= Solution
The join has matching positions and first <derivatives>:
$$
F(1)=E(0)=(0,0),\qquad
F'(t)=(4-4t,2-4t),\qquad E'(t)=(4t,-2+4t),
$$
$$
\boxed{F'(1)=E'(0)=(0,-2)\ne(0,0).}
$$
Because both pieces have unit parameter intervals, these are also the one-sided <derivatives> of $S$ at $u=1$. Thus $S$ has first-order <parametric continuity>, and hence first-order <geometric continuity>, with no corner or cusp. Its nonzero tangent <vector> points vertically downward.
The second <derivatives> do not agree:
$$
F''(1)=(-4,-4),\qquad E''(0)=(4,4).
$$
So this is $C^1$ but not $C^2$ <parametric continuity>. The signed planar <curvature> $\kappa=(x'y''-y'x'')/(x'^2+y'^2)^{3/2}$ has one-sided values
$$
\kappa_F(1)=\frac{-8}{8}=-1,\qquad
\kappa_E(0)=\frac{8}{8}=1.
$$
The unsigned <curvatures> agree, but the bending switches sides; there is no second-order <geometric continuity> through a regular orientation-preserving reparametrization. The join is a reversal of bending with a curvature jump, rather than a smooth zero-curvature inflection.
For the “gradient” compute $dy/dx=y'/x'$ away from the join:
$$
\frac{F_y'(t)}{F_x'(t)}=\frac{2-4t}{4-4t}\longrightarrow-\infty\quad(t\to1^-),\qquad
\frac{E_y'(t)}{E_x'(t)}=\frac{-2+4t}{4t}\longrightarrow-\infty\quad(t\to0^+).
$$
Hence
$$
\boxed{\text{no finite }dy/dx\text{ at the join; the tangent line is }x=0.}
$$
The zero horizontal <derivative> is not a stationary <point> of the <parametric curve>, since its vertical <derivative> is $-2$. It means that the locally regular <curve> has a vertical <tangent line> and cannot have a finite graph <derivative> as $y(x)$ there.
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