Q^2: The Quintessence of Quadradics
Introduction
In this project, we learned about parabolas, quadratic equations, and how to use them in different situation. We started by learning about one of the things quadratics can be most helpful in: Kinematic equations. The initial problem stated that HTH was launching a rocket from a platform, we were asked to find the time and height of the rocket's apex, and the total time from launch that the rocket hit the ground. To solve this problem, our class created an equation with an input of 'time from launch' and an output of 'the height of the rocket'. To do this we thought about the physics using geometry. If we make a graph with a y of velocity and x of time, we find that the area under whatever graph we make will be the distance because of the physics equation d=vt. An initial velocity can be expressed as the y intercept, and an increase of velocity (aka acceleration) can be expressed as the slope of the line. To find the area under the line, we can split the graph into a rectangle and a triangle. The area of the rectangle will be the initial velocity multiplied by time. The area of the triangle will be half of the increase in velocity multiplied by time. Increase in velocity is expressed as acceleration multiplied by time, so writing this out we get "initialV*t + 0.5a*t*t" If we just call initialV v, and simplify, we can make the equation vt+0.5at^2. To further generalize we can add in the initial distance, so our final equation is d+vt+0.5at^2. The equation took into account the initial height of the rocket (since we're launching from a 160ft platform), the initial upwards velocity of the rocket (92m/s), and downwards acceleration of the rocket caused by gravity (-32m/s/s). The initial physics equation was d+vt+0.5at^2, after substituting out values the final equation was h=160+92t-16t^2. Upon rearranging the equation, it became apparent that we were actually dealing with a quadratic equation in the form of y=ax^2+bx+c. From here all we had to find is the vertex coordinates, and the positive x intercept of the equation.
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Exploring Vertex form
The equation we had made was certainly a start, but It didn't make any of the values we wanted very easy to get. we set the rocket problem aside and started learning about quadratics, with the intent to return to our initial problem at the end of the project. To get more information from the equation, it was necessary to convert to vertex form (explained further down the page). Vertex form is written as y=m(x-h)^2 +k, with the vertex of the equation being at (h,k). To explore how these values affected the equation we used handout packets and an online graphing calculator called desmos (seen in the picture to the left). When we input our equation into desmos, we realized that quadratic equations create parabolas when you graph them.
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Other Quadratic Forms
Throughout the project we learned about all three forms of writing quadratic equations, not just vertex form. The other two forms are called Standard and Factored form.
Standard form was the first type of quadratic equation we encountered (see the top of the page) and is written as ax^2 + bx + c. The a constant affects how "squished" the graphed parabola is, and the c constant is the y-intercept of the parabola. Unfortunately, the b constant doesn't affect the parabola in any easy-to-understand way. To seek why and prove that the b constant is so weird, you have to look at other forms of quadratics: turns out the b constant is where all the weird stuff from other forms gets stuffed when you're converting to standard form (which explains a lot). In conclusion, while the b constant isn't very helpful, standard form is still useful if you want to find the y-intercept or slope of the parabola. Factored form (sometimes called intercept form) is written as a(x-n)(x-m). The constants n and m are the x intercepts (aka the zeros) of the graph, and the a constant is (as usual) the "squished-ness" of the parabola. This form is useful for finding the zeros of a parabola, which turns out to be really useful for most problems involving quadratics. |
Converting between forms
Knowing what constants mean what is useful. But you will rarely be given an equation in the correct form. This means that if you want to get anything out of quadratics you need to know how to convert between forms. The short explanation on how to do this is "Algebra", the long explanation is as follows:
vertex to standard Whenever you're converting something to standard form, a good practice is to start simplifying until you get something you want. First we want to get rid of the ()^2 part. We do this by writing out the equation as it actually is and multiplying every number by every other number. We then use the distributive property to take care of the "a" constant and simplify again. standard to vertex To convert between standard and vertex we do the previous conversion backwards. First we factor out the "a" constant from the first two terms. Now we need to make the value inside the parentheses equal to some squared binomial. If you think about it, the second term should be 2 times the value of the constant other than x, times x; the third term should be the constant other than x squared. We already have the second term, and logic tells us that the third needs to be that second term divided by 2, then squared. We can't just add something to the equation, so we also have to subtract the same value and then multiply that subtraction out of out parenthesis (remember to multiply it by your "a" constant when you do so). From here we just simplify the inside of the parenthesis to it's squared binomial form. A word of advice when converting between standard and vertex: Stay Organized. It's easy to try and remember all the numbers in you're head, but it's a lot easier if you write everything down. factored to standard Converting factored form to standard form is just another matter of simplification. Multiply every number by every other number, then simplify. standard to factored In my opinion, this is probably the hardest conversion to do. Working backwards, we know we want out n and m values to multiply together to equal the "c" constant. We also know that if the n and m terms are added together they should equal the "b" constant, except that before they are added they need to be multiplied by values such that the values they are multiplied by are equal to the "a" constant. If you're lucky you might be able to factor the "a" constant out of the equation and reintroduce it later, but this isn't always the case. To make things worse, unless you use the quadratic formula (which we didn't cover in this project), you basically just have to make an educated guess and check to find the values you need. In other words, you need to Conjecture and Test. Speaking of Habits of a Mathematician, another useful one might be to "be confident, patient, and persistent", because this process could take some time. Once you find the values, just write out the factored equation and give yourself a pat on the back. |
Solving problems with quadratic equations
Quadratics show up in a lot of problems, here are a few examples we covered over the coarse of the project:
Kinematics For an example of a kinematic equation being solved using quadratics, we need look no further that the central problem of the whole unit: the victory celebration. At the beginning of the project, we converted the physics equation into quadratic standard form. Since we want the vertex coordinates and the positive x-intercept value, all we need to do is convert to vertex and factored form. Unfortunately our main challenge here isn't knowing what to do, it's doing the arithmetic with the numbers given. As it turns out this problem quickly forces you to deal with long decimal numbers. I used a calculator to convert to vertex form, which was easy enough, and then took the h and k constants as the time of maximum height and the maximum height respectively. Converting to factored form stumped me for a while (calculators can't really help with that type of logic problem). I eventually came to the conclusion that the n and m values I was looking for might be decimals. This would make the conversion nearly impossible, so I decided to try another method. I started from Vertex form, and since there was only one x in that form, all I had to do was algebra. Once again the numbers started getting ridiculous so I used a calculator for the arithmetic. As expected, I ended up with two answers (this occurred because I calculated for the square root of a number, which will ways be either the positive or negative version of some value). Me decision on which answer was correct was made easy because one was positive ant the other was negative. It wouldn't make sense that the rocket landed before it was launched, so the positive answer was clearly the one I was looking for. I checked my answer by making a graph in desmos, and both the graph and my worksheet can be seen to the left. Geometry We covered a lot of geometry problems during this unit. Most of them involve a situation where you need to maximize the area of a pen from a given amount of fencing. The catch is that one side of the fence is "free" (usually because you're sharing it with another fence-builder). This creates a quadratic equation because the area of a rectangle is base * width, so if the side length of the side of the pen perpendicular to your neighbor's fence is x, and the given total fence length you have available is F , the area of the fence will be x(F-2x). You can re-write this as a quadratic equation in factored form: x(-2+F). Economics Economic problems usually involve quadratic equations because, generally, the more expensive you make something the less people will buy. So, if you want to know the best price of an object you need to take into account the price of the object and how many you will sell, but knowing how many you'll sell depends on the price of the object. This will result in an equation in which the same variable appears twice, which almost always results in an x^2 term, which is a good sign you're dealing with a quadratic equation. |
Reflection
I came into this project thinking that I knew all there was to know about quadratics and parabolas. Instead of doing the regular project handouts, I spent most of my time doing the challenge options. Through those challenge options I realized how wrong I was. It turns out that no matter how many rules you know, you're never really prepared for a tough math question because the 'math' part of those questions are rarely the difficult part. In most of the problems I attempted, Most problems required you to Describe and Articulate some relationship, but the main challenge was figuring out what was even going on in the scenarios described. To solve these problems, I mainly used two habits of a mathematician: Taking apart and putting back together, and starting small. Instead of figuring out the optimal angle for a baseball shot all at once, I made separate equations for the time-in-air, and the horizontal speed of the ball, then used them both to find the answer. I'm also proud of my ability to look for patterns during this project. One problem asked me to find all y=1 values of a trinomial to the power of another trinomial. At first I was stumped, but then I realized that any positive number to the power of 0 is 1, and 1 to the power of any number is 1. All I had to do was solve each equation individually, and then use both answers. I also did a lot of generalizing during this project. One challenge option was to generalize the relationship between vertex and standard form. To do this, I had to convert between them, but using letter constants rather than numbers. The challenge problems were hard, but I collaborated with my teacher, Dr. Drew, whenever I got thoroughly stuck. In the end, I was able to complete every challenge option but two, this was because both problems involved a lot of guess and check that was to daunting for me to even start. I feel If I had been more systematic and dedicated I would've been able to complete these as well.