Answer
The graph of $U(x)$ does not have a vertical tangent at the origin.
Work Step by Step
$U(x)=0$ for $x\lt0$ and $U(x)=1$ for $x\ge0$.
The graph of $U$ has a vertical tangent at $x$ when
1) $U$ is continuous at $x$
2) $$U'(x)=\lim_{h\to0}\frac{U(x+h)-U(x)}{h}=\pm\infty$$
Here, we can see that $$\lim_{x\to0^+}U(x)=\lim_{x\to0^+}1=1$$
$$\lim_{x\to0^-}U(x)=\lim_{x\to0^-}0=0$$
Since $\lim_{x\to0^+}U(x)\ne\lim_{x\to0^-}U(x)$, $\lim_{x\to0}U(x)$ does not exist.
This means $U(x)$ is not continuous at the origin. The graph of $U(x)$ does not have a vertical tangent at the origin then.