In the problem of the week, I would occasionally throw in some quadratics, or problems with squaring involved. If you’ve seen my performances, you might have noticed I like things that have to do with squaring, and quadratics is definitely one of them. Quadratics is normally done in the form f(x) = ax^2 + bx + c, which looks nice, but isn’t so useful. However, there is a form that is pretty commonly used that is almost magical in a way.
This form is the form f(x) = a(x - h)^2 + k. First, I will show you how to get into this form, then we will look at its properties.
The technique we will use is called “completing the square.” The first thing you do is factor the a out of the equation. Let’s use 1/2x^2 + 3x + 5 as an example.
1/2x^2 + 3x + 5
1/2(x^2 + 6x + 10)
Next, plug that new b term into b^2/4. What I find easier is to divide the b term by two and then square it, so the division doesn’t get too messy.
6/2 = 3
3^2 = 9
What this means is that the equation (x^2 + 6x + 9) is square, or a “perfect square trinomial.” It is in fact (x + 3)(x + 3). To figure out that three, all you have to do is divide that b term by two.
There is only one problem though. We have x^2 + 6x + 10, not x^2 + 6x + 9. However, we can put the (x + 3)^2 there, but we must add one to the (x + 3)^2, which ends up getting multiplied by the 1/2 to get a constant of one-half at the end. So, we have:
1/2(x + 3)^2 + 1/2
And that is vertex form. Let’s look at what is cool about it.
First off, both a’s happen to be equal. This is not coincidence, as we factored out the a in order to switch forms.
What I find really cool is those two terms we couldn’t control, h and k. In this case, they are -3 and 1/2 (the equation is a(x - h)^2 + k, not a(x + h)^2 + k, so h is -3). What does that have to do with anything? Take a look at the graph of this equation.
If you’ll notice, the vertex of this parabola (graph of a quadratic function) is in fact (-3, 1/2). In fact, you can actually graph a quadratic equation using only vertex form, just with this fact and a. I think that is cool that there is a format out there that can do that.
Bonus: Say that instead of doing a(x - h)^2 + k, you did (bx - h)^2 + k. b is unfortunately not the same as the other b, but much more useful. This form is messier, but there is a cool thing about it. In this case, the vertex is actually (bh, k), which is interesting.
What the a factor did was it told us the factor for the “vertical stretch/compression” of the parabola, or the number you multiply every single y value by. In the case before, every y value of plain (x + 3)^2 + 1 was multiplied by 1/2 to create the graph.
For this form, there is actually a “horizontal stretch/compression,” which isn’t common in quadratics. This form gives you a stretch/compression of 1/b. You might see it as just b, but it is usually written like this:
((1/b)x - h)^2 + k)
Either way, it is pretty cool.