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Here's a question one of my precalculus students asked me, paraphrased for clarity:

You know how if you have the line $y=x$, and you want to reflect the graph of a function $f(x)$ across it, you can just switch the $y$ and $x$ in the equation of the function (since you're just finding $f^{-1}$ if it exists). What if you wanted to reflect a graph about something like $y=2x$ instead? Or $y=2x-3$? Is there some trick analogous to "switch the $x$ and the $y$" to find the equation of this reflected curve?

My trouble is that I don't have an explanation that would be particularly good for a precalulus student. Is there an elegant way to explain to a precalculus student how to do this that looks analogous to the "switch $x$ and $y$" trick? Here are the approaches to reflecting a curve about a line that I know, that I think are a bit too heavy for this student.

  • Do some vector-calculus looking stuff: To reflect the graph of $f$ across $y=mx+b$, you translate the $y$ values by $b$ to make the line go through the origin, look at the projection of a point on the translated graph onto the normal vector $\langle -m,1 \rangle$ of the line, use the projection to reflect the graph, then translate back by $-b$. I can calculate this for the student, but I don't think the formula will look memorable, and certainly won't be a clean "trick." Right now I'm leaning towards showing the student a nice picture of this without any calculations. I'll probably type this up in detail as an answer sometime unless someone has a better suggestion.
  • Talk about reflection matrices: a reflection about a line in the $xy$-plane with an angle of $\theta$ with the $x$-axis is given by multiplication by the matrix $$\begin{pmatrix}\cos2\theta&\sin2\theta\\\sin2\theta&-\cos2\theta\end{pmatrix}\,,$$ then the case where $\theta = \pi/4$, where you're reflecting about the line $y=x$, corresponds to the matrix $$\begin{pmatrix}0&1\\1&0\end{pmatrix}\,,$$ which just switches the $x$ and the $y$ coordinates of the graph. But then introducing matrices as linear transformations, and explaining why that matrix corresponds to reflections, would be tough.
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  • $\begingroup$ Good question, and it seems you have answered it yourself! I like the matrix approach - perhaps you can use it as a hook to encourage the student to then take a first course in Linear Algebra? $\endgroup$ Nov 20, 2018 at 20:05
  • $\begingroup$ @bounceback Thanks. :) I was shying away from the second approach because, off the top of my head I don't know how to justify that matrix being a reflection matrix. :P But maybe I should just bite the bullet and figure it out. And the student has learned about the sum-of-angles formulas (rotations matrices in disguise), so maybe that does provide a reasonable lead-in. $\endgroup$ Nov 20, 2018 at 20:10
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    $\begingroup$ You might be able to derive the reflection matrix by the requirement that $(r \cos \alpha, r \sin \alpha)$ must be sent to $(r \cos(2\theta - \alpha), r \sin(2\theta - \alpha))$ and then expanding using the cosine and sine of differences formulas, then grouping together $r \cos \alpha$ as $x$ and $r \sin \alpha$ as $y$ again. $\endgroup$ Nov 20, 2018 at 22:54
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    $\begingroup$ You could also do some "sanity" calculations such as that the square of the matrix gives the identity, the matrix is orthogonal (though that might take more explanation why that's relevant and why you'd expect it to be true), it fixes any point $(r \cos \theta, r \sin \theta)$, it negates any point on the normal $(-r \sin \theta, r \cos \theta)$, etc. $\endgroup$ Nov 20, 2018 at 22:58

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I don't suppose that there are many methods other than the two you suggested, but here's what you could do which is closer to your first one. The following requires no knowledge of vectors or matrices.

  • Given a line $M:y=mx+k$ and a point $A(a,b)$ on the curve $f(x)$, the line perpendicular to $M$ through $A$ is $P:(a-x)/m+b$. The point of intersection of $M$ and $P$ is $I([m(b+k)+a]/[m^2+1],[m^2(b+k)+ma]/[m^2+1]+k)$.

  • Suppose that after the reflection of $(a,b)$, the new point is $B(c,d)$, or $B(c,(a-c)/m+b)$. Then this requires $AI=IB$, which is a quadratic in terms of $c$. Solving this and choosing the correct root gives $c=g(a,b)$, and hence $d$. The algebra may become quite fiddly though.

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At the most general level, if we have a graph described implicitly as the level set $F(x,y)=0$ and have some invertible coordinate transformation $\phi:(x,y)\mapsto(x',y')$, then the transformed graph has the implicit equation $(F\circ\phi^{-1})(x',y')=0$. It shouldn’t be too hard to get this general principle across: you essentially solve for $x$ and $y$ in the transformation equations and substitute into the equation of the curve. For an affine transformation, this is a simple matter of solving a pair of linear equations. A reflection is its own inverse, so in that special case the inversion just involves replacing $x$ with $x'$ and $y$ with $y'$ in the transformation equations.

You still have the problem of constructing the reflection through an arbitrary line, but here’s an alternative viewpoint that might work without introducing too many new concepts. Observe that the equations of the coordinate axes are hidden in the implicit equation $F(x,y)=0$. That is, every term that involves $x$ is in a sense talking about the $y$-axis ($x=0$) and every term that involves $y$ is talking about the $x$ axis. When you perform the substitution described in the preceding paragraph, you’re replacing the equations of these axes with the equations of their preimages. This reduces to exchanging $x$ and $y$ for a reflection in the line $y=x$. You can illustrate this with parabolas (and other conics): if the equation of the parabola’s axis is the linear equation $f(x,y)=0$ and the tangent at its vertex $g(x,y)=0$, the parabola’s equation can be factored into the form $f(x,y)^2=k g(x,y)$.

So, reflecting a graph in an arbitrary line can be reduced to finding suitably normalized equations for the images of the coordinate axes. Those can be found with straightforward geometric constructions if you don’t want to introduce vector calculus-looking stuff. You can’t completely avoid introducing some vector concepts since you have to be careful to choose the signs in the equations of the transformed axes so as to preserve the positive axis direction, but it seems like that can be done without delving too deeply into that.

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