Quadratics
By Steven J
What is a Quadratic relationship?
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!
Quadratics Part 1
The connection part 1
How to discover if a relation is a quadratic, linear or non linear with First Differences
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.
Vertex form
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
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.
Quadratics Part 2
The connection part 2
The Expanding of Polynomials
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
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 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
Factoring by Grouping
Factoring Simple Trinomials
*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
Quadratics Part 3
The connection
Word Problem
a) Write the formula in factored form.
b) When will the rock hit the water?
Answers:
a)
Quadratics Part 3
Method 1: Factoring the Equation to solve the equation.
An example to be factored will be: x^2+5x+6.
SAMDEB
Example: y=2(x-5)^2-50
Solving Equations using the quadratic formula
Example equation : 0=3x^2+4x+5
Word Problem #1
First, lets make and equation and assign variables and isolate.
Word Problem #2
To solve, we will assign variables and then make an equation and isolate.