Introduction |
This unit was focused and specified on quadratic equations, there relevance and application to the modern world, and the multiple ways of solving them. Improving our understanding on the mathematical concepts behind the equations and defining what a quadratics is. This project was centralized around a problem which involved finding the height of a rocket at the very peak of it's destination, the time to reach the ground, and the time it reaches it ultimate peak. Focusing on the main components behind a quadratic equation geometry, algebra, and the habits of a mathematician are what helped me gain the confidence throughout this unit. Using both vertices and x-intercepts we were able to solve for the equations. First beginning with kinematic equations also the distance formula: h(t) = d0 + v0 · t + 1/ 2 a · t^2
Using the information we already knew we were capable of plugging in different variables throughout the equation. Such as, the starting height, acceleration factored upon gravity, and the rockets initial velocity. When they are replaced within the equation our outcome transitions into : h(t) = 160 + 92t - 16t^2 After completing this equation with no previous knowledge we continued on to understand how the equation worked. Continuing throughout this project we worked with algebraic symbols to see how to plug in for the variables. This helped us to find the placement of each independent vertex coordinates , and the x-intercept of the equation. |
There are a total of three different type of quadratic forms; vertex, standard, and factored. The benefits of vertex form is it allows you to identify parabolas because your able to find the exact coordinate for the vertex. Standard from found in the right left hand corner is (y=ax^2+bx+c), where a, b, and c can be any integer. A representing the "slimness" and "thickness" of the graphed parabola. C representing the y-intercept of the parabola. Factored form is written out as: y=a(x-w)(x-m). W and M are the x-intercepts of the graph and A is again the the "slimness" and "thickness" of the graphed parabola. The benefits of this form is how it allows you to vividly see the x-intercepts.
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Vertex Form to Standard FormUsing an area diagram, you multiply out the squared terms, distribute a, and combine like terms.
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Vertex to Standard: you take the expression (x+3)^2 and break it down to (x+3) two times since thats what the expression "^2" is asking of you. Then you carry down the "+3" and then multiply 3x3 which=9 and carry it down with the "+9". Then you have the two X's, which can be boiled down to X^2.
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Factored to Standard: You are given the expression (x+3) (x+3), and to get to standard form you would carry down the "+0" and then multiply 3x3 which=9 and add 3x and 3x to =6x. Then you have the two X's, which can be boiled down to X^2.
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Factored Form to Standard FormMultiply the terms in the parenthesis, combine like terms, and distribute a
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Standard Form to Vertex FormTo convert to vertex form you need to put the ax^2 and bx terms together and take a out. Then you need to complete the square by filling in the missing terms. Then take the subtracted term out of the parenthesis by multiplying it by a. Finally rewrite the parenthesis as a product of the square and combine like terms outside of the parenthesis.
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Standard to Vertex: To get from standard form to vertex form you have to solve the equation. You take x^2+6x+9 and break it down to (x+3) (x+3), then your'e simplifying the equation to (x+3)^2 +3
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Standard to Factored: To convert from standard to factored, you have to un-factor the equation. In this case, you would take x^2+6x+9 and break down 6x into a number that when multiplied together will = 6. Also, when added together equal 6. In this case, it's 3, and your equation looks like (x+3) (x+3).
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Standard Form to Factored FormTo solve for factored form we have to work backwards. You know you want n and m values to multiply together to equal the c constant. You also know that if the n and m terms are added together they should equal the b constant, except that before they are added they need to be multiplied by values such that the values they are multiplied by are equal to the a constant. Then you are able to factor the a constant out of the equation. Then you have to guess and check to find the values you need. When you find the values, just write out the factored equation.
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Quadratics are/can be useful to the modern world. The following problem is an example we completed over this math course.
Kinematics An important math problem involving kinematics would be the main math problem this project was centered around, The Victory Celebration. First by converting this kinematics equation into a quadratic equation. Solving this problem by identifying the vertex coordinates and positive x-intercept. With this problem involving many detailed numbers, I used a helpful source, a calculator for my conversions. Taking the H and K constants for the time of peak and the maximum hieght of the arch. When solving for this problem it is most useful to use the guessing check method for the variables of W and M. Over time this showed me that the number was going to be in decimal form due to the fact of finding no solution dealing with whole integers. Starting once again but this time from vertex form, the process was shown because of the one x used in the equation. As the numbers again started to become a bit more challenge, I relied on my calculator. Which lead me to finding two answer one being positive and the other being negative. When looking at both number and considering the option it only made sense of the positive number to be correct due to the fact of the rocket would land after being launched. I was able to check my work by placing it into demos, revealing the answer as positive. Geometry Throughout this unit there was plenty of practice with geometry. Many and most of them dealing with maximum area of given spaces. An example being a pen with with a given amount of fencing or variable for perimeter. With one of the side being shared in the problem it already converts itself into a quadratic equation. If the side length of one side of the pen is perpendicular to the neighbor's fence representing as the variable of x, and the total amount of fence's length available is known as F. The area of the fence will be x(F-2x). Which can rewritten as x(-2+F) a quadratic equation in factored form. |