The quintessence of quadratics (q^2)
I. Introduction:
Within this unit of Mathematics, we covered all sorts of topics ranging from kinematic equations, area and volume, and of course quadratic equations. The end goal for this project being for us as students to be able to better understand how physical motion can be measured and represented through graphs and equations. Our first task that introduced us into kinematics and soon into quadratics, was a hypothetical problem where a rocket was being launched off of the roof of a building as a part of a celebration. The goal for this problem was to find the peak of the rocket and how long it would take for it to reach its vertex, so that the firework would detonate at the proper time. We would calculate these values understanding that there were opposing forces and calculating the rate of deceleration along with forces such as gravity. Since this was the introductory problem, as students we weren’t sure about how to approach the problem and we began to input values for the variables that we knew within the equation to see if it would provide results. Usually I would collaborate with my peers to see if they had any idea of what to do or what were some options.
Afterwards we would learn that this was not only related to kinematics but to something called the displacement equation, which was exactly what we were working with. The displacement formula is as such: d=v0t+12at2. The way we got to this equation was by using some geometry and algebra. As you can see in the image aside, the displacement formula can be produced by looking at the area in acceleration over time. Within A1, the area of that section can be determined by multiplying the initial velocity by time. Then to calculate for the second area, area A2, we would be using the distance formula which is the base, by the height, by 1/2, or in our case, acceleration by time squared, by 1/2. When you combine the equations for both A1 and A2, you would get a formula that can be used to calculate the distance that an object travels while incorporating any initial displacement. This formula would serve a very important role in the future and was the start of our journey into quadratics.
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II. Exploring the Vertex Form of the Quadratic Equation:
After having moved on from kinematics a little, we began to cover area and how it correlates to things such as quadratics. Our first example was one where we would have to find the maximum area that could be produced from a corral where two sides of the fence would be equal and we would have a total of 500 feet of fencing to use. The equation that we would generalize at the end would be A=(500-2x)x, and we would reach this conclusion after looking for patterns and seeing similarities between the formula and the outcome. This was similar to the work that we would be doing, only that it was written in a different format than we were used to. So what does the equation for a parabola look like?
Although it can be written in various forms to give specific information, the one that is best for sketching a parabola would be Vertex form. When the equation is in this form, it is easiest to identify what the parabola will look like and where it will be placed. The equation is as follows: y=a(x-h)2+k. This equation can be manipulated in various ways, and heavily relies on what the values for a,h, and k are.
A: What this variable represents is how wide or narrow the parabola will be and dictates whether or not the parabola is concave up or concave down. The greater the value for A, the narrower the parabola will be, and vice versa. If the value for A is a negative number, then the parabola’s opening will be concave down.
H: This variable will dictate the placement of the vertex of the parabola and is responsible for the X-coordinate. (The vertex of a parabola can be written as (h,k)) If the value for H is negative then the parabola will be on the positive quadrants of a coordinate plane, this is because since the equation is squared, the value would always be positive. If the value for H is positive, then the parabola’s vertex would be placed on the left side of a coordinate plane.
K: The variable K is responsible for the placement of the y coordinate for the vertex of a parabola. What this means, is that the K just moves our parabola up and down on the y - axis, and sometimes there is no value for K since it isn’t a part of the equation.
Although it can be written in various forms to give specific information, the one that is best for sketching a parabola would be Vertex form. When the equation is in this form, it is easiest to identify what the parabola will look like and where it will be placed. The equation is as follows: y=a(x-h)2+k. This equation can be manipulated in various ways, and heavily relies on what the values for a,h, and k are.
A: What this variable represents is how wide or narrow the parabola will be and dictates whether or not the parabola is concave up or concave down. The greater the value for A, the narrower the parabola will be, and vice versa. If the value for A is a negative number, then the parabola’s opening will be concave down.
H: This variable will dictate the placement of the vertex of the parabola and is responsible for the X-coordinate. (The vertex of a parabola can be written as (h,k)) If the value for H is negative then the parabola will be on the positive quadrants of a coordinate plane, this is because since the equation is squared, the value would always be positive. If the value for H is positive, then the parabola’s vertex would be placed on the left side of a coordinate plane.
K: The variable K is responsible for the placement of the y coordinate for the vertex of a parabola. What this means, is that the K just moves our parabola up and down on the y - axis, and sometimes there is no value for K since it isn’t a part of the equation.
Using this example, we can visualize how the variables have an effect on the parabola’s shape and location. For the purposes of this demonstration, I decided to keep the A variable as positive one, therefore it won’t have an effect on the shape. In the parabola that is shown, we can see that the coordinates of the vertex are (3,2), what this tells us, is that the the value of H is -3, and the value of K is 2. From here we can formulate the equation for the parabola, keeping in mind that the value of A is 1. Our vertex form of the equation would look something like this: y=1(x-3)2+2, although we are inputting values for all of the variables, usually the 1 would be excluded since it doesn’t contribute to the final solution. This is what the vertex form of the equation of a parabola would look like. And although it is great to describe and articulate, it sometimes is easier to understand when you have a manageable graph. Therefor a very important tool that we used throughout this unit called DESMOS, allowed for us to visualize the effects of a variable in real time.
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III. Other Forms of the Quadratic Equation:
The reason to why I continued to explicitly say that the parabola’s equation was in vertex form, is because there are other ways of expressing the same parabola but in different formats that can be used to calculate for specific information in the parabola. The two other forms that the equation can be written as are Standard form and Factored form. In the parabola to the side we see that it has it’s vertex at (-1,-9), and it has two x-intercepts. From this we can write the vertex form of the equation which would be y=(x+1)2-9. If we were to write the equation of the parabola using Standard form, we would rewrite it as such: y=x2+2x-8. And if we were to write the equation to the same parabola in Factored form, we would get: y=(x-2)(x+4).
The average equation for the Standard Form would be, y=ax2+bx+c. The benefits of writing the equation this way is that it provides the value for A which is a recurring value and like I said earlier, dictates how wide/narrow the parabola is and in which direction it concaves. Another reason as to why writing the equation form is beneficial is because it can be manipulated easily and can be converted into either Factored or Vertex form. Our formula for a Factored equation is: y=a(x-s)(x-t), and has our variable A. The value of A will determine the width of the parabola and, the other variables, S and T, are intercepts of the slope. The great thing about Factored form is that in the equation itself, the values for the x-intercepts of a parabola are represented by S and T. When you have these values, visualizing becomes much more simpler, and Factored form becomes extremely convenient. |
IV. Converting Between Forms:
In this section I will demonstrate how one would convert between the various forms of the equations. Of course we will start small using lesser values to easily convert between the forms and to better grasp the content.
Vertex Form to Standard Form:
When we have the equation in vertex form it is very easy to extend the equation and to begin solving and combining like terms using PEMDAS. Our first step is to square what’s in the parentheses and to solve within. Then we distribute the two that’s on the outside and incorporate the 3 at the end to have our equation in standard form. |
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Standard Form to Vertex Form:
In this equation we have a value for A, in the form of a coefficient and our goal is to convert this equation into vertex form. Our first step is to separate the coefficient thus affecting the values that will go in the parentheses. Next, our goal is to complete the square, therefore we will have to divide 16 by two and then square it to get the rest of the equation. This explains the positive and negative 64 because the positive 64 is created as a byproduct of dividing and squaring the 16 and then the negative 64 is there to counteract the value and reduce it to 0. In the last steps of the equation we have to distribute the coefficient 2 to the quadratic and to the 64 to combine the last like-terms together. |
Factored Form to Standard Form:
When we have an equation in Factored form, all we have to do is multiply the values in the parentheses and then distribute the coefficient if there is one. This process allows for us to convert the equation between Factored and Standard form in a quick and easy process. Of course, it’s best if one stays organized throughout the process so that values aren’t changed and you don’t lose track of the work that you have done. |
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Standard Form to Factored Form:
When converting between Standard Form to Factored form, sometimes it will take a couple of tries and of lot of conjecturing and testing so see if it is correct. In the example that I provided, my first step was to find the common factors of 8, and then see if those factors would add up to 9. Luckily, 1,8 worked and so we replace the previous value of 9x with the factors to rewrite our equation. Our next step is to make sure that our equations have a common factor and then write in the coefficients into one of the cells, to finish off writing the equation in Factored form. A problem that might arise while trying to factor the equation along the way is that you have to be systematic and see which values will work, and which ones won’t. |
V. Solving problems with Quadratic Equations:
Although parabolas and quadratics may be an important part of our unit to learn, they aren’t just for the purposes drilling equations into our heads. They have some real-life equivalents that we learned about and that we now understand a little better. One of the first real world scenarios that we can connect quadratics to, would of course be Kinematics. In the same way that we know objects gain and lose velocity, this can be mapped and the results are exactly what a parabola is. For example, going back to the beginning of this unit, we began the project with the launch of a rocket that would have acceleration, velocity changes. Not only could we predict the path of the rocket, but we could find the the height of the rocket at any given moment throughout its entire trajectory. The equation that we originally started with, would actually turn out to be the Standard form of an equation for a parabola, something we hadn’t known until we came back to the original problem. In the image aside, by manipulating the original form and converting it into Vertex form, can answer the questions much more easily, and can find the x-intercepts that are beings asked for.
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Another great example of how we are able to apply our knowledge of quadratics is in calculating the maximum area of a certain rectangle or even using variables for unknown values to find the length of a side of a triangle. One problem that would require the usage of quadratics would be Leslie’s Flowers, what this problem asks is to find the total area of a triangle when we are missing important values. In order to solve the problem we had to set up the problem so that we would have a quadratic equation and then solve what was being asked. The real challenge in these types of problems is to take apart and to put back together, because if you can't do this, then it becomes very difficult to solve the problem.
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Not only do quadratics, apply to kinematics and area, but they can relate even to economics. This is because of how the price of an object has a proportional relationship with the quantity of sells. The cheaper the object, the more one will be able to sell of said product, and vice versa. When writing these types of problems out, the price will be a recurring variable which will result in a quadratic equation.
VI. Reflection:
Entering this unit at first, was worrying because I wasn’t sure about what we were going to do, and I felt that maybe I wouldn’t be able to understand the unit. And this was true when we started with Kinematics, because I was lost and didn't know how this would relate to our math unit. But I stayed confident, and patient, and soon enough it would be revealed on how this related, and what we would be learning about for the next month or so. As we progressed through the unit, it became much more easier to do the work and I was able to constantly keep up with the work that was being given. I didn’t want to just write down the answers, but to seek why they were as such and convince myself of how they worked.. Before having covered this section I had seen similar problems, but since I didn’t know what they were talking about, I would most likely answer them incorrectly. This makes me feel that we did learn something because I understand this portion of quadratics and I would be confident to answer similar problems that would require the understanding of the equations. As for the next stages in my life, moving onto 11th grade, and taking the SAT, I do believe that I am much more prepared than when I originally started the year. I have grown academically, and I would like to believe that I am slightly more capable of the work that is to come.