Parabolas and the Quadratic Formula

#1

1 Any parabola can be written in the form $y = a(x−x1)(x−x2)$ why?

• We could define a parabola by its two solutions, $x1$ and $x2$. Then it could be described by $f(x) = (x-x1)(x-x2)$. What’s the intuitive reason? We are setting up a two-part system which takes into account x’s horizontal distance from each solution. The closest x is to the solutions, the smaller the value of y. Think about a plane about to “land” on the x-axis.

• However, only having two points isn’t sufficient, as we can draw an infinite number of parabolas trough them. This is where a comes in. I think about it as the “slope” of the parabola. It tells you how fast the parabola will grow (either in the positive or in the negative direction, depending on a’s sign) thus defining the parabola completely. However, and this will be crucial for later, it doesn’t affect the solutions, which stay the same as before Its only effect is “stretching” the parabola (and perhaps flipping it when negative).

• In the real world, however, you don’t usually get to see situations like this very often. It is more likely you will get something like $f(x) = ax^2 + bx + c$. What do we do now? One way is eyeballing it and coming up with factors by trial and error. While this is sometimes the best option, I will explore the “autobreak” (as defined by Kalid). That is, the quadratic formula.

2 Rewriting the Quadratic Formula.

• This is the key insight: rewrite the formula as $$x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a} = -\frac{b}{2a}\pm\frac{1}{2}(\frac{\sqrt{b^2-4ac}}{a})$$ Starting to make sense? I highly suggest to draw a parabola right now and follow the explanation. What the formula is telling us is “To find intersections of parabola with x-axis: run along the x-axis until you find the axis of symmetry of the parabola (-b/2a). You are now in between the two solutions, and you have a choice: you can either go right or left by half the distance between the two roots, and you’ll find a solution!”

• In this explanation the distance between the two roots is said to be $\frac{\sqrt{b^2-4ac}}{a}$. How does this even begin to make sense? And why is the axis of symmetry $\frac{-b}{2a}$?This bugged me until I found a way to work around it. Rewrite the formula again! $$x = \frac{1}{a}(\frac{-b}{2}\pm\frac{1}{2}(\sqrt{b^2-4ac})$$ You might ask: “how does this help at all?”. Remember how I pointed out that $f(x) = a(x-x1)(x-x2)$ and $f(x) = ax^2+bx+c$ contained the same amount of information? You might not see the information in the second function right away, that’s because we first have to “mine it” by comparing it to the first one.
$$f(x) = a(x-x1)(x-x2) = a(x^2-(x1+x2)x+x1x2) = ax^2-ax(x1+x2)+ax1x2$$
This is where we understand the purpose of 1/a: in a sense, a “contaminates” the function by “linking” to each term. When the function is in the form $f(x) = ax^2+bx+c$ a is “mixed up” with the rest and we can’t isolate it, so we do the whole axis of symmetry operation normally, and then divide by a!
Now we’re on a roll: comparing the two functions above we see that b is the negative of the sum of the two solutions. To understand this intuitively, we have to look at the interactions when multiplying out $(x-x1)(x-x2)$

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#2

Please comment if you have any feedback!

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#3

This is some great stuff! I find parabolas very fascinating as well. However I was wondering wether it would be useful to point out that the last equation written in the form ax2-ax(x1+x2)+ax1x2 is solvable with the quadratic formula, and thats where they resemble each other. The explanation is very clear up to the pont of contaminating the function. All Best

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