The answer is: **yes**, it is always quasi-convex! I'll show this by first proving a stronger characterization, from which the other facts follow. Please bear with me as I first make a few definitions.

Let convex $S \subseteq \mathbb{R}$ and a function $f:S\to \mathbb{R}$ be given. To avoid existence of derivatives, let $f'(v)$ refer to any subgradient of $f$ at $v$, and say $f$ is convex if for any $x,v \in S$, $f(x) \geq f(v) + f'(v)(x-v)$. (This is an equivalent formulation of convexity, and when $f$ is differentiable, gives the 'first-order' definition of convexity.) Note critically that for $u,v\in S$ with $u\leq v$, it follow that
$f'(u) \leq f'(v)$. (This is sort of like the mean value theorem, though not exactly since those subgradients are technically sets; I think all of I've said so far may appear in the thesis
of Shai Shalev-Shwartz.) Define
$$b_x(v) = f(x) - f(v) - f'(v)(x-v)$$
to be the Bregman divergence of $f$ at the point $x$, taking the linear approximation at $v$. By the definition of convexity, if follows that $b_x(v) \geq 0$ for all $x,v\in S$.

**Fact:** $b_x(\cdot)$ is decreasing up to $x$, exactly zero at $x$, and increasing after $x$.

*Proof.* $b_x(x) = f(x)-f(x) - f'(x)(0) = 0$. Now consider $u\leq v \leq x$; we'd like to show $b_x(u) \geq b_x(v)$. To start, write
$$
b_x(u)-b_x(v) = f(v) + f'(v)(x-v) - f(u) - f'(u)(x-u).
$$
Now, using $f(v) \geq f(u) + f'(u)(v-u)$ yields
$$
b_x(u)-b_x(v) \geq f'(v)(x-v) + f'(u)(v-x) = (f'(v) - f'(u))(x-v),
$$
and $b_x(u)-b_x(v)\geq 0$ follows since $f'(v) \geq f'(u)$ and $x-v\geq 0$. To
show the last case, that $x\leq v\leq u$ gives $b_x(u) \geq b_x(v)$, the proof is
analogous. QED.

Some remarks:

- To see that this means $b_x(\cdot)$ is quasi-convex, take any $y\leq z$ and any
$\lambda \in [0,1]$. Then the point $w:=\lambda y + (1-\lambda)z$ lies on the line
segment $yz$, and $b_x(\cdot)$ must be increasing in the direction of at least one of
these endpoints.
- This also gives a strong idea of how convexity breaks down for $b_x(\cdot)$. In
particular, let $f= \max\{0, |x|-1\}$ (a 1-insensitive loss for regression). Then the
function $b_0(\cdot)$ is 0 on $(-1,1)$ and 1 everywhere else except $\{-1,+1\}$ (those points
are different since, by using subgradients, these functions have sets as output; but if you
took a differentiable analog to this loss, something like a Huber loss, you'd get basically the same effect, and $b_0(\cdot)$ is a vanilla continuous (non-convex) function).