The Quintessence of Quadratics
Q to the Power of 2
Approaching this project about working and articulating the quintessence of quadratics, my goals were to learn the many ways to use quadratic functions. Also I wanted to develop more knowledge in graphing calculators, since I had never learned fully about using a graphing application, and overall exploring quadratic functions. Also another concept I wanted to grow in was learn how to articulate a quadratic equation, when given a word problem. When growing and getting stronger at these habits, I expected to feel comfortable or be proficient with rewriting quadratic expressions either in factored form or in vertex form. The project was launched with the upbringing of graphing a given equation, and be able to accurately describe the effects of changing the equation on the graph, and learning to use a calculator to create graphs of quadratic functions. Also to get a full understanding of quadratic functions and their representations, which was one of my expectations, we started with projectile motion, areas and volumes, and then got into more depth with parabolas. We cross roads and strengthened our knowledge of the quadratic equation by familiarizing ourselves with areas and shapes, this consequently led to the larger subject, which was deriving the quadratic function. A fun fact I learned was Babylonian mathematics as early as 2000 BC, could solve problems in quadratics relating the areas and sides of rectangles.
We obtained a series of handouts that had us exploring the A, H, and K parameters on the graphing application Desmos and how they affect the individuality of a parabola. (Such as the shape) The threshold of learning the reasoning behind a parabola was when I discovered the closer A is to 0, the wider the parabola and when A is greater than 0 or when A is less than 1, its below the X axis. So from this information I determined that, the A in the equation affects the width of the parabola. All this information was obtained while exploring Desmos and after consecutive tries, testing different substitutes for the variables, I learned H in the equation is the vertex coordinate on the x-axis. That led to me being resolute about X affecting how wide or narrow a parabola will be. Then I learned more about the variable X, which was graphs with a number substituted for H greater than 0 are on the positive X axis. Adding to the K variable, I learned that value for it determines the parabolas position on the y axis. This led to me determining that a negative X value gives you a facing down parabola, or known as "concave down".
We obtained a series of handouts that had us exploring the A, H, and K parameters on the graphing application Desmos and how they affect the individuality of a parabola. (Such as the shape) The threshold of learning the reasoning behind a parabola was when I discovered the closer A is to 0, the wider the parabola and when A is greater than 0 or when A is less than 1, its below the X axis. So from this information I determined that, the A in the equation affects the width of the parabola. All this information was obtained while exploring Desmos and after consecutive tries, testing different substitutes for the variables, I learned H in the equation is the vertex coordinate on the x-axis. That led to me being resolute about X affecting how wide or narrow a parabola will be. Then I learned more about the variable X, which was graphs with a number substituted for H greater than 0 are on the positive X axis. Adding to the K variable, I learned that value for it determines the parabolas position on the y axis. This led to me determining that a negative X value gives you a facing down parabola, or known as "concave down".
There are other forms to a quadratic equation, which we explored in this project, which are standard and factored form. Standard form of a quadratic equation is ax^ 2 + bx + c = 0 and it has a coefficient for the x^ 2 and a constant term alone. A cannot be equal to 0, but A and C can. Factored form is y=(x-a)(x-b) and the benefit from going to factored form from standard form is when you graph the standard equation, you can find out the x intercept, thats not provided in the standard form equation. But once you convert it to factored form, you can obtain the x intercept (s) and have it in your equation (Shown on the graph as well), contrasting to standard form (But does show on the graph). For example if you had the standard form equation x^ 2 +8x+16=y, then wanted to convert it to factored form, you could easily find the x intercepts. So when you make the square table to factorize it, you achieve factored form y=(x+4)(x+4). Some benefits from the quadratic equations are:
Vertex form allows you to easily find the vertex.
Factored form allows you to easily find the x-intercepts.
Standard form allows you to easily find the y-intercept. Also, if it is not factorable, then it is in the correct form to apply the quadratic formula to find the x-intercepts.
Vertex form allows you to easily find the vertex.
Factored form allows you to easily find the x-intercepts.
Standard form allows you to easily find the y-intercept. Also, if it is not factorable, then it is in the correct form to apply the quadratic formula to find the x-intercepts.
The first equation on DESMOS is a factored form equation, the second is standard form, and the third is vertex form.
Converting Between Forms
1. Vertex form to standard form.
Vertex form to standard form would consist of our vertex form equation, y=(x+4)^2 and we want to achieve the equation ax^ 2 + bx + c = 0.
So you would first square (x+4) and get y=(x+4)(x+4). Then, you would complete the square and multiply getting you y=(x^2+4x+4x+16).
You would finally combine like terms ultimately getting the standard form equation, y=(x^2+8x+16). Also 16 would be your Y intercept!
3. Factored form to standard form.
Lets say we had the factored form equation y=(x+4)(x+4) and we wanted to convert it to standard form. We would need to distribute and multiply X and 4, X and 4, along with 4 and 4, to achieve the perfect square. So we would end up with y=(x^2+4x+4x+16). Then once we combine like terms, we achieve our factored form equation to standard form, y=(x^2+8x+16).
4. Convert from standard form to vertex form.
Lets say we have the standard form equation x^2+8x+16. You would then make a square and factorize everything from the square method. So we got the final term in vertex form from finding in the square to include 4^2 and 16 because we need everything to in the equation to be equivocal to the square. So from x^2+8x+16, it would go to x^2+8x+4^2+16-16.
Then we achieve the final value by factoring everything.
So the standard form to vertex form equation is y=(x+4)^2.
5. Convert from factored form to vertex form.
Suppose we had the factored form equation y=(x+4)(x+4) and we wanted to convert it to vertex form, which is y=a(x=h)^2+k and the first thing to do is distribute the values in the two brackets and then combine like terms. Then, to accomplish a perfect square with the square method. Then that would result in 8x getting divided by 2, which equals 4^2 and then that would be 16.
y=x^2+4x+4x-16
y=x^2+8x-16-16 (From this you would half 8x then square it, receiving 16)
Then finally you would get the equation, after working with the values in PEMDAS order-
y=(x^2+8x)-32
The area diagram definitely helps me through this process and helps me quickly determine in my mind to figure out the perfect square when given an equation. I always remember now how to find “distribution of multiplication over addition”, and that achieves“completing a perfect square” and I have a instant thought now when seeing a equation to “factor the quadratic” instantly. Its essential for others to see this importance so they could have that reoccurring consistent thought that will be there to help you when using quadratic equations in the future.
Vertex form to standard form would consist of our vertex form equation, y=(x+4)^2 and we want to achieve the equation ax^ 2 + bx + c = 0.
So you would first square (x+4) and get y=(x+4)(x+4). Then, you would complete the square and multiply getting you y=(x^2+4x+4x+16).
You would finally combine like terms ultimately getting the standard form equation, y=(x^2+8x+16). Also 16 would be your Y intercept!
3. Factored form to standard form.
Lets say we had the factored form equation y=(x+4)(x+4) and we wanted to convert it to standard form. We would need to distribute and multiply X and 4, X and 4, along with 4 and 4, to achieve the perfect square. So we would end up with y=(x^2+4x+4x+16). Then once we combine like terms, we achieve our factored form equation to standard form, y=(x^2+8x+16).
4. Convert from standard form to vertex form.
Lets say we have the standard form equation x^2+8x+16. You would then make a square and factorize everything from the square method. So we got the final term in vertex form from finding in the square to include 4^2 and 16 because we need everything to in the equation to be equivocal to the square. So from x^2+8x+16, it would go to x^2+8x+4^2+16-16.
Then we achieve the final value by factoring everything.
So the standard form to vertex form equation is y=(x+4)^2.
5. Convert from factored form to vertex form.
Suppose we had the factored form equation y=(x+4)(x+4) and we wanted to convert it to vertex form, which is y=a(x=h)^2+k and the first thing to do is distribute the values in the two brackets and then combine like terms. Then, to accomplish a perfect square with the square method. Then that would result in 8x getting divided by 2, which equals 4^2 and then that would be 16.
y=x^2+4x+4x-16
y=x^2+8x-16-16 (From this you would half 8x then square it, receiving 16)
Then finally you would get the equation, after working with the values in PEMDAS order-
y=(x^2+8x)-32
The area diagram definitely helps me through this process and helps me quickly determine in my mind to figure out the perfect square when given an equation. I always remember now how to find “distribution of multiplication over addition”, and that achieves“completing a perfect square” and I have a instant thought now when seeing a equation to “factor the quadratic” instantly. Its essential for others to see this importance so they could have that reoccurring consistent thought that will be there to help you when using quadratic equations in the future.
Solving Problems with Quadratic Equations
While taking part in this project and exploring quadratic equations I learned there are three types of problems we can solve using quadratic equations.
Kinematics which is projectile motion and working with motion.
You will need to use quadratic equations in kinematics when the kinematic question involves a parabolic problem that requires a quadratic equation to find the solution. Whenever you are asked to describe the motion of an object without worrying about the cause of that motion, you have a kinematics problem, which the involvement of quadratics.
You will need to use quadratic equations in kinematics when the kinematic question involves a parabolic problem that requires a quadratic equation to find the solution. Whenever you are asked to describe the motion of an object without worrying about the cause of that motion, you have a kinematics problem, which the involvement of quadratics.
Geometry which is triangle problems and rectangle area problems.
We can apply quadratic equations to Geometric Applications because they can be used to solve those problems that involve area by its formula cooperating with values given.
Economics which is maximizing revenue/profit or minimizing expenses/losses.
Quadratic equations are used in economic equations to represent both the production cost function and the revenue function. Also quadratic function helps us articulate the parabola of the data between cost, revenue, and demand.
We can apply quadratic equations to Geometric Applications because they can be used to solve those problems that involve area by its formula cooperating with values given.
Economics which is maximizing revenue/profit or minimizing expenses/losses.
Quadratic equations are used in economic equations to represent both the production cost function and the revenue function. Also quadratic function helps us articulate the parabola of the data between cost, revenue, and demand.
Leslie's Flowers- Problem
In this problem that we used in our project, it asked us to recall the pythagorean theorem and when using that formula, we would have to solve a quadratic equation. The problem then discussed that architect Leslie needs to know the area of the triangular flower bed. (Image to the left)
One part of finding that area involved figuring out where the altitude meets the base of the triangle, in other words finding the value of x. But first we needed to write an equation that allow us to find x and that equation was h^2= 13^2-x. However that equation was only part of the actual equation, and I found that part by utilizing the theorem and splitting the flower bed in half. Then I took the values for each triangle and made two parts of the equation for each of the 2 parts of the triangle bed. We wanted to achieve an equation that replicated this, h^2=h^2.
a^2+b^2=c^2
h^2+(14-x^2)=15^2 Then we distribute.
x^2+h^2=13^2
Then equalling,
h^2=13^2-x^2
h^2=15^2-(14-x)^2
h^2=h^2
13^2-x^2=15^2-(14-x)^2 We have now achieved our equation from combined the two parts from the flower bed that was all derived from the theorem.
So from this equation-
13^2-x^2=15^2-(14-x)^2
We must then factorize all the values and then distribute.
169-x^2=225-(x2-28x-196)
Then add x^2 from both sides. Canceling them out from both sides of the equation.
169=225+28x-196
Adding like values.
169=29+28x
Minus 29 from both sides. 169 then changing to 140.
140=28x
Once you divide 140 by 28x, you derived the value, 5, for x!
X=5
This picture's significance is to show how I split the flower bed into two sections. To achieve h^2=h^2.
Reflection
I believe that working with patterns, a habit of a mathematician, definitely applies and has helped me with quadratic equations, along with completing the square every time. I feel as if I've grown with this constant pattern of checking to see if my equation is equivalent to the square. Also being systematic, along with starting small while converting between formulas has been key and apparent during this project. I believe that I learned a lot of subsequent knowledge that leads me into 11th grade and I have also grown my work ethic because of how determined and set I was on being absolutely familiar with the math I was learning. I have realized the importance of your work ethic especially with valuable equations that are critical to learn for your future. Now I can be working independently with these equations and articulating what they mean. When making this DP reflection, I am extraordinarily relieved and happy I stayed organized and saved all of my notes and hand outs that were a crucial part in my learning. Within my notes were key elements to this project I wrote down that I wanted to make sure I put in my reflection and because of this habit, I was able to describe and articulate everything I learned with the quintessence of quadratics project. Conjecture and test was a huge part of this project as well because of how many times I thought I knew the right answer to complete the square, and other times I would rush and believe I knew exactly the correct value to variable, even though I was sometimes incorrect. However I know this habit was blatant because there were many times I conjectured while converting quadratic equations, but I learned from my failures and successes. From conjecturing and testing while solving and converting quadratic equations, the habit stay confident derived from that process of the project because of how many times I grew from my errors. Another important habit I remember from this project was starting small because it was prudent to start small during the square method, or else you would overcomplicate or rush, that would only lead to a subpar answer. Also collaborating and listening benefited me tremendously when I was so overwhelmed by a stump in my process of solving an equation, and once I collaborated my process was improved, along with my future work. Then, generalizing was a prudent component when articulating the overall concept of quadratics because it made each equation seem more approachable and I was more keen towards noticing crucial patterns that each equation differentiated with. Overall, I believe I had a huge work ethic that was thriving while we experimented with our handouts and tackling quadratics. I have grown so much as a mathematician, especially with organizing my thoughts and work. This consequently enables me to work harder for the SAT and strengthen my habits for college readiness.