Simple But Evil #4 — Find The Vertex Of A Parabola

Welcome to another edition of “Simple But Evil” where we talk about finding the vertex of a parabola. Since this is “Simple But Evil”, I’ll introduce a few problems that are accessible, “simple”, but generally not covered in an Algebra or pre-Calculus course. The two most common methods for finding the vertex of a parabola are as follows:

  1. When we have a quadratic written in the form of \(y = ax^{2} + bx + c\), then the \(x\)-coordinate of the vertex is \(\frac{-b}{2a}\) and the y-coordinate is found by substituting \(\frac{-b}{2a}\) back into the equation.
  2. Complete the square to obtain a quadratic written as \(y = a(x – h)^{2} + k\) and the vertex can be directly read off.

Then, in a Calculus course, students learn about derivatives and can easily find the maximum or minimum of a quadratic.

Now, here are two problems that I think should find their way into an instructor’s repertoire.

The Sum of Quadratics

Ask this question:

Find the vertex of $$y = (x – 3)^2 + (x – 5)^2$$

Here are one or more of a few things will happen — maybe other things will happen, too.

  1. Students will say that there are two vertices.
  2. Students, knowing that this parabola should have a minimum by way of the leading coefficient being positive, will choose \(x = 3\) as the \(x\)-coordinate of the vertex because it is “smaller”.
  3. Students won’t know what to do.
  4. Students will try to expand the right hand side and botch the simplification.
  5. Students will correctly answer that the vertex is \((4,2)\).

If (1), then clearly we have some problems. Go back to definitions. If (2), well at least there was some reasoning using all the buzzwords, but we still need to get back to basics. If (3), I actually prefer (3) over (1) or (2) because this tells me that they at least know not to just attempt this problem out of reflex. For (3), they probably need a nudge or two to coax it out of them to expand the expressions. If (4), then we have to probe the Algebra steps. Negative signs and multiple second-degree and first degree terms give enough room for error. If (5), then great! That leads us to the next set of questions.

Find the \(x\) coordinate of $$y = (x – 3)^2 + (x-4)^2 + (x – 5)^2$$

Here, the \(x\) coordinate is \(4\) and they should have found it via \(\frac{3 + 4 + 5}{3}\). Curious? Looks like an average?

Find the \(x\) coordinate of $$y = (x – 3)^2 + (x-4)^2 + (x – 5)^2 + (x – 6)^2$$

Here, the \(x\) coordinate is \(4.5\) and they should have found it via \(\frac{3 + 4 + 5 + 6}{4}\). Curious? Looks like an average????

Find the \(x\) coordinate of $$y = (x + 3)^2 + (x + 4)^2 + (x – 5)^2 + (x – 6)^2$$

Here, the \(x\) coordinate is \(1\) by averaging -3,-4,5,6.

Then you can try to generalize, either formally or by hand-waving (if you’re into that sort of thing) with something like $$y = \sum_{n=1}^{N}(x – a_{n})^{2}$$ has the \(x\)-coordinate of the vertex at the average of the \(a_{n}\).

Why is this path worth going down? First, what we’re doing is minimizing a sum of squares. This shows up when we are trying to find a line (or curve) of best fit. A typical function to measure the goodness of fit is the sum of squares. It is nothing more than Pythagorean distance in \(N\) dimensions.

You may say, “But my students aren’t going to get this. It’s way too advanced.” That’s probably true, but my counter arguments are that (a) they should still be able to find the vertex of a parabola written in a form different from what they are used to seeing — that is relevant and (b) we can forget regression for the time being, what we are showing is why we compute average the way we do. Virtually no student knows why we calculate average as “sum of all the numbers divided by the number of numbers”. They just know that’s how we do it. Here, for the first time, we can pose a question in which the solution yields this formula they already know. The point is, they didn’t have to invoke “average”. Average is what came out. THAT in and of itself is worth the existential discussion. It tells students that these formulas weren’t pulled out of thin air. They are the result of trying to solve another problem. In this case, minimizing the sum of squares, which is a measure for what kind of error your fit has.

Axis of Symmetry

When parabolas are introduced, there is discussion about axis of symmetry. But it is often brief and involves just drawing a vertical or horizontal line through the vertex, depending on the parabola’s orientation. Try this question:

We have a parabola with roots at \(x = 5\) and \(x = 12\). Find the \(x\)-coordinate of the vertex.

It’ll be impressive if any student gets this correct and can explain why. One way to answer this question is to rest on the fabled quadratic formula, which gives us that the roots are at \(x_{1,2} = \frac{-b}{2a} \pm \frac{\sqrt{b^{2}-4ac}}{2a}\). Thus, the \(x\) coordinate of the vertex is the mid-point of the roots. We don’t explore this formula enough. A course may discuss the discriminant as a segue into complex numbers, but rarely is there a focus on the \(\frac{-b}{2a}\) part. We can probably blame how we write the quadratic formula: $$\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$ where the \(\frac{-b}{2a}\) part gets lost in the shuffle.

Well, I hope you enjoyed this edition of Simple But Evil! If you have a topic you’d like for me to discuss, leave a comment, send me a tweet, or shoot me an email. Until next Sunday (or tomorrow if you want to know what the puzzle is!)!

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