basis for this unit, but this help me practice the material I learned in class. And you know how they say, practice makes perfection.


Throughout this unit, I was always able to do well on questions in class and other activities we did in class, but I was never able to get the marks I wanted on test and quizzes. I was not doing completely horrible, but I except way more from myself. I was expecting to get around a 95% at the end of this unit, but right now I only have 87%. I think this is because of the lack of momentum I had on studying. In the beginning part of this unit, I felt this is too easy so I did not study that hard for the test. Also I did check my work properly during the test. I lost around 5 to 8 marks this unit, just because I did not check my work properly. But, now I have learned to check my work and this learning experience will really benefit me in the future.


All in all, for me this was not the best unit in terms of the marks I got, but It was a great learning experience. I learned essential math concepts that are necessary for me in the future and also I learned to check my work over and over again before handing it in. This key things that I have attain from this unit will really help me succeed in the future. I think that the only reason I was able to understand this unit so well is because of my teacher, Ms.Johal. She help me whenever I need it and also told me where I was going wrong. Her support and teaching was the only reason I liked this unit. Thanks Ms. Johal, you are great!!

Other websites to check out!

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