$\newcommand{\vec}[1]{\mathbf{#1}}$The reflected curve is not generally a graph, but it's easy to obtain a parametric description. For generality, let's say the graph $y = f(x)$ is to be reflected across the line $\ell$ with symmetric equation $ax + by + c = 0$ ($a$ and $b$ not both zero). Dividing the equation of $\ell$ by $\sqrt{a^{2} + b^{2}}$, we may assume $(a, b)$ is a unit vector.
Pick a point $(x_{0}, y_{0})$ on $\ell$, and make a translational change of coordinates so this point is the origin: $(u, v) = (x - x_{0}, y - y_{0})$. The graph becomes $v + y_{0} = f(u + x_{0})$, and the line is $au + bv = 0$.
The (unit) vector $\vec{p} = (-b, a)$ lies on $\ell$, and $\vec{n} = (a, b)$ is orthogonal to $\ell$. Each point $\vec{x} = (u, v)$ is uniquely represented as
\begin{align*}
\vec{x}
&= (\vec{x} \cdot \vec{p}) \vec{p} + (\vec{x} \cdot \vec{n}) \vec{n} \\
&= (-bu + av)(-b, a) + (au + bv)(a, b).
\end{align*}
The image of $\vec{x}$ under reflection across $\ell$ is
\begin{align*}
R_{\ell}(\vec{x})
&= (\vec{x} \cdot \vec{p}) \vec{p} - (\vec{x} \cdot \vec{n}) \vec{n} \\
&= (-bu + av)(-b, a) - (au + bv)(a, b) \\
&= \bigl((b^{2} - a^{2})u - 2abv, -2abu - (b^{2} - a^{2})v\bigr).
\tag{1}
\end{align*}
(If we write $(a, b) = (\cos\theta, \sin\theta)$, the coefficients in the preceding expression are $\pm\cos(2\theta)$ and $-\sin(2\theta)$.)
The graph is parametrized by
$$
(u, v) = \bigl(t + x_{0}, f(t + x_{0}) - y_{0}\bigr);
$$
substituting these functions into (1) gives a parametrization of the reflected graph.
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.
Best Answer
Your hunches are on the right track and it sounds like you should just review some definitions.
When you reflect across a line, any vector on the line (a subspace of dimension 1) is unchanged (i.e. multiplied by 1), and any vector in the plane perpendicular to the line (a subspace of dimension 2) is turned into its opposite (i.e. multiplied by -1).
In general:
An eigenvector of a transformation $T$ is a nonzero vector $v$ with the property that $Tv$ is a scalar multiple of $v$, i.e. $Tv=\lambda v$ for some scalar $\lambda$. We call $\lambda$ the eigenvalue associated with $v$. For a particular eigenvalue, the corresponding eigenvectors (with 0 included) form a subspace. We call such a subspace an eigenspace.
If our overall vector space is finite-dimensional (like $\Bbb R^3$), then the equation $\det(T-\lambda I)=0$ (which expresses the fact that $Tv=\lambda v$ for some nonzero vector $v$ - see this video for an explanation) gives us a polynomial equation whose roots are the eigenvalues of $T$. It turns out that the multiplicity of such a root is equal to the dimension of the corresponding eigenspace. So when we list eigenvalues with repetition, the multiplicity corresponds to that of these polynomial roots as well as to the dimension of the eigenspaces.