Introduction//
During this project we used our knowledge of projectile motion areas and volumes, the Pythagorean theorem, and economics. These skills were used to better develop our understanding of quadratic functions and their representations, a well as methods for solving quadratic equations. There were various handouts with problems to solve, they all corresponded to what we were learning that day. Once every week we would get an SAT warm-up that served as more practice apart from the practice we were already getting from the worksheets. What helped kick start our project was an overview in the beginning regarding kinematics which was,
(d= distance, v= velocity, t= time)
d=vt
v=d/t
t=d/v
Our class worked together to complete and help each other understand the various problems which would soon give us the answers to the essential question, What is a quadratic Equation?
Exploring the Vertex Form of the Quadratic Equation//
The handouts started off with a graphed parabola and a question asking to find the equation that makes up the design. Each day we would get a new worksheet that would challenge us more than the day before. Not only did we find the equation for each graph but we also worked on finding out what made the equation either positive or negative. For example, (y =(x + 1)² + 5) has the vertex above the x-axis, the parabola is concave up, and has no x-intercepts. We used a website called Desmos to help us classify if the vertex was above, below, or on the x-axis, the direction of the concave, and the number of x-intercepts the graph has. Vertex form has the power that gives you the information about a parabola in terms of a, h, k. Each letter represents a different thing.
a= steepness of parabola
h= intercept of the vertex on x axis
k= intercept of the vertex on y axis
Other Forms of the Quadratic Equation//
Other than vertex form there are two other forms of a quadratic equation known as standard form and factored form. Standard form is found by using the distance formula. An example of a quadratic equation in standard form looks like, 3x + 2x2 + 4 = 5. Next we looked at quadratic equations in factored form. Factored form is lined up as this… Y=a(x-p)(x-2)
Factored form on the other hand is very important because it gives you the x-intercepts of the parabola immediately. As you can see here,
y= -8(x-3)(x-7)
0= -8(x-3)(x-7)
x-3=0 x-7=0
During this project we used our knowledge of projectile motion areas and volumes, the Pythagorean theorem, and economics. These skills were used to better develop our understanding of quadratic functions and their representations, a well as methods for solving quadratic equations. There were various handouts with problems to solve, they all corresponded to what we were learning that day. Once every week we would get an SAT warm-up that served as more practice apart from the practice we were already getting from the worksheets. What helped kick start our project was an overview in the beginning regarding kinematics which was,
(d= distance, v= velocity, t= time)
d=vt
v=d/t
t=d/v
Our class worked together to complete and help each other understand the various problems which would soon give us the answers to the essential question, What is a quadratic Equation?
Exploring the Vertex Form of the Quadratic Equation//
The handouts started off with a graphed parabola and a question asking to find the equation that makes up the design. Each day we would get a new worksheet that would challenge us more than the day before. Not only did we find the equation for each graph but we also worked on finding out what made the equation either positive or negative. For example, (y =(x + 1)² + 5) has the vertex above the x-axis, the parabola is concave up, and has no x-intercepts. We used a website called Desmos to help us classify if the vertex was above, below, or on the x-axis, the direction of the concave, and the number of x-intercepts the graph has. Vertex form has the power that gives you the information about a parabola in terms of a, h, k. Each letter represents a different thing.
a= steepness of parabola
h= intercept of the vertex on x axis
k= intercept of the vertex on y axis
Other Forms of the Quadratic Equation//
Other than vertex form there are two other forms of a quadratic equation known as standard form and factored form. Standard form is found by using the distance formula. An example of a quadratic equation in standard form looks like, 3x + 2x2 + 4 = 5. Next we looked at quadratic equations in factored form. Factored form is lined up as this… Y=a(x-p)(x-2)
Factored form on the other hand is very important because it gives you the x-intercepts of the parabola immediately. As you can see here,
y= -8(x-3)(x-7)
0= -8(x-3)(x-7)
x-3=0 x-7=0
As you can see, the two x-intercepts are x=3 and x=7.
Converting Between Each Form//
Vertex to standard: The equation y= 2(x+1)² -5 is in vertex form but you need to have this equation in standard form. What you do is start off by writing down the squared part of the equation out, y= 2(x+1)(x+1)-5. What you do next is distribute the first x to the other x and to the 1, then you distribute the first 1 to the second x and second one for example (x+1) distribute to --> (x+1). What you will then see is, y= 2(x² +2x +1) -5, in this next step you are going to be distributing the 2 that is outside the parentheses to the inner part of the equation. 2 times x²= 2x2, 2x2= 4x, 2x1=2, 2-5=3, once you plug in these numbers your equation should look like this,
y= 2x² + 4x -3 and your work here is done, this is now standard form.
Standard to vertex: You see a standard form equation such as y= 3x² + 18x -1 and want to turn it into vertex form. What you do is is take 3x² and separate the 3 from x squared and divide 18x by 3 which is 6x. It the ends up looking like, y= 3(x² + 6x ). Now don’t be fooled by thinking the (-1) isn’t in the equation because it still is and you will see in a minute. So going on you need to divide what is inside the parentheses, 6, by 2 and square the answer. You do this to make a perfect trinomial. Continuing on, when you divide 6 by 2 you get 3 and you have to remember to square 3 which is 9. You add 9 to the equation and now have y= 3(x² + 6x + 9) but you still need to multiply 3 by nine which is 27, but in order to balance the equation it becomes -27. This is where the -1 comes back in, technically if we stuck with it it would be on the outside of the parentheses which is where the -27 is going to go so instead of having both on the outside you subtract them, and -27 - -1 = -28. Now you have y= 3(x² + 6x + 9) -28, but you use half of x and 6x which means you have y= (x +3) but you need to square it because you actually have two of them, so it is y= (x+3)² -28, this is now vertex form.
Factored to standard: You start off by seeing the equation, y= -8(x-3)(x-7). What you do now is multiply x by x which is x squared or x to the second power. This is because there are two x’s. Then you multiply the farthest number, which in this case is seven, by x, which ends up being 7x. You then do the same to 3, which then becomes 3x. Finally, you multiply -3 by -7 and get 21. Your equation now looks like, y= -8(x² + 7x + 3x +21). Now your next step is to add 7x and 3x which is 10x. The equation is y= -8(x² + 10x+21) and now you need to distribute the -8 to x², 10x, and 21. Once done distributing your equation should look like y= -8x² + 2x + 13. There you have it, y= -8x² + 2x + 13 is your standard form.
Standard to factored: You start off with an equation in standard form such as the equation, y= x² + 6x + 8 and you want to get it to factored form. What you do first is find factors of +8 that multiply to get 8 and add up to get 6. You know that the factors of +8 are 4, 2 and 4 + 2 is 6. So what you do now is write out your factored form which is the equation, y=(x+4)(x+2).
Solving Problems with Quadratic Equations//
In our daily life we come across real life problems that can be solved with quadratic equations but don’t know it. These problems are like kinematics, economics, and geometry. When we look at kinematics we think of physics, acceleration, velocity, and so on. Although a lot of problems can be solved by using quadratics. Let's say you have a kinematic based question asking about a rocket launch. It tells you the initial velocity, the angle it was launched as, how it was launched. The question asks you how far away from the launching pad, the rocket would land. One way you can solve this is through quadratic equations. You can create an equation with the information given to you in the beginning. Once you have your equation you can convert it into one of the three forms such as, standard, vertex , and factored form. From here you can solve the problem using algebra.
The next real world problem we come across is in economics. Imagine you are a succeeding executive of a donut shop and your shop sells about 300 of your local favorite donuts. You see how well these donuts are selling and want to make more of a profit than you are already making, the better profit. You originally say you want to sell them for 10 dollars each, but another shop is selling them for 5. You are a bit competitive and want to sell the most donuts compared to the other shop. This is where you bring out the quadratics equation. You would use the different forms while conjecturing and testing. From here you can find the numbers.
The last real world problem you could use quadratics is in geometry. An example is in area and perimeter. If a customer was trying to get their dream house built with a large dimension on a large piece of land they would want to know the how much space will be on the inside and out. If the customers already had some sort of idea of what it was going to look like, such as, 5,000-x for the width and 1,200 ft for the length. The constructor would have to use the numbers and solve for the inside dimension. From there you would conjecture and test based off your results. You could then switch from standard form to vertex and find the greatest dimensions.
Reflection//
This project took a lot of thinking and paying attention and I enjoyed learning about quadratic equations. Of course, just like when learning a new subject within a subject, everything starts off with needing to ask for help and understanding what you need to do but, once you get it you just go into every worksheet knowing what to do and feeling like you know what you are doing. Although there were many worksheets I managed to not lose any of my papers and keep them together. I kept my work organized and that falls under “keep your work organized” habit of mathematician. There were times where I did not quite understand which papers I needed to turn in but I managed to talk to my teacher and get back on track and getting all my papers turned in. It was quite fun getting to learn something haven’t learned, something new and I really appreciate all the help I got from my peers and teacher. Of course like anything else I have learned, there is area to improve on, not just quadratics but just everything that I have learned this year. I hope to keep on working with quadratics next year.