# Would this be a good way of teaching about symmetry?

Symmetry is an important concept in mathematics and it has a built-in aesthetic appeal. The following shows how different types of symmetry relate to one another. Start with the equation $$f(x) = -x^2 + bx + c$$. Without having to complete the square, we can find the axis of symmetry. Rewrite the equation as $$f(x) = x(b - x) + c$$. Now find $$f(b - x) = (b-x)x + c = f(x)$$.

Since $$f(b-x) = f(x)$$, we know that for every point $$(x,y)$$ on the curve, there is a point $$(b - x,y)$$ on the curve. The midpoint of these two points on the curve is $$(b/2, y)$$. The line $$x=b/2$$ is an axis of symmetry. We can generalize this by noticing that what made this work is that the equation for $$f(x) = x(b-x) + c$$ is symmetric in $$x$$ and $$b -x$$. We can get similar results for $$f(x) = x^2 + bx + c$$. $$f(x) = -x(-b - x) + c$$. Reasoning as before, it follows that $$x = -b/2$$ is an axis of symmetry.

Consider the function $$f(x) = \arctan(x) + \arctan(6-x)$$, which is symmetric in $$x$$ and $$(6 - x)$$. We can say immediately that the line $$x=3$$ is an axis of symmetry. One of the reasons that this works is that the function $$g(x) = b - x$$ is self-inverse, $$g(g(x))=x$$. If we take any function $$f(x)$$ that is symmetric in $$x$$ and a self-inverse function $$g(x)$$, we can say that $$f(x) = f(g(x))$$, although this will not usually result in an axis of symmetry.

Finally, there is a way of talking about self-inverse functions in terms of symmetry. The equation $$y = b - x$$ does not at first appear to have any symmetry, but if we write it as $$x + y = b$$, it is clear that it is symmetric in $$x$$ and $$y$$. The function for getting $$y$$ in terms of $$x$$ is the same as the function for getting $$x$$ in terms of $$y$$, causing $$g(g(x)) = x$$. For every point $$(x, g(x))$$ on a self-inverse curve there is a point $$(g(x), x)$$ on the curve. The midpoint is $$((x+g(x))/2, (x+g(x))/2)$$, which is on the line $$y=x$$. The line $$y=x$$ is an axis of symmetry for all self-inverse curves.

• What's the question here? I suppose you have it in the title of the post, but it is worth reiterating what exactly you're asking, along with some important information: Who are the students you're teaching to? What is the context? – Brendan W. Sullivan Oct 11 '20 at 18:10
• It sounds like you are enjoying thinking about symmetry this way. That's cool. I don't know if this would help someone else understand it or not. – Sue VanHattum Oct 11 '20 at 18:55
• To me this seems more of a way of summarizing symmetry notions after a student has worked with them a lot in specific cases, spread out over at least several days (and don't toss every symmetry notion at them on the same day), than how you would introduce the topic. However, for that particular purpose this might work well, but you'll probably want to make it more into a guided instruction summary or a student-guided-discovery summary worksheet, than simply throwing all these assertions and connections at them like this. – Dave L Renfro Oct 11 '20 at 18:55
• This was not intended as a single lesson. I was thinking that the material would be suitable for a ninth grade math class. I was hoping that looking at the equation for a parabola in a fresh way would catch the students' attention and act as a motivator. Surely at least some of the students would be impressed with how the plot of y=arctan(x)+arctan(6-x) is so clearly split in half by the line x=3, – user1153980 Oct 11 '20 at 22:08
• arctan in 9th grade? – JTP - Apologise to Monica Oct 12 '20 at 12:22

I think that most teachers have discovered first-hand that there are consequences for skipping steps along that path. Probably fewer teachers have learned from those mistakes, because it's easy to blame students for not comprehending a lesson that was hard to understand and had no purpose. For myself, I remember teaching a similar lesson the first month I taught Calculus when I thought I had prepared my students to understand difference quotients. Lo and behold, I asked them what $$f(x+h)$$ was and they pretty much replied $$fx+fh$$ because they had never seen an expression inside function notation but they did know the distributive property.