Answer
The graph of $U(x)$ does not consist of a vertical tangent at the origin.
Work Step by Step
1. The graph of $U(x)$ has a vertical tangent at $x$ when $U(x)$ is continuous at $x$
2. The graph of $U(x)$ has a vertical tangent at $x$
when $U'(x)=\lim\limits_{h\to 0}\dfrac{U(x+h)-U(x)}{h}=\pm\infty$
Also, $\lim\limits_{x\to0^+}U(x)=\lim\limits_{x\to0^+}(1)=1$ and
$\lim\limits_{x\to0^{-}}U(x)=\lim\limits_{x\to0^{-}}(0)=0$
Since $\lim\limits_{x\to0^+}U(x)\ne\lim\limits_{x\to0^{-}}U(x)$, thus, $\lim\limits_{x\to 0} U(x)$ does not exist.
This implies that $U(x)$ is not continuous at the origin and the graph of $U(x)$ does not consist of a vertical tangent at the origin.