Quadratic Relationships

Everything you need to know about quadratics.

Quadratics Overview

After learning about quadratics for several weeks, there are many aspects to what this unit was about. The topics involved in quadratics are:


- Parts of a Parabola

- Second Differences

- Quadratic Transformations

- Equations in Vertex Form

- Equations in Standard Form

- Equations in Factored Form

- Multiplying Polynomials

- Common Factoring

- Special Products

- Difference of Squares

- Factoring Perfect Square Trinomials

- Going From Standard Form to Factored Form

- Completing the Square

- The Quadratic Formula

- -b/2a and c-b^2/4a Method

- Partial Factoring


By learning about each of these individual subtopics, we were able to fully grasp the entirety of what quadratics is about. By learning about these subtopics, if a problem was given to us about quadratics, we would be able to take what we know and apply it to solve the problem.

At the end of all these lessons, I will be working through several word problems that test everything you've learned.

Quadratics Basics

Second Differences

The main difference between quadratic relations and linear relations is the direction of the line.


When a linear relation is shown, you know that it is linear because the line on the graph is straight (left graph). When a quadratic relation is shown, you know it is quadratic because the line on the graph is curved (right graph).

However, when a chart of numbers is given and it is up to us to decide whether or not the data shown is quadratic or linear, it is not as clear. If this were to occur, we would use the method of second differences to decide if the relation is quadratic or not.


The second differences method is a simple method used when a table is given and someone wants to determine whether or not the table displays a quadratic relation. To use this method, first find the difference between the numbers in the left side of the chart by subtracting the first number from the second number. Repeat this with several sets of numbers. Then, with the differences just found, find the differences of those numbers by subtracting the first number from the second once again. If the second set of differences comes out to be the same number, the data displays a quadratic relation. If the second differences are not the same, chances are the data is linear, or neither linear or quadratic.

If this concept is still confusing, here's a video.
Linear, Quadratic or Neither using First and Second Differences Tables

Parts of a Parabola

Before quadratics finally begins, everyone should know parts of a parabola. Besides the parabola itself, there are different parts of a parabola that tell us different things.

Vertex: the point on the parabola that depicts the highest/lowest value and at what point. This is written in coordinate form [(x,y)].

Axis of Symmetry: Shows at what point in the graph the parabola will turn back upwards/downwards. This is written with y=...

Optimal Value: The highest or lowest value on the parabola. If the parabola is opening up, it will have a minimum value. If the parabola is opening down, it will have a maximum value. This is written with x=...
Zeros: When a parabola intersects with the x axis, it has x-intercepts known as zeros. A parabola can have none, one, or two zeroes.

Quadratic Transformations

A quadratic relation, when shown on a graph, is represented by a curved line called a parabola. A parabola can be described on a graph by its transformations. The types of transformations that can describe a parabola are what factor it is stretched or compressed by, how many units up/down and right/left it has been translated, if it is opening up or down, and if it has been reflected. When stating translation, know that ever parabola is being compared to a parabola with the equation y=x^2.


First let's begin with finding out what factor a parabola has been stretched/compressed by. The slope of a quadratic equations can tell us what factor a parabola has been affected by.


When looking at a quadratic equation that has a slope between -1 and 0 or 0 and 1, the parabola will be vertically COMPRESSED by a factor whatever the slope happens to be.
For example, if the equation y=1/3x^2 were given, we would say that it is vertically compressed by a factor of 1/3, and this will be proven when graphed.

When looking at a quadratic equation that has a slope between greater than 1 or less than -1, the parabola will be vertically STRETCHED by a factor whatever the slope happens to be.
For example, if the equation y=4x^2 were given, we would say that it is vertically stretched by a factor of 4, and this will be proven when graphed.


Keep in mind that when describing how vertically stretched/compressed by a factor something is, the factor is always positive (even if the slope is negative, make it positive).

The second transformation to talk about is how many units left or right the parabola is. A person can tell how many units left or right a parabola has moved by looking at the value a quadratic equation has within its brackets.

For example, with the equations y=(x+3)^2, the number inside the brackets is 3. When we know the number inside the bracket, we have to multiply it by -1, and you will get how many units the parabola should be shifted. So in this case, the parabola will be shifted -3 units, which is 3 units to the left. Now we can say that this parabola is shifted 3 units to the left in relation to y=x^2.
We can apply these steps to other equations. For example, the equation y=(x-6)^2, the number is -6. When multiplied by -1, -6 becomes 6. Therefore, this parabola is shifted 6 units, which is 6 units to the right. Now we can say that this parabola is shifted 6 units to the right in relation to y=x^2.
Proof of this concept can be seen through graphing.
As for finding out how many units up or down a parabola has been translated, it is found by looking at the number that is being added or subtracted to x^2.

If a number is being added, the parabola has been shirted up. If a number has been subtracted, the parabola has been shifted down.

For example, if we were given the equation y=x^2-5, this would tell us that the parabola has been shifted 5 units down compared to the equation y=x^2. For the equation y=x^2+3, this would tell us that the parabola has been shifted 3 units up in relation to the equations y=x^2. This is proved when graphed.
Next is stating whether or not a parabola is opening upwards or downwards when we are given a quadratic equation. We can tell when a quadratic equation is opening up or down by looking at the slope.

If the slope is positive, the parabola will be opening upwards. If the slope is negative, the parabola will be opening downwards.

For example, the equation y=3x^2 would be opening upwards (because of the positive slope). But for the equations y=-3x^2, the parabola would be opening downwards (because of the negative slope). When graphed, these statements are proven.

The last transformation to talk about is whether or not a quadratic relation has been reflected or not. We know if a quadratic relation has been reflected if the parabola is opening down and has been translated up.
If a quadratic equation fits these two requirements, you must say that the parabola has been reflected along the x-axis.
Now putting this altogether, we can describe any transformations a quadratic relation has gone through when given an equation.

For example, the equation y=-3(x-2)^+2. A description for its transformations would be: In relation to the y=x^2, the parabola is opening downwards, is vertically stretched by a factor of 3, is shifted 2 units right and 2 units up, and has been reflected across the x axis. This is proven in the picture below.

Different Types of Quadratic Forms

Equations in Vertex Form

An equations written in vertex form always follows the formula structure of y=a(x-h)^2+k.


In this equation, a is the slope, h is the x value, and k is the y value. From learning earlier about parabola transformations, we can easily tell where a parabola is located when looking at vertex form (think back on what we learned about slopes (opening up/down, compressed/stretched), the parabola being translated left or right (remember opposites), or being translated up and down).


This equation directly gives people a parabola's vertex thus its name of vertex form.

To prove this, we can graph it.


For example, if given the equations:

1. y=2(x-6)^2-6

2. y=-0.5(x+2)^2+3
For equation one, the vertex would be (6, -6), and for equations two, the vertex would be (-2, 3).

Here is a video of seeing a parabola on a graph when an equation is given in vertex form.
5.1 Graphing Quadratic Equations in Vertex Form

Equations in Standard Form

A quadratic equation given in standard form follows the mathematical form of y=ax^2+bx+c.

We will talk more about this form later, however, you should be able to tell a quadratic equation is in standard form when given an equations in standard form (i.e. y=2a^2+3b+6).

Equations in Factored Form

A quadratic equation given in factored form follows the mathematical form of y=(x-/+ a number)(x-/+ a number). This equation form, unlike vertex form, tell the person the x-intercepts of a parabola.

We can tell what the x-intercepts of a parabola are easily with this form, all we need to do is make to solve for x.
For example, if given the equation y=(x-6)(x+2), all we need to do is isolate for x by making the equation equal 0.
First, you would take (x-6) and make it x-6=0. In doing so, you would bring -6 to the other side to isolate and you will end up with x=6.
Next, you would do the same for the other term, (x+2). After making the equation equal 0 and isolating by bringing the -2 over, you get x=-2.
And so, you end up with -2 and 6, which are your x-intercepts! This is proven when graphed as seen below.
This method works with any set of numbers.
Let me prove this by making the equation more tricky.
For example, the equation y=(2x-8)(3x-24).
You can find the x-intercepts of this equation by isolating once again.
For the first term, you would bring -8 over to the other side and the equations would become 2x=8. Then, you would divide both sides by two so that x can be isolate. In the end, you would get x=4.
For the second term, you would do the same as the first. Bring -24 over and you'll get 3x=24. Divide both sides by 3 and you should end up with x=8.
So, your x-intercepts would be (4,0) and (8,0). Below you can see the equations graphed.
If solving in factored form to find x-intercepts, here's a link that talks more in depth about what factored form is and how it is used to find the x-intercepts of a parabola.

Types of Factoring

Multiplying Polynomials

You multiple polynomials when you are given an equation in factored form. To multiply polynomials, you simply expand and simplify the given equation. You should multiply polynomials when you want an equation in factored form to be in standard form.


For example, let’s do the equation y=(3x+2)(x+4). The first step to multiple the first term in the first bracket by every term in the second bracket. The next step is to multiply the second term in the first bracket by everything in the second bracket.

Once you have finished multiplying everything, you would simplify what you got by adding like terms.

And you're done! Your finished solution, if done correctly, should give you your original equation in standard form. In this case, the equation you should have got after all the steps is y=3x^2+14x+8. Simple, right?

Let me just reiterate the steps:
1. Multiply the first term in the first bracket by everything in the second bracket.

2. Multiply the second term in the first bracket by everything in the second bracket.

3. Collect like terms and solve to get your original equation in standard form.


Multiplying polynomials will be an easy and quick concept once you get enough practice doing it.


If you still need help understanding what multiplying polynomials is used for or how to do it, watch the video or go to the site I've linked below.

Algebra I Help: Multiplying Polynomials

Common Factoring

Common factoring is method used to simplify an equation or expression. This method involves finding like terms between terms and taking it out of the expression/equation.


For example, let’s look at the expression 3x^2y^4+6x^4y^2+3xy^5. A like term between all three terms is 3xy^2. You would take this term and divide each term by 3xy^2.

The finished quotient you get should be 3xy^2(xy^2+2x^3+y^3). This is the simplified form of the equation.
Here are the steps once more:
1. Find a term that is common between all terms in the equation to divide by.
2. Divide every term in the equation by the term that you found.
3. Re-write the statement with the new terms in brackets and your main term as the brackets' coefficient.


If you need more help or want to know more about common factoring, visit the site I've linked below or watch the video I've included.

Common Factoring Tutorial

Special Products

Special products is a concept in which you multiply a phrase by itself. We see this happening in quadratics when dealing with vertex form. You can tell if an expression is a special product if it is written in the form of (a-c)^2. When expanded, the expression would look like (a+c)(a+c).


When given a special product, you would want to use a special method to convert it into standard form rather than using the multiplying polynomials method.


The special way we convert special products directly into standard form is easy. To do so, you should always keep in mind the form y=a^2+2ac+c^2. This form will be the basis of working through any special product.


Let’s use the equation y=(2x-6)^2. The first step is to input the given numbers into the form I just gave you (remember y=(a)^2+2ac+(c)^2).

In the next step, you would whole square the second and third terms and multiple the second term accordingly (this step is included in the special product form from earlier).

After all the multiplying, there you have it! You know have the equation in standard form. This method can be used on any equation as long as it follows the special product form. Notice that I simplified the equation by common factoring.


Here are the steps again:
1. First see if the equation is a special product.
2. Input the numbers in the equation into the special product solving form ((a)^2+2(a)(c)+(c)^2).

3. Solve and you are done!

If you are still confused or would like to practice/know more on special products, visit the link or watch the video I've included below.

Special Products of Binomials

Difference of Squares

The difference of squares method is used when we are given an equation in the form of y=a^2-b^2 number. When given an equation like this, we know the difference of squares method can be used if both terms have whole square roots.


For example, the equation y=x^2+36 can be factored using difference of squares because both terms have whole square roots. For the first term, the square root would be x, and for the second, the square root would be 6. In the end, after using difference of squares, the equation should be y=(x-6)(x+6). The end product for difference of squares should always be in factored form.


However, this method can’t exactly be used for equations like y=5^2+8 because the square roots for the terms aren't whole.


Here are the steps if you can't remember:
1. Determine if the equation can be solved using difference of squares by seeing if the equation's terms both have whole square roots. If it does have whole square roots, continue on.
2. Find the square roots of the terms.
3. Put the newly found terms in factored form format.

Below I've included an extra link and video if you need more help or understanding on difference of squares.

Factoring the Difference of Two Squares - Ex 1

Factoring Perfect Square Trinomials

You know when you should be factoring using the perfect square trinomial method if you are given an equation in standard form and the first and last terms have whole square roots. After factoring using this method, you will get your equation in standard form put into factored form.
factor perfect square trinomial

Solving Quadratic Equations

Factoring From Standard Form

Factoring from standard form allows someone to get change a standard form equation to a factored form equation. This means the equation is changing from y=ax^2+bx+c to y=(x+a number)(x+a number).


To use this method, there are certain steps. When given an equation in standard form, for example, y=3x^2+5x-2, you must divide the b value into two separate values. The values that you split it up into must add up to the product of a and c multiplied together (Note: the numbers found must be factors of the product of a and c). In this case, a(3) and c(-2) multiple together is -6. You will use the -6 to find b. After working through the possibilities, you will find that 6 and -1 are the numbers you must split 5 into. 6 and -1 are the right numbers because added together they equal 5, and when multiplied, equal -6.

Your equation at this point should look like y=3x^2+6x-x-2. You will know have 4 terms in your equation. The next step is to common factor the first two terms together and the last two terms together. The common factor between the first set of terms is 3x and the common factor between the second set of terms is -1.
If the statements in brackets match, then you know you have factored correctly. The last step is to simplify. In this method, you would simplify by putting outer bracket numbers together and inner bracket numbers together.
The finished equation should be y=(3x-1)(x+2). And there you have it! How to go from standard form to factored form.


Here are the steps once again:
1. Find two numbers that when added together equal the middle term, and when multiplied, equal the product of the first and third term.
2. Split the middle term into the two numbers you found.

3. Common factor the first set of terms together and the last set of terms together.

4. Group the outer bracket terms and put them into a bracket, and then group the inner bracket terms together.

I've included an extra learning source below.

Completing the Square

So far, we only know how to go from standard to factored form and from factored to standard. The perfect square trinomial method is used to go from standard form directly into vertex form.


To showcase this method, let’s use the example y=x^2+2x-1. The first step is to separate the first and second terms from the third with brackets. The second step is to take the b value, in this case 2, and divide it by 2 and then square it. After doing so, you’ll get 1. You will then take this 1 and add it into the bracket as well as subtract it.

Next, you will take the -1 out of the bracket so that the statement in the bracket can truly be in standard form.
Then, you will add the numbers outside of the bracket.

Once you have done all these steps, you can now put this equation into vertex form. Remember that vertex form follows the format of y=(x-h)^2+k. The x value will be whatever the ax value is, the h value will be the opposite of half of the b value, and the k value will be the number outside of the bracket. The finished answer is shown above.


The equation we just did together has an a value of 1. When we have an equation in standard form that has an a value other than 1, there are a few additional steps to be done. For example, the equation, y=2x^2+2x+3. The first additional step would be to common factor the first two terms.

You would then continue on as usual until you get to the step where you bring the negative c value outside of the bracket. As you bring the c value outside of the bracket, you must multiply the negative c value and the bracket by the number you have common factored out of the bracket. Now the factored out number will become to coefficient of the bracket and the negative c value will be doubled.
From there you would continue on as usual with all the normal steps and then put the equation into vertex form.
Here are the reiterated steps:
1. Common factor the first two terms if needed.

2. Find the square root of the second term in the bracket and multiply by 2. Add and subtract the product inside the bracket.

3. Take the negative c value outside of the bracket.
4. If necessary, multiply the negative value just taken out of the bracket and the bracket accordingly.
5. Add the outer terms together.
6. Input the numbers into vertex form. The a value will be the coefficient of the bracket, the h value will be half of the b value in the bracket, and the k value will be the number on the outside of the bracket.

Below I've included extra help and learning links.

Writing Quadratic Functions in Vertex Form by Completing the Square

The Quadratic Formula

The quadratic formula is used when an equation is given in standard form and the terms in the equation are either: not whole numbers, not factorable, or are really big. The quadratic formula allows people to find the x-intercepts of an equation without standard form.


What is the quadratic formula? It is x=-b+-Öb^2-4ac/2a.


In order to use this formula, you would plug the a,b, and c values into the formula, much like the special product technique, and then solve up until a certain point.


Let’s do a question together. Let’s do the equation 0=5x^2-7x+2. First off, the is the perfect equation to use the quadratic formula on because it is not factorable. Okay, so the first step is to plug all the values into the equation. Do whatever the equation tells you to do (remember BEDMAS). When you get to the point where it is a number plus minus another numbers divided by another number, that’s where you must solve for two equations. The first equation involves you solving by adding the second number to the first number and then dividing, and the second equations involves you subtracting the second number to the first number and then dividing. The numbers you get from each equations are the x-intercepts for the equation. In this case, the x-intercepts will be (1,0) and (0.4,).

When using the quadratic formula, there is a way to know how many x-intercepts the equation has (1, 2, or none). The discriminant of the formula tells us this. When the discriminant is 0, the equation has 1 zero, when the discriminant is negative, the equation has no zeros, and when the discriminant is positive, the equation has 2 zeros.

Here are the steps again:
1. Input the numbers into the quadratic formula.
2. Solve up until you only have 3 terms left.
3. Make two equations in which one adds the terms, and the other subtracts the terms.


I've included some helpful links related to the quadratic formula below.

Solving Quadratic Equations using the Quadratic Formula - Example 3

-b/2a and c-b^2/4a Method

The -b/2a method allows you to find the h value in vertex form, and the c-b^2/4a allows you to find the h value in vertex form. You can use this method when you are given an equation in standard form. You use this method by inputting the numbers into the two formula.


For example, the equation y=x^2+7x+10. When you input the numbers accordingly in the first equation and solve correctly, you would get your h value. Then, when you input the numbers accordingly in the second equation and solve correctly, you should get your k value. Now all you have to do is put the numbers inside the vertex form structure, and you’re done! This method allows you to easily go from standard to vertex form (much like the completing the square method).

Partial Factoring

Unlike previous methods that focus on finding the vertex or x-intercepts of an equation, the partial factoring method allows users to find points on a graph where y values are the same. From there, you can go and find the vertex of an equation easily.


Let’s use the equation 0=x^2+6x-10. The first step would be to remove the c value.

Next, you would common factor the remaining terms.

Lastly, you would solve for x for the term outside the brackets and the terms inside the brackets. The answers you got are the x values in which -10 will be the y coordinate (in this case (0,-10) and (-6,-10).

With these x values, you can find the midway between them and then find the vertex from there.


This goes for any set of x values that you get from partial factoring.

Quadratic Word Problems

Word Problem One

In this problem, they ask us to find the maximum height of the rocket. The maximum height can be found by finding the vertex of the rocket's path. To find the vertex, I used the completing square method. When I converted the equations from standard to vertex form, the k value was my answer.

Word Problem Two

Word Problem Three

When writing the equation for this word problem, you have to make x represent number of increments and y represent the total revenue. After writing the equation, you would solve for x to find the x intercepts. You would then add those numbers and divide by two so you can get the x value for the vertex in order to find the max/min (in this case max) revenue.

Word Problem Four

To solve this equation, I had to use the formula for area which is A=lw. The question asks to write two different equations that would end up being the same expression that would get the area of the figure. So, I made two equations with one focuses on subtracting from the overall general area, and the other adding from parts of the area.

Quadratic Reflection

Overall, I think the quadratics units, for me, has had its ups and downs. I found graphing and factoring to be really easy concepts. I think I probably struggle mostly with word problems but other than that, I've learned from the multiple mistakes I've made here and there on tests. Throughout the unit, I often attempted to find solutions to problems on my own rather than ask for help which made me have a lesser understanding of some concepts. However, I more often ask for help from peers and teachers when I need it. I wish I had learned the quadratic formula earlier in the unit because that formula has made my life 10x easier while solving quadratic equations. Now that the unit is done, I feel like I have a good grasp on everything we've learned in quadratics (factoring, changing quadratic equations forms, word problems, etc.), and if I were given any of those tests again, I could probably ace all of them.