Introduction
We started this project with the goal in mind to learn about not about quadratic equations but their relation to what we’ve already learned (Geometry) and about how they can be used to describe motion in physics and related fields. We kicked of the project by creating graphs of distance, time, and velocity. It showed how the area underneath the lines on the graphs showed meant different relationships.
Knowing the area relations and using rate equations like for example: Area of Square=Length(Width) We were able to find the Displacement Equation or d=(v)t+1/2a(t*t) The (V)T is the original starting point’s area so if on a velocity time graph rectangle (red) while the rest from the triangle(blue)
Last Problem
The last problem we were presented before fully exploring the topic was about a rocket being launched. The problem was that a rocket was launched from a platform 160 ft in the air. We were tasked with porting it’s path and then finding the answers to 3 questions. What was it’s maximum height, how long would it take to reach that point? And, How long until it hits the ground. We got as far as to getting an equation that plotted that path. H=height T=time h(t)=-16t(t)+92t+160. However we did not have the knowledge about quadratics we now have that would allow us to manipulate the equation to get our answers.
Vertex Form of Quadratic Equations
Taking a break from the rocket problem, we wanted to start looking at quadratic equations and what different parts on an equation effect. So we started with the simplest quadratic equation y=x^2. (Graphed Below)
The Variables
(of Vertex Form)
For a more complex but use full quadratic equation is one where you can find the vertex of the parabola, this quadratic equation however has three variables commonly known as H, K, and A. In an equation they look like: Y=A(X-H)^2+K. To discover what each variable does to the parabola we solved a series of problems were we looked at the variables as well as used a graphing website called desmos that allowed use to create sliders and directly edit the variables values to show us what lowering and raising the numbers would do.
H and KGrouping these variables together makes sense as compared to A they have very similar effects on the parabola. Through our experimentation we found that H is the x coordinate for the vertex of the parabola, this means by increasing it or decreasing it shift the entire graph. K works in a similar way, affecting the Y value of the vertex up or down, which in turn affects all of the points. First graph below shows changing H. Second shows changing K.
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AThe variable A effect not the position of the vertex, but instead the concavity of the parabola. However the relationship between how large A and the width of the parabola is inverse. This means as the A gets larger and larger the narrower the parabola is. There is one rule to A, and that is that it can’t equal zero because if it does then the parabola has no arch and is actually just a line. But if the A is a negative number then the parabola is reflected and instead of arching upwards to the positive Y axis it extends into the -Y axis. In the image below the purple, blue, and yellow show different A values. The Green line shows a negative A
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Other Forms of Quadratic Equations
While having the vertex form of an equation is incredibly useful, it is not the only equation. There are two other main forms for an equation. These are Standard form and Factored form. Each look a little bit different and serve a different purpose.
Factoredy=a(x-p)(x-q) |
StandardY=ax^2+bx+c |
Advantages
The advantage to this form is similar to the use of the vertex form. While vertex forms tell you the location of the vertex the Factored form shows you the the X-intercepts of the parabola. However if the graph doesn’t have the an x intercept or only one then you can’t use factored form. To find the X-intercepts, you look at the (q) and (p) in the equation. If the equation can use factored form then you know the location of the two intercepts will be (p,0) and (q,0).
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The advantage to this form is different then factored and vertex. It’s first use is that it allows you to convert between vertex and factored forms allowing you to find both the x intercepts and the vertex allowing you to easily graph the equation. It also allows you to use the axis of symmetry to find different x coordinates in the form of x=-b/2a.
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Below is an image showing how an equation in all three forms still give the same graphed line.
Converting Between Forms
Listed below are four step by step tutorials to switch between the forms using example problems.
Vertex to Standard |
Standard to Vertex |
Standard to Factored |
Factored to Standard |
Factored to Vertex or Vice Versa
Simply Convert to Standard Form and Fallow Steps From There.
Problems You Use This For
Although the amount of real problems these equations can apply to are almost limitless they can be summed up into three distinct groups.
Kinematics: Projectile motion, for example you want to know about the path of a baseball flying through the air.
Geometry: Dealing with area’s and triangles this could be used to find the given length of a triangle's side or how to maximize area given a certain perimeter.
Economics: These equations are used to maximize revenues and minimize losses, for example given a certain amount of product setting price points for the products while seeing what would maximize profit.
Now for a detailed example of a problem. I will be using a problem called Leslie’s Flowers For this problem we are given a triangle with all three sides however we don’t the height. Diagram below.
Kinematics: Projectile motion, for example you want to know about the path of a baseball flying through the air.
Geometry: Dealing with area’s and triangles this could be used to find the given length of a triangle's side or how to maximize area given a certain perimeter.
Economics: These equations are used to maximize revenues and minimize losses, for example given a certain amount of product setting price points for the products while seeing what would maximize profit.
Now for a detailed example of a problem. I will be using a problem called Leslie’s Flowers For this problem we are given a triangle with all three sides however we don’t the height. Diagram below.
Knowing Pythagoras theorem A^2+B^2=C^2 we can make a system of equations that looks like:x^2+B^2=13^2 and (14-x)^2+B^2=15^2.
The next step for me was using algebra to rearrange the first equation to solve for B^2. This changed it from :x^2+B^2=13^2 TO (13^2 or 169)-x^2=B^2
Then I changed (14-x)^2+B^2=15^2 from vertex form which it is into Standard form as well as simplifying the 15^2. This gave me x^2-28x+196+b^2=225 I then subtracted 196 by both sides giving me; x^2-28x+b^2=29
For the next part I started by substituting the b^2 in the second equation for 169-x^2 giving me, x^2-28x+169-x^2=29. I then simplified the equation by getting ride of the x^2 and the -x^2 as they cancel each other out. That gave me -28x+169=29 from which I subtracted 169 from both sides. The equation is now -28x=-140. The final step is to divide each side by -28 giving me x=5. With x=5 I am able to plug it into the x^2+B^2=13^2 giving me 13^2-5^2=B^2 or 169-25=b^2 So 144=B^2 and by square rooting both sides you get B=12. Last but not least to find the area of the triangle you take l/2 the base times the height or ½(14)(12)=area So the area is equal to 84 units.
The next step for me was using algebra to rearrange the first equation to solve for B^2. This changed it from :x^2+B^2=13^2 TO (13^2 or 169)-x^2=B^2
Then I changed (14-x)^2+B^2=15^2 from vertex form which it is into Standard form as well as simplifying the 15^2. This gave me x^2-28x+196+b^2=225 I then subtracted 196 by both sides giving me; x^2-28x+b^2=29
For the next part I started by substituting the b^2 in the second equation for 169-x^2 giving me, x^2-28x+169-x^2=29. I then simplified the equation by getting ride of the x^2 and the -x^2 as they cancel each other out. That gave me -28x+169=29 from which I subtracted 169 from both sides. The equation is now -28x=-140. The final step is to divide each side by -28 giving me x=5. With x=5 I am able to plug it into the x^2+B^2=13^2 giving me 13^2-5^2=B^2 or 169-25=b^2 So 144=B^2 and by square rooting both sides you get B=12. Last but not least to find the area of the triangle you take l/2 the base times the height or ½(14)(12)=area So the area is equal to 84 units.
Reflection
Personally for me this wasn’t a project it was a long and well thought out lesson. I say this for one main reason as there was no real product as all projects have some form of product. This does not meen that I didn’t learn from this. This was a topic I was not well versed int but essential to higher math concepts. So as a building block I feel like it has provided a stable footing to move forward into later math courses. I feel like this whole unit was build around 3 base habits with others sprinkled in such as organization, and working as a group. These were to find patterns, To visualize those pattern, and to Generalize those problems. And we did this in a variety of ways starting by starting small at the simplest quadratic equation then slowing and systematically made our equations more and more complex. Discovering what certain variables did and didn’t do through a combination of visualization and algebra, allowing us to generalize our own rules. I personally feel that while some parts were boring or felt repetitive, I truly learned this unit and will most likely use its core principles to better understand future units and courses I take.