Intro to Quadratics
A quadratic equation is an equation involving the term x^2. It can be written as a binomial and is used in kinematic equations. We started off the project with a kinematic equation that described the height of a small rocket launched off the roof of a building as a function of time. The kinematic equation presented for this was h(t)=-16t^2+92t+160, which is, of course, a quadratic equation of the form y=ax^2+bx+c. After further analysis you can surmise that in the kinematic equation the coefficient A describes the force of gravity on the rocket, B describes the rocket's initial vertical velocity and C describes the rocket's initial height. Our original goals in this project were to find how high the rocket would go, how long it would take for it to get there, and how long until the rocket crashes into the ground. To solve these problems we would have to convert the equation into other forms.
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Vertex Form
The vertex form of a quadratic equation is written as y=a(x-h)^2+k, where the vertex of the parabola is (h,k). It is very useful for finding the vertex of a parabola, because otherwise you would probably have to find the x value in between the two x intercepts and solve from there, but sometimes there are no x intercepts so you would have to deal with imaginary numbers which are needlessly complicated, so it's usually best to just convert to vertex form and save yourself a headache.
Other Forms (Standard and Factored)
The standard form of a quadratic is y=ax^2+bx+c. It is useful for finding the y intercept of the parabola, which is always C and, as shown earlier, it is useful when creating kinematic equations because you can describe forces using the other coefficients.
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The factored form of a quadratic is y=a(x-p)(x-q), where p and q are the x intercepts. We know they are the x intercepts because in a x intercept the y value is 0, so either (x-p)=0 or (x-q)=0, therefor, the x intercepts are x=p and x=q. You can also find the vertex from factored form by finding the center point between the x intercepts (the x value of the vertex) and using that value to find y.
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Converting Between Forms
Factored to StandardConverting a factored quadratic to a Standard form quadratic can be done by distributing one of the factors into the other (this is sometimes called FOIL for binomials). For an equation y=(a+b)(c+d) you would multiply every term by each term in the other factor, resulting in y=ac+ad+bc+bd. When doing this with a quadratic in the form y=a(x-p)(x-q) you get
y=ax^2-axq-axp+apq |
Standard to VertexTo convert a quadratic of the form y=ax^2+bx+c to y=a(x-h)^2+k you need to complete the square for the standard form. Leaving the c term alone, you factor out the A coefficient from ax^2+bx and then add and subtract a value from the factored equation so your equation looks like y=a(x^2+bx+d-d)+c where x^2+bx+d can be written as a squared term (x-h)^2. You then multiply the
-d by a and add that to c to get k. Now the equation is in Vertex form y=a(x-h)^2+k |
Standard to FactoredYou can factor a standard quadratic in 2 main ways, by guessing and and experimenting with random potential factors or by using the quadratic formula. For an equation y=x^2+bx+c you might be able to guess the factors (x+p)(x+q) by recognizing that p*q=c and p+q=b. However, if you don't feel like doing that you can use the quadratic formula
(-b+-sqrt(b^2-4ac))/2a, where a, b and c are the same constants from y=ax^2+bx+c and the equation is equivalent to the x intercepts of the parabola. You can prove this equation is correct by describing the factored form and standard form of a quadratic with the same constants and using equalities to write the intercepts in terms of those constants. |
Vertex to StandardTo convert from vertex form y=a(x-h)^2+k to standard form you just expand the equation so it contains no parentheses. You do this by "FOILing" the (x-h)^2 term and multiplying that by a. Combine like-terms and the equation should now be in the form y=ax^2+bx+c
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Solving Problems
Problems involving quadratic equations come in all shapes and sizes, but the larger word problems we did in class were mostly about kinematic equations, geography/geometry or economics.
Kinematic ProblemsA good example of a problem involving a kinematic equation is the one I covered earlier, where we had to find the maximum height, the time until maximum height and the time until landing for a rocket launched off the roof of a building whose height follows the kinematic equation h(t)=-16t^2+92t+160. We know that the function is a parabola, so we can simplify the question to know that we need to find the coordinates of the vertex and the positive x-intercept. We can do this by converting the equation to vertex form to find the vertex, and factored form to find the x-intercept. By completing the square we know that, when written in vertex form, the equation is
h(t)=-16(t-2.875)^2+292.25, so we know that the maximum height of the rocket is 292.25 and the time until it reaches that height is 2.875. The equation can't be factored into nice looking numbers, but by using the quadratic formula we can find that the positive x intercept is roughly equal to 7.14883. |
Geography/Geometry Problems
A lot of the problems we did involved some sort of geometry, ranging from finding the distance of a boat of a coast based on it's distance from two lighthouses to finding the area of a shave a variable side lengths. An example of this would be the problem shown to the right, where we were tasked with finding various bits of information about a cattle pen built against a preexisting fence. The person building the fence has a set amount of fencing they can use, and we are tasked with using equations to find the best way they can set up their corral to have the most area. You can do this by writing an equation describing the area of the corral based on a side length and then converting it to vertex form to find the maximum area.
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Economics Problems
In one of the problems we did we had to find the best price for "Widgets" based on a prediction of how many will be sold based on their price. If the company started with 1,000 widgets, it is predicted that they will sell 1000-5d widgets where d represents dollars. This means that the amount of money earned by the company based on the cost charged for each widget is M(d)=d(1000-5d) or M(d)=-5d^2+1000d. To find the best value to price the widgets at, we again re-write the equation in vertex form
M(d)=-5(d-100)^2+50,000 so we know we should sell 500 widgets at 100 dollars each for a total of 50,000. In this question we were also asked what we thought of the prediction, I think that it is oversimplified; the amount of widgets bought based on the price is probably not linear and the equation doesn't include any information about the price of manufacturing each widget.
M(d)=-5(d-100)^2+50,000 so we know we should sell 500 widgets at 100 dollars each for a total of 50,000. In this question we were also asked what we thought of the prediction, I think that it is oversimplified; the amount of widgets bought based on the price is probably not linear and the equation doesn't include any information about the price of manufacturing each widget.
Geometrical definition of a Parabola
To define a parabola in geometry you have to start with single point, called the focus, and a line, the directex. The line can have any slope, but if you want the parabola to represent a function it must be horizontal. The focus and directex can be anywhere on the coordinate plane. A parabola is defined as all the points that are an equal distance from both the focus and the directex. You can easily convince yourself that the point directly in between the focus and directex is the vertex of the parabola, but it is a little bit harder to prove that the geometrical definition of a parabola is the same curve described by y=ax^2+bx+c. You can do this by writing a new equation for a parabola based on the geometric equation, and simplifying from there. On the paper to the right, I proved that the a term in standard form is always equal to 1/(4f) where f is half the distance between the focus and directex.
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Reflection
I took Pre-Calculus 1 last summer, so I already knew a lot of the stuff we learned in this project. However, there were a few proofs and identities that I did not learn until this project, such as the proof of why the quadratic formula works. I had also somehow managed to go through all of Algebra and Pre-Calculus without learning anything about Vertex form, so I'm glad I learned that. For most of the project though, I already knew what was being taught, so I instead did all of the challenge problems which involved much larger polynomials. I think I worked pretty hard in this project, I managed to finish all of the provided challenge problems and some extra proofs as well as participate in a lot of the in-class work. For this project I defiantly exercised the Habit of a mathematician "Stay organized". When doing proofs with large polynomials sometimes a single version of an equation could take up the width of a page, so I had to stay organized so that I wouldn't loose track of any bit of work. I think a habit I could of worked more on is "describe and articulate". There were some instances where a classmate would ask me a question about a problem and while I knew the answer to their question I couldn't describe the process in an understandable way or I would go into superfluous detail that they wouldn't get anything out of.