It is useful to think about this by using vectors. Suppose that your points are $p,q,a \in \mathbb{R^2}$.
Then $x = p + (q-p)t$ is a point on the line $pq$. We wish to find the projection of $a$ on the line. To do this we require
$\|x - a\|^2 = \|x\|^2 + \|a\|^2 - 2 (x \cdot a)$ to be minimal. That is we have to minimize
$$\|p + (q-p)t\|^2 + \|a\|^2 - 2 (p + (q-p) t) \cdot a)$$
w.r.t $t$. To do this we write this as
$$\|p\|^2 + \|q-p\|^2 t^2 + 2 t p \cdot (q-p) + \|a\|^2 - 2(p \cdot a) - 2 t (q-p) \cdot a.$$
This is a quadratic in $t$ with minimum at the tip of the parabola:
$$t = \frac{-2 p \cdot (q-p) + 2 (q-p) \cdot a}{2\|q-p\|^2} = \frac{(q-p) \cdot (a-p)}{\|q-p\|^2}.$$
Thus the projection is given by
$$x = p + (q-p) \frac{(q-p) \cdot (a-p)}{\|q-p\|^2}$$ and the reflection is then just $x + (x-a) = 2x-a$.
This method doesn't have problems with infinite slopes.
$\newcommand{\vec}[1]{\mathbf{#1}}$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.
Best Answer
Write $t= 2x$, then we have to find a reflection of $y= {4+2t\over 4+t^2}$ across $y=t$. So replace $y$ and $t$ in given equation and express $y$ in this new equation: $$ t={4+2y\over 4+y^2}$$ So $$ 4t+ty^2=4+2y\implies ty^2-2y+4t-4=0$$ so $$ y= {2\pm 2\sqrt{1-4t^2+4t}\over 2t} = {1\pm \sqrt{-x^2+2x+1}\over 2x}$$ so non of the offered.