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The Quintessence of Quadratics
This project was designed to help us understand mathematical models, quadratics, and their applications.
We started with kinematics problems that required deriving and using the displacement formula.
We were able to identify the area underneath a velocity/time graph as the displacement and derive this equation.
It required taking apart and putting back together the shapes made underneath the line.
We started with kinematics problems that required deriving and using the displacement formula.
We were able to identify the area underneath a velocity/time graph as the displacement and derive this equation.
It required taking apart and putting back together the shapes made underneath the line.
The displacement formula is a quadratic, thus beginning our exploration of how to solve, represent, and interpret
equations where the variable is squared. Our class moved from simple motion problems through using standard,
vertex, and factored forms to solve quadratic problems involving geometry and economics. Each assignment was a building block for the next, so we had to stay organized and remain confident, patient, and persistent.
equations where the variable is squared. Our class moved from simple motion problems through using standard,
vertex, and factored forms to solve quadratic problems involving geometry and economics. Each assignment was a building block for the next, so we had to stay organized and remain confident, patient, and persistent.
Exploring Vertex Form
Our class was prompted by a series of worksheets to use a graphing app to understand the effects of
a, k, and h on a parabola in the form y=a(x+h)^2+k. We had to look for patterns as we assigned different values for the constants. Looking at one variable at a time was a helpful way to be systematic and start small. We found that vertex form allows the viewer to
quickly identify the coordinates of the vertex. We were asked to make posters to help us describe and articulate the meaning of each constant. These posters are shown below.
a, k, and h on a parabola in the form y=a(x+h)^2+k. We had to look for patterns as we assigned different values for the constants. Looking at one variable at a time was a helpful way to be systematic and start small. We found that vertex form allows the viewer to
quickly identify the coordinates of the vertex. We were asked to make posters to help us describe and articulate the meaning of each constant. These posters are shown below.
THE FORMS
Quadratics can be expressed in vertex form, standard form, and factored form. Each highlights different
key information. Our project involved solving problems that required eall of these forms.
key information. Our project involved solving problems that required eall of these forms.
Vertex | y=(x+4)^2-4
Standard | y=x^2+8x+12Standard form is a compact way of expressing a quadratic that allows the viewer to identify the coefficient. It is written as y=ax^2+bx+c .
|
Factored | y=(x+2)(x+6)Factored form represents the two x coordinates of the x intercepts. It is written as
y=a(x-r)(x-s) . |
Conversions
Vertex to Standard
I participated in a challenge option. A partner and I were asked to find a formula we could use to
convert vertex form to standard form. We did this by converting to standard form without substituting numbers for the constants, this helped us to generalize. We collaborated and listened to each others ideas. He eventually found the form by conjecturing and testing and I had to seek why and prove the result. I was also able to support this algebraically. The following shows how it can be used.
convert vertex form to standard form. We did this by converting to standard form without substituting numbers for the constants, this helped us to generalize. We collaborated and listened to each others ideas. He eventually found the form by conjecturing and testing and I had to seek why and prove the result. I was also able to support this algebraically. The following shows how it can be used.
Standard to Vertex
Factored to Standard
Factored to Vertex
Applications
Kinematics - Projectile Motion
The first word problem we were given involved a rocket that had to be launched from a platform. We were given the quadratic equation h(t)=-16t^2+92t+160 to represent it's height as a function of time. A parabola illustrates the motion of a projectile when the force of gravity is upon it. The path of the rocket looks like this;
The parabola begins before the start of recorded time because of the starting platform height, 160 feet. It goes up to it's peak, the vertex, before returning to the ground. We were tasked with finding the the vertex (the rocket's highest point) and the last x-axis (the point when the rocket returns to the ground. We were able to find the vertex by converting the equation to vertex form. Additionally, the x-axes can be identified in factored form.
Geometry - Triangles and Rectangle Area Problems
Quadratics also have geometric applications. We looked at a situation where we were asked to use a quadratic to find the maximum area of a pen with 500 feet of fencing for use on 3 sides. The basis of this problem is the equation for the area of a rectangle, A=l*w . I started by writing length in terms of width. 500-2w or -2w+500 The result is having to multiply the width variable by this expression, which makes it a quadratic equation. This is how we developed the equation A(w)=-2w^2+500w . We knew that the point where w = l (forming a square) would maximize the area of the pen, and were able to prove this using the equation.
We originally solved this through educated guesses about the most effective rectangle. Now that I know how to convert to vertex form, I went back to apply this method. I used the equation that my partner and I developed in the challenge option. Once converted, h and k represented the x and y coordinates of the vertex. This shows that the maximum area the pen can have is 31250 as a result of a width of 125.
Economics - Maximizing Profits
The economics applications we were introduced to involved maximizing profits from a product. The business had developed an equation to find the ideal balance between making a good profit per item and making the cost as reasonable for the public as possible. The equation was P(d)=d(1000-5d). It makes sense to represent the situation as a quadratic because if the company doesn't charge enough, they will have too low a profit. If they charge too much, no one will buy the product. The vertex represents the ideal price where they can get the most money and sell the most. The graph below shows that the ideal price is 100 dollars.
Reflection
I am proud of the work I did in this project. I collaborated with others, challenged myself, and learned to show my work in a clear and understandable way. I practiced self advocacy by speaking up when I was confused, and I am now confident in my ability to solve problems involving quadratics. This will help me with the SAT, as I have already solved SAT practice problems on Khan Academy that I was unable to solve before. I regret not keeping better track of my work throughout the semester. That would have helped me with this DP update immensely. As I move into 11th grade, I will remember the importance of keeping papers organized, even after they are graded. Overall, I am confident that I will be a better 11th grade math student because of this project.