Quadratic Systems

Dinesh's Weird Problems With Parabolas

Table of Contents

1.Introduction to Quadratics
2.Introduction to Parabolas
3.Steps of Solving Parabolic Problems
4. Expanding and Simplifying
5.Example of Parabolic Problems
6.Understanding The Use of Parabolas
7.Conclusion And Other Information Links

8.Part Two Factoring Quadratic Equations

9.Introduction to Factoring

10.Types of Factoring

11.Polynomial Factoring

12.Simple and Complex Trinomial Factoring

13.Word Problems With Factoring

14.Conclusion and Other Informational Links

15.Part Three Maxima and Minima

16.Introduction to Maxima and Minima

17.Introduction to Solving Quadratic Equations

18.Solving Equations

19.Solving Using The Quadratic Formula

20.How To Find The Vertex


22.Informational Sites On Quadratics

Introduction to Quadratics

Other than linear systems there is another graphing method used for graphing lines that are in the shape of hilltops. This particular method is called quadratics. The other lesson in quadratics is also about factorizing quadratic equations. Before we start just remember there are three types of forms of an quadratic equation. Those three are Standard Form, Factored form, and vertex form.

Introduction to Parabolas

The line in quadratics is called parabolas. They turn out to be hilltop shapes because the equation has 2 Xs, the first x is to the power of 2 and the second one is just x. You can find quadratic problems when you have a problem that has to be simplified. For example when you get an answer of x squared + x + 2, it is the quadratic equation. There are labels for each part of the parabola and here is the description. The vertex: the point of the parabola where it starts and the top/ bottom of it. The X-Intercepts: It is the point where the parabola passes through the x axis. The Y-Intercept: Where the parabola passes through the y axis. NOTE its not really possible for all parabolas to have an y-intercept. The Axis of Symmetry: The line that is right in the middle from the vertex point and down. Your axis of symmetry line is your x value in your vertex for example (4,-5) your line would be x=4. Optimal value:It is the highest point where your parabola cannot move above or below that point. Parabola Zeroes: It is the two points where the parabola touches the x axis not the x-intercept, Example (2,0) (-2,0)

Steps of Solving Parabolic Problems

There are steps to solving quadratic problems. But the most effective is,
1. Graphing the line.

Graphing parabolas is the best way to solve all systems involving lines. In quadratics to graph the parabola first you need to create a table of value chart. When your done the second step is to add the x and y variables onto the chart. The third step is to multiply the x numbers on the chart by its self twice because the x in the equation is squared so when you do that, the answer you get is the y intercepts.

Expanding and Simplifying Quadratic Equations

When you have a form like (x+2) (x-2) what you have to do is expand it and simplify it. How well when your expanding you don't have to multiply every term with every term in the brackets, instead you take (x+2) and multiply both terms with both terms in the other brackets. After you'll end up with x2 - 2x + 2x - 4, now all you have to do is simply add the two like terms that are left ( - 2x + 2x), which will give you zero. Now your equation is x2 - 4.
What will you do when you have a factored form like this? 2(x+3) (x+4). Basically what you would do is leave the 2 outside the brackets and just multiply the first to terms to the second brackets. You will have 2(x2 + 4x + 3x + 12) now you will add the two like terms in the brackets. After you have done that you will have 2(x2 + 7x + 12) the actual answer is what you have to do here. The whole point of this is to expand and simplify so just multiply everything in the bracket by 2. The equation will be 2x2 + 14x + 24.

Example of Parabolic Word Problems

1. An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where is in meters. When does the object strike the ground ?

2. A picture has a height that is 4/3 its width. It is to be enlarged to have an area of 192 square inches. What will be the dimensions of the enlargement?

3. A basketball has been shot. The ball was shot 10m high and was 5m low before being scored. How high was the ball before it reached the basket?

These are word problems that are solved by graphing, factoring, solving for x-inters, and using the vertex form and find vertex.

Conclusion and Informational Links on Parabolas

The main reason of quadratics is graphing parabolas and now that you know how to graph and know what a parabola is, lets get started on the second step which is how to expand a quadratic equation and how to create an expanding form, from isolating the equation.

Part Two: Factoring Quadratic Equations

- Introduction to Factoring

What is factoring?
It is basically isolating a quadratic equation to find two numbers for a quadratic problem. For example
Expanding (x+3) (x-3) into x2 - 9 and Factoring x2 - 9 into (x+3) (x-3).

How many ways are there to factor?
1. Common Factoring
2. Factoring Simple Trinomials
3. Factoring Complex Trinomials
4. Factoring By Grouping
5. Differences of Squares
6. Perfect Squares

1. Common Factoring

To factor a number or expression and turn it into an multiplication format. In this case we are using factoring to factor quadratic equations into multiplication. In some cases were you have all numbers that can't be divided by something, you call those not factorable expressions.

For Example: x2-2x-8 factor find the two factors of c that gives you c when multiplied and b when adding. two numbers 2,-4. into (x+2) (x-4)

2. Factoring Simple Trinomials

This method of factoring is the same thing as common factoring, these are expressions that have no numbers infront of the x2

For example: x2+2x+6 there is no number infront of the x2 or no a value.

3.Factoring Complex Trinomials

Relating to simple trinomials this is a little different. This typical form of factoring has an a value or there is a number infront of the x2.

For example: 2x2-10x+12 then find GCF of all terms 2(x2+5-6) and find the two numbers that give you c value when multiplied and b value one adding 2(x-2) (x-3)

4. Factoring By Grouping

This type of factoring is complicated compared to the others. It is used only when you have four terms in a quadratic expression and from that you turn it to a multiplication format without the two numbers from the b and c values.

For example:

xy-5y-2x+10 into xy-5y -2x+10 then you find GCF of it y(x-5) -2(x-5) Then take the factors and the x-5 in both expressions and put it into one expression. (x-5) (y-2)

5. Differences of Squares

Is one all terms (two) are squared and when you factor them it should be (number- another number) (same number+same second number)

For Example: you have a2-b2 you have to make them into ( ) ( ) so what you do is add the two A's into the two brackets and the two B's into two brackets. (a,b) (a,b) in order for you to have a negative expression a2-b2 you have to add in one bracket and subtract in another. (a+b) (a-b)

6. Perfect Squares

This type is similar to differences of squares but in this you only have one bracket with a number that gives you the c value and the b value.

For example: x2+10x+25 find the two numbers that gives you b and c well guess what you get the same number which is 5. Why 5x5= 25 and 5+5=10 so in this case the answer will be (x+5)2 because 5 gives you both the numbers 10 and 25 and you don't have to write (x+5 (x+5) instead this is a short cut to that way of writing the answer.

Part Three: Maxima and Minima

Maxima and Minima are the values of the parabola when graphed. How to find this is that if I have my vertex in positive and the parabola's values are negative it means that my parabola's maximum value is the vertex because the vertex value can't be changed. But that also means my x intercepts have to be negative. In this part we are going to learn about how to solve quadratic equations. We are also going to learn about the quadratic formula and what its used for.

1.Finding The C Values

How this works is you have an equation like this x2 + 4x + c and you don't know how to find c. Now that's why I'm here to help. So what you do is to find c you always divide the b value by 2. If you don't know what I'm talking about let me show you. Here is the label for standar form. ax2 + bx + c = 0 and what you do to find the c value is to divide your b value by two. So it would be 4/2. Your answer will be c=3 and your equation will be 0= x2 + 4x + 2.

2. Solving Equations

By solving quadratic equations I mean to say finding the x-intercepts for the equation. There are two ways to solve these equations but right now lets focus on the this way that I'm going to tell you. Okay lets take a look at an factored form equation (x+3) (x+5) now if you say to solve this you have to expand it YOU ARE WRONG!!!!! What you have to do is remove the brackets and place it in this form x+3=0 x+5=0 and move the like terms to the other side of the equal sign and you'll have this x= -5 x= -3. This is the basic way of finding it.NOTE: if you have an simplified equation then just factor it and solve it.

3. Quadratic Formula

The quadratic formula is the formula of finding both x-intercepts and what you have an equation in standard form (expanded and simplified). The formula is

so in this case your equation is ax2+bx+c=0 and all you do is add the values for each variable from the equation. Sometimes you can use this to check if your previous way of finding the x-intercepts and see if you get the same x-intercepts in this too.

4. Vertex Form: To Find Vertex of The Parabola

As you know the vertex is the top point of the parabola where the line starts. This will help you find the vertex of the parabola in order for you to graph it using the vertex, and the x-intercepts. The formula is y = a(xh)2 + k. Lets say you have an equation like this y= x2 + 4x - 5 basically what you have to do is put the brackets between the x2 and 4x so it will be something like this. Know you will have y= (x2 + 4x) - 5 now what you do is find the c value which is 4/2 and equals 2squared so it would be y= (x2+4x+2squared - 2squared ) -5 and then it would be y= (x2+4x+4) -4 -5. Now finally it would be y= (x+4)2 -9 and your vertex will be (4,-9). Now you can graph the parabola. If have a question like why do you have to square it to two? Here is the reason why... When we were finding c we were doing that for the c value outside of brackets, but here we are dividing a square root so we would have to have the number squared.
Big image


Know you know everything about quadratics have fun graphing factoring and solving equations and parabolas!!!!!!!
Big image

Informational Links

Top 3
1. www.khanacademy.com
2. www.purplemath.com
3.and My Website