#### Answer

$a.\quad $yes
$b.\quad $yes
$c.\quad $yes
$d.\quad $yes

#### Work Step by Step

The graph of f(x) contains points $(x,f(x))$, where x belongs to the domain of f.
$a.$
The point $(-1,0)$ belongs to the graph of f,
so $f(-1)=0$
( $f(-1)$ exists)
$b.$
As x approaches the value $-1$ from the right, f(x) approaches $0$.
The right-sided limit exists, $\displaystyle \lim_{x\rightarrow 1^{+}}f(x)=0$,
$c.$
$\displaystyle \lim_{x\rightarrow 1^{+}}f(x)=0$= $f(-1)$
(they are equal)
$d.$
The domain of f contains the half-closed interval $[-1,0)$.
The result of (c) implies that f is right-continuous at the left (closed) border, (see the first definitions of the section, and the discussion below them, referring to continuity over closed and half-closed intervals).
So we say that f is continuous at $x=-1$.