Introduction
When beginning this project, we used a variety of contexts like; Projectile motion, areas and volumes and the Pythagorean Theorem. We also learned about economics to develop our understanding of the quadratic functions and their representations. Our focus point when beginning this unit was to learn different methods for solving these equations. The central problem that really kicked off this project involved finding the height of a rocket at the peak of its journey, the time it reaches the peak, and the time it takes to reach the ground. We started out trying to solve this problem by reviewing kinematic equations and distance formula: h(t) = d0 + v0 · t + 1/ 2 a · t^2 . We plugged into the equation all of the information that was given in the project sheet, Such as the starting height, the acceleration (due to gravity), and the rocket’s initial velocity. Then we combined all our like terms with the resulting equation as: h(t) = 160 + 92t - 16t^2 Once we got to this point of the problem we really didn't have much previous knowledge to continue on solving the problem so we paused the problem in order to acquire the skills to continue solving the problem. Throughout the rest of the project we learned how to work with algebraic symbols to see how different algebraic representations could relate to the problem. We were able to see how writing quadratic equations in 3 ways; Vertex form, standard form, and factored form. Each form can help us find different parts of different problems( x & y intercepts and the vertex.) Later on, we used a graphing app Desmos, that helped us visualize and understand how the equation affects the parabola and how different parts of the equation controlled the directions the parabola goes in.
Exploring the Vertex Form of the Quadratic Equation
Vertex form is one of the ways the equation for a parabola can be written, it is the form we used when we were first introduced to this unit because it is one of the simpler ways to solve a problem. We worked on several different worksheets and Desmos ( online graphing calculator) in order to explore all the different parameters that affect the parabola. We also broke up the equation y = a (x - h)^2 + k into smaller variables, studying a, h, and k, and how the values affect the parabola. When we used Desmos we were able to see how adding A to an equation like y=x^2, turns to y=Ax^2. We also noticed how adding the A value effects the shape of the parabola. For example, a higher sum of A creates a narrow curve, and a lower sum of A creates a wider curve. Then we began adding K into the equation, and K is equal to the y-coordinate of the parabolas vertex when it is put in the equation. As we began adding values to the equation it started to become more complex and it created the new equation y=Ax^2+K. When we were looking at H we found that it was similar to K since it controls the x-coordinate for the equation. The final variable we added was h. H is equal to the x-coordinate value in the vertex of a parabola. Once figured that out, we found that our final vertex form equation was: y=a(x-h)^2+k. So, if we decided to plug in numbers into the equation, for example y= (x-2)^2 + 4, the h value would be 2. and the vertex would be (2,4). That completed our vertex form which created: y=A(x-H)^2+K. Slowly going through this process helped us understand what it meant to solve an equation in vertex form. We learned that if an equation is in vertex form you can usually automatically know what the values for h and k are, and if it will concave up or down.
By learning how to find the vertex form helped us find the peak of the rocket.
Other forms of the Quadratic Equation
The Standard form for a parabola is (ax^2 + bx + c = y). It's the most commonly known way to find for the parabola. It also solves for y-intercept value C.
The Factored form is a(x - p)(x - q) = y. This form is simpler to use because it's faster to solve for. When solving you know that the the x-intercept coordinate must have a y value of 0. You also know that one of the parenthesis must equal zero to make to equation work
Example: y = 2(x - 6)(x - 10) rewritten as 0 = 2(x - 6)(x - 10).
x = 6 and 10.
Converting between Forms
1. Vertex form to standard form
Using an area diagram, you multiply out the squared terms, distribute a, and combine like terms. When converting from vertex to standard form you can use an area diagram to help multiply out the squared terms, distribute a, and make sure to combine like terms.
2. Standard form to vertex form
Group the ax^2 and bx terms together and take a out; complete the square, filling in the missing term (but make sure to subtract the added term so the equation stays the same); take the subtracted term out of the parenthesis by multiplying it by a; rewrite the parenthesis as a product of the square; combine like terms outside of the parenthesis.When converting from standard form to vertex form you have to factor out a from the first terms and make sure to group the ax^2 and bx terms together and factor a out. Then we have to complete the square.
3. Factored form to standard form
You have to first multiply the terms in the parenthesis. Then combine like terms and distribute a. When you're going from factored to standard form you can use an area diagram to help visualize the factoring being done, but really all you need to do is foil (first, outside, inside, left) the terms in the parenthesis and combine all the like terms. Then distribute the a value to the whole equation to complete the equation.
4. Standard form to vertex form
Complete the square, write in new form. When you convert from standard to factored form, you can just complete the square, or in other words use the foil method in order to create the new and required standard equation. It's a relatively easy topic to master especially with enough practice.
When beginning this project, we used a variety of contexts like; Projectile motion, areas and volumes and the Pythagorean Theorem. We also learned about economics to develop our understanding of the quadratic functions and their representations. Our focus point when beginning this unit was to learn different methods for solving these equations. The central problem that really kicked off this project involved finding the height of a rocket at the peak of its journey, the time it reaches the peak, and the time it takes to reach the ground. We started out trying to solve this problem by reviewing kinematic equations and distance formula: h(t) = d0 + v0 · t + 1/ 2 a · t^2 . We plugged into the equation all of the information that was given in the project sheet, Such as the starting height, the acceleration (due to gravity), and the rocket’s initial velocity. Then we combined all our like terms with the resulting equation as: h(t) = 160 + 92t - 16t^2 Once we got to this point of the problem we really didn't have much previous knowledge to continue on solving the problem so we paused the problem in order to acquire the skills to continue solving the problem. Throughout the rest of the project we learned how to work with algebraic symbols to see how different algebraic representations could relate to the problem. We were able to see how writing quadratic equations in 3 ways; Vertex form, standard form, and factored form. Each form can help us find different parts of different problems( x & y intercepts and the vertex.) Later on, we used a graphing app Desmos, that helped us visualize and understand how the equation affects the parabola and how different parts of the equation controlled the directions the parabola goes in.
Exploring the Vertex Form of the Quadratic Equation
Vertex form is one of the ways the equation for a parabola can be written, it is the form we used when we were first introduced to this unit because it is one of the simpler ways to solve a problem. We worked on several different worksheets and Desmos ( online graphing calculator) in order to explore all the different parameters that affect the parabola. We also broke up the equation y = a (x - h)^2 + k into smaller variables, studying a, h, and k, and how the values affect the parabola. When we used Desmos we were able to see how adding A to an equation like y=x^2, turns to y=Ax^2. We also noticed how adding the A value effects the shape of the parabola. For example, a higher sum of A creates a narrow curve, and a lower sum of A creates a wider curve. Then we began adding K into the equation, and K is equal to the y-coordinate of the parabolas vertex when it is put in the equation. As we began adding values to the equation it started to become more complex and it created the new equation y=Ax^2+K. When we were looking at H we found that it was similar to K since it controls the x-coordinate for the equation. The final variable we added was h. H is equal to the x-coordinate value in the vertex of a parabola. Once figured that out, we found that our final vertex form equation was: y=a(x-h)^2+k. So, if we decided to plug in numbers into the equation, for example y= (x-2)^2 + 4, the h value would be 2. and the vertex would be (2,4). That completed our vertex form which created: y=A(x-H)^2+K. Slowly going through this process helped us understand what it meant to solve an equation in vertex form. We learned that if an equation is in vertex form you can usually automatically know what the values for h and k are, and if it will concave up or down.
By learning how to find the vertex form helped us find the peak of the rocket.
Other forms of the Quadratic Equation
The Standard form for a parabola is (ax^2 + bx + c = y). It's the most commonly known way to find for the parabola. It also solves for y-intercept value C.
The Factored form is a(x - p)(x - q) = y. This form is simpler to use because it's faster to solve for. When solving you know that the the x-intercept coordinate must have a y value of 0. You also know that one of the parenthesis must equal zero to make to equation work
Example: y = 2(x - 6)(x - 10) rewritten as 0 = 2(x - 6)(x - 10).
x = 6 and 10.
Converting between Forms
1. Vertex form to standard form
Using an area diagram, you multiply out the squared terms, distribute a, and combine like terms. When converting from vertex to standard form you can use an area diagram to help multiply out the squared terms, distribute a, and make sure to combine like terms.
2. Standard form to vertex form
Group the ax^2 and bx terms together and take a out; complete the square, filling in the missing term (but make sure to subtract the added term so the equation stays the same); take the subtracted term out of the parenthesis by multiplying it by a; rewrite the parenthesis as a product of the square; combine like terms outside of the parenthesis.When converting from standard form to vertex form you have to factor out a from the first terms and make sure to group the ax^2 and bx terms together and factor a out. Then we have to complete the square.
3. Factored form to standard form
You have to first multiply the terms in the parenthesis. Then combine like terms and distribute a. When you're going from factored to standard form you can use an area diagram to help visualize the factoring being done, but really all you need to do is foil (first, outside, inside, left) the terms in the parenthesis and combine all the like terms. Then distribute the a value to the whole equation to complete the equation.
4. Standard form to vertex form
Complete the square, write in new form. When you convert from standard to factored form, you can just complete the square, or in other words use the foil method in order to create the new and required standard equation. It's a relatively easy topic to master especially with enough practice.
Solving Problems with Quadratic Equations
There are many different connections we can make with real world problems, we used a couple different topics during this project that showed examples of that. As a class, we covered Kinematics (projectile motion), Geometry ( triangle problems and rectangle area problems), and Economics (maximizing revanue/profit or minimizing expenses/losses). One type of problem we were able to solve using a quadratic equation was a geometry problem, Emergency at Sea. In this problem, we needed to find the distance between a boat's landing point on shore and the closest lookout tower. The problem created a large triangle between the boat and the two lookout towers, then divided it into two smaller triangles with the path the boat takes to shore. To start this problem, I labeled the two triangles' sides with a, b, c and A, B, C. I then wrote Pythagorean's theorem out (a^2 + b^2 = c^2) because I knew I would have to use it and plugged in known variables. Then I ran into a problem because I was still missing two variables. However, I realized that since the a/A was the same length I could isolate them in the equations for the two triangles and make them equal to each other. So I used c^2 - b^2 = a^2 for both triangles which came out to be 550^2 - x^2 = a^2 and 800^2 - (1000 - x)^2 = a^2 for both of the equations. I renamed b as x since that was the variable I needed to find, and substituted B for 1000 - x, since I knew that b + B = 1000. Once I had those two equations, I set them equal to each other as 550^2 - x^2 = 800^2 - (1000 - x)^2 and solved for x. It's important for me to stay organized (habit of a mathematician) so that I didn't get confused with the numbers or with my own writing.
Reflection
This project went by pretty quickly and I felt that we were busy all throughout the project. I liked how there were many different parts to it and how it all came together in the end. Kind of like analyzing a painting, part by part and then looking at it whole makes sense to all the small parts. I feel that it was slightly challenging trying to go slowly because of the fact I just wanted to take it in all at once, and for that reason I had to be patient (habit of a mathematician). However this system was very helpful for me because it built on itself and allowed me to really understand where everything came from which helped when tackling more difficult problems. I had to collaborate and listen (habit of a mathematician) a lot during this unit, both working with others to figure out a difficult problem. I feel that these strategies I used for this unit will help me when taking the SAT because patience will keep me from making a mistake. I will continue to practice these strategies for years to come.
There are many different connections we can make with real world problems, we used a couple different topics during this project that showed examples of that. As a class, we covered Kinematics (projectile motion), Geometry ( triangle problems and rectangle area problems), and Economics (maximizing revanue/profit or minimizing expenses/losses). One type of problem we were able to solve using a quadratic equation was a geometry problem, Emergency at Sea. In this problem, we needed to find the distance between a boat's landing point on shore and the closest lookout tower. The problem created a large triangle between the boat and the two lookout towers, then divided it into two smaller triangles with the path the boat takes to shore. To start this problem, I labeled the two triangles' sides with a, b, c and A, B, C. I then wrote Pythagorean's theorem out (a^2 + b^2 = c^2) because I knew I would have to use it and plugged in known variables. Then I ran into a problem because I was still missing two variables. However, I realized that since the a/A was the same length I could isolate them in the equations for the two triangles and make them equal to each other. So I used c^2 - b^2 = a^2 for both triangles which came out to be 550^2 - x^2 = a^2 and 800^2 - (1000 - x)^2 = a^2 for both of the equations. I renamed b as x since that was the variable I needed to find, and substituted B for 1000 - x, since I knew that b + B = 1000. Once I had those two equations, I set them equal to each other as 550^2 - x^2 = 800^2 - (1000 - x)^2 and solved for x. It's important for me to stay organized (habit of a mathematician) so that I didn't get confused with the numbers or with my own writing.
Reflection
This project went by pretty quickly and I felt that we were busy all throughout the project. I liked how there were many different parts to it and how it all came together in the end. Kind of like analyzing a painting, part by part and then looking at it whole makes sense to all the small parts. I feel that it was slightly challenging trying to go slowly because of the fact I just wanted to take it in all at once, and for that reason I had to be patient (habit of a mathematician). However this system was very helpful for me because it built on itself and allowed me to really understand where everything came from which helped when tackling more difficult problems. I had to collaborate and listen (habit of a mathematician) a lot during this unit, both working with others to figure out a difficult problem. I feel that these strategies I used for this unit will help me when taking the SAT because patience will keep me from making a mistake. I will continue to practice these strategies for years to come.