Quadratics Relations Website

By: Vithusan Vijayakumar


Quadratics is one major unit in grade 10. In this unit, you will learn about the three different forms of equations and how to solve word problems, or graph using these equations. Quadratics may seem a bit complicated at first since it is a new unit, and hard to understand. However, by looking through this website, you will find many tutorials, videos, and examples of Quadratics to get you back on track. Please keep scrolling to learn more.

Unit 1: Vertex Form

Let's start off with Vertex Form. Over here you will learn more in-depth about the parabola and the important parts of a parabola. I will also be showing and explaining many different methods of solving a vertex form equation.

Parabola In-Depth

Parabola: graph of a quadratic equation (a symmetrical curved line).

Vertex: is the maximum or minimum point of the parabola and is the point where the axis of symmetry and parabola make contact

Optimal Value: is the y-coordinate of the vertex (if parabola opens upwards, then it is a minimum value while if parabola opens downwards, it will be a maximum value

Zeros: when the parabola passes the x-axis (x-intercepts)

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Learning Goals

1. Using finite differences to analyze quadratic relation

- Create a table of values of the graph

- Find the first differences of the y-values

- If all the first differences are not constant, then continue to find the second differences.

If the first differences of the y-values are equal, then the equation is linear.

If the second differences of the y-values are equal, then the equation is quadratic.

If neither the first or second differences are constant, then it is neither a linear or quadratic relation.

In other words, the first differences of the table determine that the line is linear, while the second differences determine that the line is quadratic.

2. Using a table of values to create a graph
Using a table of values to create a graph is simple.

First of all, locate the first (x,y) values of the table.

After, locating, plot the points on the graph.

Continue to follow this process until you have plotted all the points from the table

Then, use the line of best fit to determine whether the graph is a linear equation or a quadratic equation.

3. Graphing the base graph of a quadratic function without technology

Create a table of values for y=x^2

If the equation is y=3x^2 all you have to do, it multiply the y value for y=x^2 by three.

After finding all the values, plot the points on the graph

"a" Value: determines whether the graph is stretched or compressed

When 0<|a|<1 vertical compression, |a|>1 vertical stretch, a<0 reflection in the x-axis

"h" Value: determines whether the graph is horizontally shifted to the right of left

h>0 shift to the right, h<0 shift to the left

"k" Value: determines whether the graph is vertically shifted up or down

when k is positive, the parabola moves up, when k is negative the parabola moves down

Solving for y-intercept: sub x=0 for the equation and solve for y

Solving for x-intercept: sub y=0 for the equation and solve for x


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Step pattern can help graph a parabola, easier than trying to find the slope to graphing a parabola, as long as you remember the numbers 1,3, and 5. The basic pattern for the step pattern is 1 over 1 under, 3 over 3 under, 5 over 5 under if your slope is to be negative. If it is supposed to be a positive then you are to go 1 over 1 up, 3 over, 3 up, 5 over 5 up.

Video To Help You Understand Even More

Graphing a parabola in vertex form | Quadratic equations | Algebra I | Khan Academy

Word Problem

The height of a flare is a function of the elapsed time since it was fired. An expression for its height is h=-5(t-10)^2+500, where h is the height of the flare above ground, in metres, and t is the elapsed time in seconds.

1. What is the maximum height of the flare?

The maximum height of the flare is 500 metres because it is the y-coordinate of the vertex, and also known as the optimal value.

2. At what time does the flare reach its maximum height?

The flare reaches the maximum height at 10 seconds because it represents the x-coordinate of the vertex, which shows the duration that the flare took to reach maximum height.

3. Determine the height at 5 seconds.

To determine the height of the flare at 5 seconds, we have to substitute t=5.






Therefore, the height of the ball at 5 seconds is 375 metres.

4. Determine the time when the height of the flare is 200 metres above the ground.

Completed in the following picture.

5. Determine the time when the flare hits the ground.

Completed in the following picture.

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Video of Graphing Using Mapping Notation

Graphing Vertex Form

Unit 2: Factored Form

Now, you will learn all about factored form and equations using factored form. I will also guide you through solving a few equations and word problems. Furthermore, you after scrolling through this website you will understand the seven different types/methods of factoring. So please, continue moving down the page to learn more about Quadratics using factored form.

Learning Goals

  • I am able to identify the different types of quadratic relations used and how to factor it efficiently
  • I am able to identify the x-intercepts using factored form
  • I am able to find the axis of symmetry using factored form
  • I am able to find the vertex using factored form
  • I am able to graph using factored form

Summary of the Unit


  • Formula for Factoring: Y=a(x-r)(x-s)
  • When finding y-intercept, you have to sub x=0, and then solve the equation to find the value of y
  • Graph the equation using the factors (x-int, y-int)
Finding Values:
  • The value of A determines the shape and direction of opening (up or down)
  • X-Intercepts= the value of R and S
  • To find the axis of symmetry, you have to find the average of the value of R and S

You can find the average by adding the two numbers, then dividing by 2.

Graphing a Factored Form Equation

Step 1: Solving for x-intercepts:

The x-intercepts can be easily solved by finding the value of R and S but reversed operations.

For Example:


the x-intercepts are: (9,0)(-7,0)

Step 2: Solving for the Axis of Symmetry

Solving for the Axis of Symmetry is simple, all you have to do is find the average of both x-intercepts.

Step 3: Solving for Y-value

After solving for Axis of Symmetry, you use that value and replace is with the x in the equation and solve for y.

Step 4: Plotting the Points

Once you have figured all the points, begin to plot them on a graph.

There are many different types of factoring:

  • Grouping (4 Terms)
  • Complex Trinomial Factoring
  • Common Factoring
  • Simple Trinomial Factoring
  • Expanding and Simplifying
  • Perfect Square Trinomials

The flowchart below can help you figure out what type of factoring is used.

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Grouping (4 Terms)

Step 1: First of all, you need to factor two terms at a time. So start off by finding the greatest common factor of the first two terms, then factor.

Step 2: Follow the same steps of step 1 and factor the last two terms.

Step 3: After Factoring, you take the two terms outside the bracket and put it with one of the two terms in the bracket.

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More examples of factoring by grouping | Algebra I | Khan Academy

Simple Trinomial Factoring

This is a very simple method, and you get the your answer in factor by grouping question form. It is like the opposite of factor by grouping. The point of it is to get 2 Binomials that multiply to get the given trinomial. It is a simple trinomial if there is 3 terms. One being x² the other being x with no power and it can have a number in in font, and the last being a number with no variable.
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Complex Trinomial Factoring

Finding the complex trinomial Factoring is somewhat similar to Factoring simple Trinomial Factoring.

Step 1: Multiply the a value and c value.

Step 2: Then, find one number that is the product of the c value and the sum of the middle value

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Factoring Complex Trinomials

Common Factoring

Common factors= finding a greatest common term and dividing the rest

Lets say we have a polynomial 8x^3y^2+3xy^3+15x^2y^2

The greatest common factor is 3xy^2.

So, we factor the whole expression by that.

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Step 1: Multiply the first term within the first bracket with the first term within the second bracket.

Step 2: Do the same step with the second term.

These are the only two main steps to Factoring

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Perfect Square Trinomial

Perfect squares are any numbers that can be squared into a whole number. However in Quadratics we can use Perfect Squares to our advantage. We can have a Binomial that is being squared, and convert it into a Trinomial. You might be confused as to how we can do this. Well, lets take an example (x+4)^2. Now to convert this Binomial into a Trinomial, we will write the equation as Binomial *Binomial { (x+4)*(x+4) }. Next, use F.OI.L and you will end up with a Quartic polynomial ( x^2+4x+4x+16 ). Finally collect like terms and you will have a Trinomial. You might be thinking "I dont wan't to do these steps, how can I make it easier?"

Well, a simpler rule would be to square the first term (in this case "x"), then add 2*the product of the first and second term (2*(4*x)), and finally add the second term squared (4^2). You will end up with the same answer but in a faster way. This rule applies in the same way if the two terms are being subtracted, but one step would change. Instead of adding 2*the product of the first and second term as our second step, we will subtract 2*the product of the first and second term.

Below is another example of Perfect Square Trinomial.

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Difference of Squares

Difference of Squares are when you have two binomials and you want to simplify it.

You will always end up with a single binomial at the end. For starters, to use Difference of Squares, you need to have two binomials with one of them being added and the second needs to be subtracted (x+4)(x-4).

Now you will use F.O.I.L and will end up with something like this (x^2+4x-4x+16). After collecting like terms you will have a Binomial (x^2-16) (note: for difference of squares you will always end up with the two numbers being subtracted).

You end up with a Binomial because the two middle terms cancel each other out. Again, you might be saying "is there anyway to make this faster?"

Lucky for you, you can just square the first number and subtract that by the square of the second diget which would be { (x^2)-(4^2) }.

Below is another example of Differences of Squares

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Video of Solving Equation Using Factored Form

Writing a Polynomial in Factored Form

Unit 3: Standard Form

Last but not least, you will find information about Standard Form Quadratics. Scroll down to see word problems and examples of Standard Form equation.

Learning Goals

  • I am able to use the quadratic formula in a quadratic equation.
  • I am able to use the standard form equation, to complete the square, and graph the parabola
  • I am able to find the x-intercepts of the equation, by using the quadratic formula.
  • I am able to find the vertex of the equation, by completing the square.
  • I am able to graph the x-intercepts and the vertex, then draw the parabola.



"a" value: represents the parabola's direction of opening and shape (stretched or compressed)

"c" value: represents the y-intercept

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The quadratic equation is used when you are given a standard form equation and are asked to find out the x intercepts.

When asked to find out how many solutions(x-intercepts) there are you can use a simple equation to find out the number of solutions.

For Example

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The Discriminant adapts the formula that is inside the square root of the Quadratic Formula (b^2-4ac). The Discriminant helps us find out how many solutions the given quadratic equation will have, without us solving the entire equation with the Quadratic Formula. If the value of “d” is less than 0, so if it’s a negative, there will be no solutions. If the value of d is greater than 0, there will be 2 solutions and if the value of d is 0, there will be 1 solution. So to sum up the Discriminant:

  • D<0- no solutions

  • D>0- 2 solutions

  • D=0- 1 solution

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Graphing Standard Form

  1. Find the x-intercepts of the equation with the quadratic formula
  2. Find the vertex by completing the square
  3. Graph x intercepts
  4. Graph vertex
  5. Draw parabola
To complete the square, divide b by two then square the quotient. Then, if a isn't 1, add and subtract that product to the standard form equation. If a were 1, add the negative value of the product. Add the negative value and c value together after multiplying it with a if a does not equal 1. Put everything with the exception of the c value in brackets and use perfect squares method to solve by factoring. Next, put the factored part and add the c value. The result will be that your successfully converted it into vertex form.
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Word Problem

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Overall, Quadratics was a really complicated and fun unit to learn. I really enjoyed learning the different types of equations and methods to solving a word problem. This unit is mostly based around graphing parabola's and it helped me solve complicated word problems precisely and efficiently.

How the Three Different Types of Equations Relate to Graphing:

Vertex Form: Using this form of equation, you will be able to find the vertex using the flipped operation of the value "h" and "k". This helps you determine the maximal or minimal point of the parabola.

Factored Form: Using this form of equation, you will be able to determine the two x-intercepts of the parabola. This will help you plot the x-values on the graph.

Standard Form: Using this form of equation, the direction of opening will be determined. As well as how stretched or compressed the parabola is.


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This picture is a test from our Quadratics: Factored Form Unit. I decided to choose this test because of many reasons. First of all, I faced a lot of problems trying to factor the equations. This kept making me loose confidence. However, I didn't stop doing more work. Second of all, I decided to do extra questions for all the lessons in this unit since I needed more help. I believe my hard work has helped me achieve better than I expected.