Priya Bhardwaj

-What is a Parabola

- Terminology for a parabola structure

-~Vertex Form~

1. Learning Goals for Vertex Form

2. Summary of the Vertex Form Unit

3. Graphing (mapping notation & step pattern

4. Word Problems

~ Factored Form ~

1. Learning Goals for Factored Form

2. Summary of the Factored Form Unit

3. Factored Form to Standard Form

4. Types of Factoring
5. How to get your x-intercepts and vertex

6. How to find a

7. Word Problems

~ Standard Form ~

1. Learning Goals for Standard Form

2. Summary of the Standard Form Unit

4. Completing the Square

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What is a Parabola?

A parabola is a graph form for a quadratic function. On a graph, it appears to be a symmetrical curve.

How can a parabola appear in real life?

Terminology for a Parabola Structure

~~We will first be learning important terminology of a parabola structure ~~

Vertex

The maximum or minimum point of the parabola. It is the point where the graph changes direction.

How do I Label this?

Use the vertex's coordinates! (x,y)

Optimal Value (Max. / Min. Value)

The vertex's height of the parabola. This is your y coordinate of your vertex!

The Maximum value is the highest point of the parabola while the minimum value is the lowest of the parabola. If the parabola has a max value, it will open downwards. If it has a min value, it will open upwards.

(x,y)

Axis of Symmetry

Where the parabola is divided into half equally. This is your x coordinate of your vertex!

(x,y)

Y - intercept and X- Intercept

Y- intercept - Where the parabola crosses the y - axis

How do I Label it?

Set your x at 0 and plug in the y value

(0,__)

X- intercept - Where the parabola crosses the x axis. A parabola can have one, two or even no x intercepts.

How do I Label it?

Set your y to 0 and plug in the x value

(__,0)

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Learning Goals for Vertex Form

1. I am able to know a parabolas transformations just by looking at the equation

2. I am able to use mapping notation to convert an equation into a graph

3. I am able to solve word problems and show all my work when doing so

Summary for the Vertex Form Unit

• In the vertex form unit, we learn that a symmetrical curve on a graph is a type of quadratic function called a parabola. We will learn how to convert an equation into a graphed parabola and how to describe it's transformation, which is the movement of the 'h' and 'k' values from the parabola's equation.
• We can convert our equation into a parabola by simply using a x and y value chart. There are two methods for graphing an equation, which is the mapping formula and the step pattern. In order to complete this, we must know the basic parabola equation which is y = x^2.

• And lastly, we will learn how to solve word problems and learn how to gather important information from the given equation.

• The vertex form equation is: y = a(x-h)^2 + k. Each variable in this formula describe a certain thing:

What does "a" represent?

1. It determines whether the parabola is vertically stretched or compressed.

• If a is more than -1 and less than 1, it is compressed (wider)

Example: y= 0.5(x-2)^2+4 ~OR~ y=-0.2(x+3)^2+4

• If a is less than -1 and more than 1, the parabola is stretched (narrow)

Example: y= 3(x+4)^2-7 ~OR~ y=-4(x-2)^3

2. It determines the direction of the opening of the parabola

• If a is negative, then the parabola is facing downwards
• If a is positive, the parabola will face upwards

What does "h" represent?

It represents the x coordinate of the vertex. This means that it horizontally translates the parabola left or right.

BUT... it does not translate left or right according to the sign! It translates to the opposite to the sign!

• If h is negative, the parabola will translate to the RIGHT!
• If h is positive, the parabola will translate to the LEFT!

What does "k" represent?

Like "h", "k" represents one thing.

It represents the y coordinate of the vertex. This means that it horizontally translates the parabola up or down.

Unlike "h", it translates according to the sign!

• If "K" is negative, the parabola will translate DOWN
• If "K" is positive, the parabola will translate UP

Our x and y in our equation do not change

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Here is a visual of the vertex form equation:

Now you try! Find the transformations for each equation!

Questions:

1) y = 4 (x+5)^2 + 4

2) y = -3 (x-7)^2 -3

3) y = 0.4 (x-5)^2 +5

4) y = -0.9 (x+4)^2 -1

1) The parabola is stretched and is facing upward. It is translated 5 units to the left and is translated 4 units up

2) The parabola is stretched and is facing downward. It is translated 7 units to the right and is translated 3 units down

3) The parabola is compressed and is facing upward. It is translated 5 units to the right and is translated 5 units up

4) The parabola is compressed and is facing downward. It is translated 4 units to the left and is translated 1 unit down

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Basic Parabola

The equation for the basic parabola is y= x^2. You must know this equation in order to graph an equation. The X and Y chart for a basic parabola is:

Since y= x^2, you will square your x value in order to find your y value

x . y

-2 . 4 -> -2^2 is 4

-1 . 1 -> -1^2 is 1

0 . 0 -> 0^2 is 0

1 . 1 -> 1^2 is 1

-2 . 4 -> -2^2 is 4

As you can see there is a pattern. The first and second values are the same as the last and second last values. This shows that parabolas go back up in a symmetrical curve.

There are two different methods for graphing! Both use the X and Y value chart.

Mapping Formula

The first method for graphing your equation is the mapping formula.

To gather information from your equation to put into a chart, the mapping formula can be used:

(x+h) , (ay-k)

Here, you plug in your h, k and a values from the given equation

For example:

y= -0.5(x+3)^2 + 3

This equation will turn into:

(x - 3) , (-0.5y+3)

*Remember that h will have the opposite sign*

How Do I Turn My Equation into a Chart?

After plugging the equation into the mapping formula, you can make your chart. But in order to do this, you must use the basic quadratic function (y=x^2). One chart will be the basic quadratic equation while the other chart can be the chart of your equation.

After doing so, plug in the mapping formula points into the given chart and solve it. x and y from our mapping notation represents the x value and y value from the basic quadratic equation.

The outcome will be your new chart to graph! Visual steps are provided below.

The video below will guide you step by step on how to graph transformations:
How to Graph Transformations for Parabolas (vertex form)

Step Pattern

The second method for graphing is the step pattern.

Here is a video explaining how to do this method:

To find whether your relation is linear or quadratic, a x and y value chart can be used. This is called using finite differences.

When we use finite differences, we look at our y column of our x and y value chart. We look at the difference between one number to the next.

First Differences

When we look at our differences between the numbers in the y column, it is called first differences. If our first differences are the same, then the relationship is a linear relationship.

Second Differences

However, if a relationship does not have the same first differences, you must move on to what is called second differences. Here, you find the differences of the FIRST DIFFERENCES. If the second differences of the relation are the same, then it is a quadratic relationship.

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Word Problems

Now we will be moving onto word problems! Here, you will learn how to solve them!

Something to Remember!

• When asked about maximum or minimum height, we refer back to the parabola's vertex. We must remember that in any parabola equation, our h and k are the vertex coordinates. h being the x coordinate and k being the y coordinate
• Remember to always write a "therefore" statement after completing any questions for a word problem!

Word Problems typically ask for the following information:

1.What was the maximum height? (meters)

2. At what time did the ball reach the maximum height? (seconds)

3.What was the initial height of the ball?

4.When did the ball hit the ground?

5.What is the height of the ball at a certain time, for example 2.5 seconds

Here is an example of a word problem:

Here is an example of a word problem:

A ball is thrown in the following path in meters:

H= -2.5 (t - 4)^2 + 30. Where t represents time and H represents the height of the ball

Just by looking at the equation, we are able to identify three things:
• The vertex coordinates are (4 , 30)
• The parabola is facing downward, since "a" is negative (-2.5)
• The parabola is stretched, since "a" is more than 1

Here is how the example's parabola will look like:

1. What was the maximum height, in meters? h= -2.5 (t - 4)^2 + 30

Since it is asking for maximum height we find our vertex. This is because our vertex gives us either the maximum or the minimum value of a parabola.

-> Find the vertex. This case our vertex is (4, 30)

-> Use the y coordinate of our vertex -- 30

Therefore, the maximum height of the ball is 30 meters

2. At what time did the ball reach the maximum height? (seconds)

-> Find Vertex

-> Use the x - coordinate of our vertex -- 4

3.What was the initial height of the ball?

Now we set our t to 0 in our equation and solve for h!

h= -2.5 (0 - 4)^2 + 30

h = -2.5 (-4)^2 + 30

h = -2.5(16) + 30

h = -40 + 30

h = 10

Therefore, the initial height of the ball is 10 meters.

4.When did the ball hit the ground?

We set our h to 0 and solve for t!

0 = -2.5 (t - 4)^2 + 30

-30 = -2.5 (t - 4)^2

-30 / -2.5 = (t - 4)^2

12 = (t - 4)^2

Now we must square root to get rid of the "^2"

3.46 = t - 4

3.46 + 4 = t

7.46 = t

Therefore the ball hit the ground in 7.26 seconds.

5.What is the height of the ball at 2.5 seconds

Set your "t" to 2.5, since t represents time.

h = -2.5 (2.5 - 4)^2 + 30

h = -2.5 (-1.5)^2 + 30

h = -2.5 (2.25) + 30

h = -5.625 + 30

h = 24. 375

Therefore, the ball was 24. 375 seconds high at 2.5 seconds.

Finding "a" when given a vertex and a point

All you have to do is plug in the vertex coordinates (h,k) and the points (x,y) into the parabola equation. And solve the equation!

Here's an example:

Determine the equation for a parabola with vertex (2,5) and a y - intercept of -3. The solution is below:

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

-3 = a (0 - 2) ^2 + 5

-3 = a (-2)^2 + 5

-3 = a (4) + 5

-3 -5 = a (4)

-8 / 4 = a / 4

-2 = a

Therefore "a" is -2. Our new equation is y = -2 (x - 2)^2 + 5

~ And that is all you need to know about the vertex form for Parabolas! ~

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Learning Goals for Factored Form

1. I know what type of factoring method to use when given an equation

2. I am able to convert a standard form equation into a factored form equation (vice- versa)

3. I am able to create a parabola based on the information provided

Summary of the Factored Form Unit

Factored form is quite different from the vertex form equation. The equation for a parabola in factored form is:

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

"r" and "s" are the parabola's x- intercepts. "a" determines whether the parabola is facing downwards or upwards and if it is stretched or compressed. Finding the x- intercepts for a parabola in factored form is important. This is because once we find out what they are, we are able to find the axis of symmetry and the optimal value. These two give us the parabolas vertex (we will learn how to solve them later one). And remember that you can get two, one or even no x-intercepts! To find the x- intercepts, we set each of our brackets in the equation to 0 and solve for x. To find our axis of symmetry we add our x-intercepts together and divide them by 2.

In this unit we also learn how to graph a parabola by finding its x-intercept, vertex y - intercept, its shape and direction of opening.

In this unit, we learn that there are several types of factoring methods and sometimes the equations can be special cases. The first step when factoring is to always find a GCF.

We have also learned how to convert our equation from standard form to factored form.

Factored Form to Standard Form

As mentioned before, the factored form equation for a parabola is y = a(x - r) (x - s). First, we will be talking about converting a factored form equation into the standard form equation. This process is called expanding and simplifying. Expanding is multiplying to remove the brackets. In order to convert a factored form equation into a standard form equation we must:

Step 1: Find the product of two binomials by multiplying each term in one binomial to each of the terms in the other binomial. Or in other words, use distributive property!

Step 2: Simplify your equation by collecting like terms.

Here is a visual for this process:

Types of Factoring

Now we will be moving onto factoring. Factoring is finding common multiples to create a bracket. This means we are converting a standard form equation into a factored form (which is the opposite of what we just learned).

There are 6 kinds of factoring. The 6 types are:

1. Monomial Common Factoring (GCF)
2. Binomial Common Factoring (GCF)
3. Factoring by grouping (4 terms)
4. Simple trinomial factoring
5. Complex trinomial factoring
6. Special Product- Difference of Squares
7. Special Product- Perfect Square Trinomial

Note: Finding a GCF (if there is one) is the first thing to do when factoring a polynomial!

1. Monomial Common Factoring

Monomial Factoring consists of finding the GCF (greatest common multiple) for the coefficients and variables. After finding our GCF, we divide each of the terms by it. When writing it, the GCF is outside the bracket while the divided terms are inside the brackets.

2. Binomial Common Factoring

Binomial Common Factoring deals with two binomials which were exactly the same. When we are turning it into a factored form equation, one bracket will consist of the terms in the binomial while the other bracket will consist of the GCFs put together.

3. Factoring by Grouping

Factoring by grouping involves four terms. When doing this, we must find anything common between the first two terms as well as anything in common with the last two terms. If there is not, rearrange the equation!

We factor the groups of two terms by finding the GCF in order to produce a binomial common factor.

4. Simple Trinomial Factoring

A simple trinomial is: ax^2 + bx + c. Where x is a variable and a,b,c are constant and a is not simple. A simple trinomial is a quadratic where a = 1. To convert this equation into a factored form equation we must:

Step 1: Find the Product and Sum

- Find two numbers if multiplied is the c value and if added is the b value

For example:

x^2 + 12x + 27

9 and 3 is a product of c and a sum of b!

9 x 3 = 27

9 + 3 = 12

Next we put plug in these numbers into our factored form equation (x - r) (x - s)

(x + 3) (x + 9)

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Complex Trinomial Factoring

A complex trinomial is similar to a simple trinomial. However, a complex trinomial is when a in the equation : ax^2 + bx + c is NOT equal to 1. In order to solve a complex trinomial, we must:

1. First always look for a GCF

2. Now we must find two numbers where we get a product of a and c multiplied, but when these numbers are added, it is the sum of b.

3. After finding the numbers, we replace our b with the two numbers. (this process is known as decomposition). You should now have 4 terms.

4. Now we factor by grouping!

Watch the video below for better understanding!

How to Factor Complex Trinomials
Something to remember for simple and complex trinomial factoring:

- When b and c are both positive that means r and s are both positive as well

- When b and c are both negative that means out of the r and s, one will be positive and one will be negative.

- When b is negative and c is positive that means both r and s are negative.

- When b is positive and c is negative that means out of the r and s, one will be positive and one will be negative

Special Products - Differences of Squares

the equation is: a^2 - b^2

How do I recognize differences of squares?

It is a binomial which both terms are perfect squares and they are subtracting!

For example: x^2 - 25

In order to factor we just simply find the square root for each of the terms. We then simplify more by having one bracket where they are adding and another bracket where they are subtracting.

For example:

x^2 - 25

x^2 = x

25 = 5^2

(x + 5) (x-5)

Special Products - Perfect Square Trinomial

To factor, we must first verify if it is a perfect square. We square root a and b and check if they are perfect squares. Next we multiply the our squared a and b and multiply that by 2. If this number is the same as the middle number then we have a perfect square! Next we use the first and the last term and create a bracket.

For example:

4x^2 - 16x +16

2x^2 - 2 (2x)(4) + 4^2

(2x - 4)^2

Note: if the middle number is negative, then your b will be negative.

How Do I Know What Factor Method to Use?

Use the flowchart below for help!

How to get your x- intercepts

Step 1: If equation is in standard form, factor it into factored form!

Step 2: Now with our factored form, we set one side as 0. -> (x - r) (x - s) = 0

Step 3: Now we set each of the brackets to 0. -> (x - r) = 0 and (x - s) = 0

Step 4: Now we would solve for x. (we isolate our x). -> x= r and x= s.

Now we set these points as coordinates! (r, 0) and (s, 0)

How to find the vertex

After finding our x-intercepts, we are able to find our vertex! To find our vertex we must find our axis of symmetry and our optimal value. In order to find them, we must:

1. Find the axis of symmetry first. (also known as our x- coordinate of the vertex) Find the middle of your two x-intercepts. In order to do that you add them together and divide them by two. Now we have found our x coordinate for our vertex!

r + s / 2

2. Now we must find our optimal value (y coordinate of the vertex). All we do is plug in our x- coordinate of the vertex into either your standard form equation or factored form equation and solve!

Once you find this information, you are able to solve!

Let's find the vertex for our x-intercept equation above:

How to find "a"

If you were given a diagram of a parabola with different coordinates, you must write an equation. To do so, you must use all information provided. You will be given something similar to the right; where you are given the coordinates to the x- intercept(s) and the vertex. You must find a in order to create an equation. The following is the answer to the diagram on the right:

Finding y- intercept

The y- intercept is when the parabola touches the y axis. In order to find our y intercept, we just simply set our x to 0 and we solve!

Word Problems

The height of a rock is thrown in the following path: h = -5t^2 + 15t + 20, where h represents height, in meters, and t represents time.

a) Determine the x intercepts

b) When does the rock hit the ground?

Note: When finding when the rock hit the ground, the answer is your positive x - intercept.

The answer for this problem is below!

~ And that's all you need to know for Factored Form for Parabolas!~

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Learning Goals for Standard Form

1. I am able to convert a standard form equation into a vertex form equation by completing the square

2. I am able to use the quadratic formula in order to find the parabolas x- intercepts

3. I am able to find how many solutions an equation has by finding its discriminant

Summary of Unit

The equation for standard form is:

ax^2 + bx + c

• a determines whether the parabola is facing upwards or if it is facing downwards. If a is negative it will be facing downwards and if it is positive it will be facing upwards. It also determines whether it is stretched or compressed.
• c represents the y - intercept of the parabola.
• In this unit, we have learned how to convert a standard form equation into a vertex form equation by completing the square, which we will be taking up later on. We also learn that quadratic formula is used to find the x- intercepts of the parabola. We will also learn about discriminant and how we are able to figure out how many solutions there are. If the discriminant is more than 0, there will be 2 solutions. If it is 0 there will be 1 solution and if it is less than 0 there are no solutions.
• When we are graphing, we must graph the x- intercepts, y-intercepts and graph the equation's vertex
• We use all of this information to complete word problems.

Completing the Square: ax^2 + bx + c ---> a(x- h)^2 + k

We use this technique in order to convert an equation from standard form to vertex form. How to do this? We make a perfect square when it is not given:

1. Bracket your a and b value, leaving your c value outside of it. We will apply everything to just a and b.

2. Divide your b value by two.

3. Square your b value. The square number is now your c value. Now we must also subtract this number.

4. Now take out the subtracted number outside of the bracket

5. Add/ Subtract the numbers outside of the bracket

6. Now turn your equation into a perfect square!

(x^2 + 12x) – 5

(x^2 + 12x / 2)- 5

(x^2 + 6x) - 5

6 = 36

(x + 12x + 36 - 36) - 5

(x + 12x + 36) - 36 - 5

(x + 12x + 36) -41

y= (x + 6)^2 -41

When "a" is more than 1

• Bracket the 'a' and 'b' value as you would normally
• Divide 'a' by the bracket values and take it out of the bracket
• Complete the square as you normally would
• When you bring the squared negative number outside of the bracket, multiply 'a' with it
• Turn your equation into a perfect square

The quadratic formula is used to determine the parabolas x- intercepts. The quadratic formula is: x = [ -b ± √ ( b^2 - 4ac) ] ÷ 2a

Steps:

1. The standard form equation ax^2 + bx + c determines the values of a, b and c in our quadratic formula. This means that you must plug in these values to the quadratic formula and solve using bedmas.

2. As you can see, in the quadratic formula there is a + sign on top of a - sign. After square rooting the number, you must branch the the + and - off seperately.

Here is a video below on the quadratic formula!

The discriminant is in our quadratic formula. It is "b2 - 4ac". The discriminant tells us how many real solutions there will be to an equation!

• If (b^2- 4ac) is more than 0, there will be two real solutions
• If (b^2- 4ac) is equal to 0, there will be one real solution
• If (b^2 - 4ac) is less than 0, there will be no real solution

Examples:

Word Problems

Something to remember:

When a question is asking for max/ min height, it is asking for the vertex. This means that you should always use the completing the square method. And when a question is asking for dimensions (rectangle questions), it is asking you to find the x- intercepts, which mean you should use the quadratic formula.

Word Problem Example

A soccer ball is kicked in the following path: h= -5t^2 + 10t + 1, where h represents height in meters and t represents time in seconds.

• What is the maximum height of the ball?
• What was the height of the ball in 2 seconds?
• At what time did the ball reach the maximum?

The answer to this problem is below!

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~ And that's all you need to know for standard form for Parabolas! ~

Reflection

I have learned a variety of things throughout the entire quadratics unit. To start off, I learned what a parabola is and how one must be graphed according to the equation provided. We began with learning the basic terminology for a parabola structure. I learned that there are 3 types of equations for a parabola which are: the vertex form, factored form and standard form.

• We started off with the vertex form equation. The vertex form equation is y = a(x-h)^2 + k. The reason as to why this branch of the unit is called "vertex form" is because the equation gives us the parabolas vertex, which is our h and k value. I learned that a determines the shape of the parabola (whether it is stretched or compressed) as well as the direction of opening. h is the x - coordinate of the vertex and k is the y - coordinate of the parabolas vertex.
• After learning vertex form, we moved on to learning about the factored equation. The equation for it is: y = (x - r) (x - s). This equation tells us where the parabola's x - intercepts are located. During this unit, we learned word problems such as area word problems involving rectangles.
• The last unit we learned was the standard form equation. This unit is my favorite from the three as it is the most straight forward and simplest form. The equation for it is: y = ax^2 + bx + c. "a" determines the parabolas shape and the direction of its opening (similar to vertex form). We learned word problems such as revenue and optimization.

After learning about the different types of equations, I have come to a realization that all of them relate to one another.

• They all have the "a" value in common, and their "a" is never 0, always 1 or more.
• Standard form and vertex form relate to one another. We are able to convert from a standard form equation into vertex form by simply completing the square. We are able to use the vertex form's vertex and the standard form y- intercept to graph the parabola.
• Standard form also relates to factored form. They relate to one another because to convert from factored form into standard form, we would just simply expand and simplify using distributive property. We convert standard to factored by factoring the equation. Also using the quadratic formula for standard form, we are able to find the x-intercepts, which we can plug into the factored form equation.

My favorite branch of the quadratic unit would most definitely be standard form. This is because it was very straightforward and i enjoyed completing the different sections of this unit such as completing the square and using the quadratic formula. As first I struggled a bit as it was a new topic to me. However, with extra practice and knowledge on the previous two branches, this branch began to get easier and easier.

My least favorite branch of the quadratic unit would have to be factored form. This was because there was a lot of information given to us such as the different types of methods and when we must use them. Due to this, I struggled a bit during this part.

Overall, the quadratic unit was interesting and quite enjoyable and I feel as though it helped me grasp more mathematical knowledge as it has helped me understand mathematically related concepts better. It has taught me many important things, some of which I will need to learn in the future.