Answer
The graph of $f$ does not have a vertical tangent at the origin.
Work Step by Step
$f(x)=-1$ for $x\lt0$ and $f(x)=0$ for $x=0$ and $f(x)=1$ for $x\gt0$.
The graph of $f$ has a vertical tangent at $x$ when
1) $f$ is continuous at $x$
2) $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\pm\infty$$
Here, we can see that $$\lim_{x\to0^+}f(x)=\lim_{x\to0^+}1=1$$
$$\lim_{x\to0^-}f(x)=\lim_{x\to0^-}-1=-1$$
Since $\lim_{x\to0^+}f(x)\ne\lim_{x\to0^-}f(x)$, $\lim_{x\to0}f(x)$ does not exist.
This means $f$ is not continuous at the origin. The graph of $f$ does not have a vertical tangent at the origin then.
