The Quintessence of Quadratics
We learned a lot over the course of the Quintessence of Quadratics project. We jumped into the world of quadratic equations by reviewing distance, velocity and acceleration. We derived the distance formula to get us warmed up for later, more complicated equations. We received our first quadratic equation through a hypothetical scenario involving a rocket, a platform, and a parabola. It was presented as a kinematics problem, referring to the deceleration due to gravity, the initial height and velocity of the rocket. All of these things would later be simplified into, we didn’t know it then, a quadratic equation in Standard form. We later started practicing with Vertex form, and later still, Factored form, both of which also form parabolas. We learned how to convert between all of them using geometry and algebra together. The connection between geometry and algebra was an integral part of this particular project.
Vertex Form
The Vertex form of an equation goes like this:
y=a(x-h)^2 +k.
X and y of course are the variables on the axis, but what about the other variables involved? The first one, a, is the most complex among them. When a is changed, the parabola becomes wider or thinner, depending on the number. A large a value results in a thinner parabola, and a smaller a value results in a wider parabola. If the a variable is made negative, the wideness of the parabola remains the same, but the parabola will be concave down (as opposed to concave up, which it is when the a variable is positive.)
H and K are a lot different, but a lot simpler to understand. H corresponds to the x coordinate of the parabola’s vertex (its peak/valley) and k corresponds to the y coordinate of the parabola’s vertex. A change in either h or k just changes the parabola’s position on the coordinate grid.
A combination of the three gives us our Vertex form general equation. This form is very useful for determining the vertex or the wideness of a parabola. Shown below is an equation written in Vertex form
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Converting Between Forms
One thing that particularly aided our journey into quadratic equations were diagrams of area. We visualised these diagrams by equating them to fenced off areas or house lots. If a square lot is already x*x, then what happens to the area of the lot when 4 feet are added to one side and 2 feet are added to the other side? The area diagram of the situation would look like this:
There are several ways to express the area of this diagram, and they’re all quadratic. Besides Vertex form, The other two forms of equations are called Standard form and Factored form.
I will use the diagram to the right as an example for moving between forms of equations to discover important information about the parabola. Before doing that, we have to find the equation in Standard form. All you have to do is combine all the like terms that I already have displayed in the rectangle. The standard form of this equation would be x^2+6x+8.
Starting with Standard form to Vertex Form. We have to take our Standard form equation and focus on a certain part of it. We have to focus only on the x^2+6x part. The adding of 8 will come back in later. We have to “complete the square” for this expression. We divide the 6x into two parts, and we’re left with a small square in the bottom right corner of the whole. It can be identified as 9 ft^2 due to the fact that when we divided the 6 into two parts, it gave us 3, and 3 squared is 9. This means that x^2+6x+9 = (x+3)^2. We can substitute this into our initial equation by subtracting the 9 out of the equation. We’re left with (x+3)^2-9+8. When we combine our like terms, we get y=(x+3)^2-1.
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Next, we go from our old Standard form equation to Factored form. The simplest way to do this conversion is to focus on the two latter terms and find two numbers that when added together, make b (6 in this case) and when multiplied together make c (8 in this case). If you observe the first area diagram, you’ll notice that 2 and 4 are applicable numbers. Thus we have our factored form equation: y=(x+4)(x+2) Much like how Vertex form tells us the vertex of the parabola in question, it's worth noting that Factored form will give you the parabola's x intercept points.
Solving Word Problms
There are several ways to use our knowledge of quadratic equations to solve word problems. Much like I mentioned in my introduction, we were introduced to quadratics through a kinematics problem. The problem revolved around answering questions about the rocket’s path. From reading the problem, we already know that the rocket launches from a 160 ft tall platform, the rocket’s initial velocity is 92 ft/s, and the rocket’s deceleration due to gravity is -32 ft/s^2. If we consider the rocket’s vertical position over time, It becomes a quadratic equation.
I mentioned in my explanation of area diagrams that we practiced with problems about area. If you imagine the edges of the area diagram as a fence or a wall, you can interpret the quadratic equation through geometry. If a rectangular fence has 5 feet added to it on one side and 3 feet added to it on the other, then what is the area? It can be expressed in a quadratic equation in standard or factored form just by looking at the area diagram.
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One kind of word problem I have yet to mention at all is economics. If a company has a certain amount of an item, then what should they sell that item for in order to make maximum profits? In order to find that out, you have to know how much money is made by the sale of their inventory. The marketing director of the company selling these items claims that if they sell them for d dollars, then they will be able to sell 1000-5d items. If we multiply the whole equation by d, we’ll get the amount of money made by the items sold. When we simplify the new equation, we’ll get a quadratic equation. The maximum amount of revenue that can be made from the sale of these items as well as the price of the items that will make it possible would be found at the vertex.
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Reflection
This project did a lot to show me about non linear equations. I knew nothing about quadratic equations before this, but now I know about three major and important forms of them. I put a lot of effort into practicing with these, and I ended up getting a lot out of it. In using resources like desmos, I’ve discovered a lot of cool things you can do with simple numbers that aren’t linear. In my experimentation, I found that having an equation where x is within the denominator of a fraction, you get two separate curves that never intersect with either axis. I experimented with square roots a bit, and discovered that a square root in a quadratic equation rotates the parabola 90 degrees. An equation that I found works a lot like vertex form goes like this:
M and N seem to work similarly to h and k, while l seems to work similarly to a. O also seems to work similarly to a, but I’m mostly unsure about it. One interesting thing about this equation is that it only produces half of the parabola. To get the other half, you have to make some numbers negative. I have no idea how any of it works, but I have a feeling I’ll find out sometime in the future. Perhaps I’ll need to know it to take the SAT. Only time will tell, but until then, I feel smart for having found something new to myself. This kind of experimentation is why I love mathematics; it only gets deeper the more you investigate. As always, I look forward to what next year has in store.
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