By Steven J

Quadratic relationships are different from linear relations because they are a "U" shaped graph instead of a straight line. Unlike Linear relations, quadratic relations can have 2 x-intercepts. In this website, you will be learning about different quadratic relationships, and learn how to transform and solve them.

The different types of Quadratic Equations are:

Vertex Form:

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

This helps you to easily notice the vertex of a parabola, (x,h).

Factored Form:

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

Easily depicts the x-intercepts.

Standard form:

y=ax^2+bx+c

the standard form to solve a quadratic equation.

This website will also be teaching you about the Quadratics 1, 2, and 3.

Also it will show you various problems and give you the skills you need to become a pro with Quadratics!

The connection part 1

In quadratics part 1, this part of the unit will teach you the basics/fundamentals of the Quadratics unit and at the end you will be able to bring them all together and become an expert at it!

How to discover if a relation is a quadratic, linear or non linear with First Differences

In the first relation, we know it is a linear relation because after doing the first differences, we have the same number which appears after the subtraction of the Y terms.

The second relation is a quadratic relation because after the duration of the first differences, we have still have a series of numbers that are not the same, so we do a second difference, and now we have the result that we want, we can see the number is 2 and that this relationship is Quadratic.

Finally, in the last relation, we can see that in the first and second differences that we still have a variety of numbers, which we do not want. This relation is neither linear or quadratic, this is a non linear relationship.

Linear, Quadratic or Neither using First and Second Differences Tables

Vertex form

The Vertex form of a quadratic relationship is y=a(x-h)2+k. When it is subbed with a series of numbers such as y=(x-1)^2+5, you will get a parabola that is positive. To graph this equation you will need to find the vertex, the vertex is within the brackets. To graph this equation, follow the steps below:

1.First, find the vertex, the vertex is the only the h variable which is 1, and then you need to get the K variable which is 5.

2.After finding the vertex, you need to plot it on the graph, but you cannot sketch a rough graph because you do not have the a variable.

3. With the equation, you should always substitute the X=0 and Y=2, plug them in and solve the equation, this should give you a -3. so A=-3. * note not all vertex formed p

4.so from there the equation is now y=-3(x-1)^2+5

5. Now you can graph the equation, and with the -3, you now know it will be a negative parabola.

Factored Form

The factored form of a quadratic equation is y=a(x-r) (x-s). When the equation is subbed with a series of numbers such as y=x^2+10x+24. To begin to solve this equation follow the steps below:

1. First, but the equation into factored form: y=(x+4)(x+6)

2.Then sub X=0 and solve it, this will give you X=4 and X=6.

3. plot the coordinates on the Graph (4,0) and (6,0). And remember, since the coordinates are positive when factored then it is switched when graphing, it will become (-4,0) and (-6,0).

4. After plotting the two points, you will see that you can roughly make out a parabola. try to sketch the parabola to get a rough image of what you will get.

5. As you can see, this parabola will be opening upwards.

The connection part 2

In Quadratics part 1 we learned how to solve equations and also how to graph them using the information we have retrieved from it. In quadratics 2 we will use the knowledge we retrieved from part 1 to help us in changing a standard equation into factored form and vice versa.

The Expanding of Polynomials

To get a polynomial (Standard form equation), you need to have an equation in factored form, and you will solve to turn the expressions into standard form.

There are two ways you can solve a polynomial to get a trinomial, the first method is the rainbow method and the second one is S.D.P.S. (better known as Sam Doesn't Pull Strings.)

Rainbow Method

To solve a polynomial via The rainbow method, follow these steps:

1. First you would multiply "x" and "x", which will give you x^2. An easier way to know this is to draw a line that connects to the second "x" in the second set of brackets.

2. Secondly, you would multiply "x" with "2", and this will give you "2x". To be sure, you can draw a line from the first "x" to the "2" in the second set of brackets and it will give you "2x"

3. Thirdly, You will repeat step 2. But this time draw a line connecting the "6" from the first set of brackets to the second set "x", this will give you "6x"

4. Next, you will multiply the "6" and the "2", and this will give you 12.

5. Finally, combine all like terms, and in doing so, it will give you x^2+8x+12.

* An example is on the side.

S.D.P.S. Method. (Sam Doesn't Pull Strings)

S- First you would square the first polynomial, so in this case, our polynomial is "3x". This will give you "9x"

D.P.- Next, you will have to double the product of your second polynomial, and in this case it is "5y", so you would use the equation 2(x)(y). when you sub in the numbers, it looks like this: 2(3x)(5y). This will give you 30xy.

S- Lastly, you will need to square again, but for the second polynomial. For example squaring "5y" from the equation will give me "25y".

*An example is on the side

Solving Perfect Squares using S.D.P.S.

To solve a perfect square equation, you need to know that you cannot use the rainbow method to solve it, but rather use the S.D.P.S. method.

To solve a perfect square equation using S.D.P.S. follow the steps below:

S- you will need to square the firs polynomial. In this case, our equation, (x+2)^2, the first polynomial is "x" You will need to square this so it will give you "x^2".

D.P.- For this part of the equation, to solve this you will need to double the product, so you will be using this equation which is 2(x)(y). When subbed, your equation should like this

2(x)(2). *Note, you will have different subbed numbers depending on your equation, and also that the 2 in front of all the equations will remain the same because you are doubling your product.

S- For the final part of the equation you will need to square the second polynomial in the equation and in this equation (x+2)^2, our second/final polynomial is "2", so squaring it will make it "4".

Factoring common Binomials

To common factor a binomial, you must first find the GCF of both of the terms to factor the equation. Once you find the GCF of both of the terms, place the GCF outside the bracket and then multiply it with they LCM, to give you the terms you had in the beginning binomials, this is also a quick way to check and see if your answer is correct.

Factoring by Grouping

To factor by grouping you will need four terms. You would need to to factor each group, and when you factor them, the numbers/variables in the brackets should be the same otherwise you have factored them wrong.

Factoring Simple Trinomials

Simple trinomials are expressions that are in standard form. (y=ax^2+bx+c).

*remember if the first coefficient doesn't have a number in front of it, always assume there is a "1" in front of it.

To factor a simple trinomial, you must find two factors, that when multiplied give you the "c" of the equation. Also, when you add up the two factors they must give you the sum of "b", if they don't, the factoring of the equation is either wrong, or the expression is not able to be factored.

Factoring complex Trinomials

Complex trinomials are not that different from simple trinomials. The only difference is that the coefficient is greater than 1. And also, you would need to multiply your "c" and "b" with "a" to get the final answers of your equation.

The connection

The connection between all three parts of Quadratics is that in part 1, it is about teaching you how to graph a basic parabola using all the 3 types of equations. In Quadtics part 2 we learned how to factor the equations from standard and into factored form. Now in part 3 of Quadratics, we will learn how to solve equations, use factoring to solve it, learn to use the quadratic formula, and also how to use completing the square to solve an equation and also how to solve it.

Word Problem

The Height of a rock thrown from a walkway over a lagoon can be approximated by the formula h=-5t^2+20t+60, where "t" is the time in seconds, and "h" is the height in metres.

a) Write the formula in factored form.

b) When will the rock hit the water?

a)

b) The rock will hit the ground at 6 seconds.

Up until now, we have been working with expressions. Now we will be working with equations, and we will be solving for the variable.

Method 1: Factoring the Equation to solve the equation.

To solve an equation using the first method, you would need to factor the equation, and in doing so will give you the x-intercepets. *Remember, the variable that you recieve from factoring should give you 0, when recieving a final answer.

An example to be factored will be: x^2+5x+6.

Check:

SAMDEB

You can also solve an equation using "SAMDEB", which means, subtraction, Addition, Multiplication, Divison, Exponents, and Brackets. You can only use this when an equation is in vertex form. SAMDEB, is essientaly BEDMAS, but in reverse. Using this will provide you with the zeros/x-intercepts. When you have that, this will give you the Y-intercept and you will be able to solve the equation. *When solving, you should get 0.

Example: y=2(x-5)^2-50

When you square root the number, and it is positive, you will get two square roots, a positive and a negative, but if the number is negative, you will get no square root and if it is zero, you will get only one square root.
Using the X-intercepts, we will choose one to use and sub that into the equation to get the Y-intercept.

Solving Equations using the quadratic formula

We have looked at solving quadratic equations by factoring. But, this method may not work for every equation. Thus, we will learn a formula that will use for cases where an equation cannot be factored. The Formula is:
You can only use this Formula when the equation is standard form so you can assign variables, otherwise you will need to convert the equation into Standard.

Example equation : 0=3x^2+4x+5

Since the Square root is negative, there will be no X-intercepts. So there is no real answer for this equation because of the negative Square root. The video in the link will be extra help for using the quadratic formula to solve an equation.

Word Problem #1

A management firm has determined that 60 apartments in a complex can be rented if monthly rent is \$900, and if that for each \$50 increase in the rent, three tenants are lost with little chance of being replaced. What rent should be charged to be maximize revenue? What is the maximum revenue?

First, lets make and equation and assign variables and isolate.

After finding the X-intercepts, you will use the forumla that finds the optimal value.
After finding the optimal value, you will sub in that number into the equation, and solve.
The maximum rent that should be charged to maximize revenue is \$950. The maximum revenue would be \$54, 150.

Word Problem #2

You have a 500-foot roll of fencing and a large field which is bordered on one side by a building. You want to construct a rectangular playground area. What are the dimensions of the largest yard? What is the largest area?

To solve, we will assign variables and then make an equation and isolate.

After isolation, we will use the A.O.S. formula to give us W value and the L value. Then we will sub it into the equation for perimeter and area.
The total area will be 31,250 ft^2. The dimensions will be W=125 and L=250