Answer
$$
f(x)=\left\{\begin{array}{ll}{2} & {\text { if } \quad x \lt 0} \\
{-x^{2}+x+2} & {\text { if } \quad 0 \leq x \leq 2}\\
{1} & {\text { if }\quad x>2}\end{array}\right.
$$
(a) graphed the function.
(b) The graph is discontinuous at $x=2.$
(c)
$$
\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(-x^{2}+x+2 \right)=0
$$
From the right, where $x$-values are greater than 2,
$$
\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}\left(1\right)=1
$$
so, $ \lim _{x \rightarrow 2} f(x) $ does not exist since
$$
\lim _{x \rightarrow 2^{+}} f(x) \ne \lim _{x \rightarrow 2^{-}}f(x)
$$
Work Step by Step
$$
f(x)=\left\{\begin{array}{ll}{2} & {\text { if } \quad x \lt 0} \\
{-x^{2}+x+2} & {\text { if } \quad 0 \leq x \leq 2}\\
{1} & {\text { if }\quad x>2}\end{array}\right.
$$
(a) graphed the function.
(b) The graph is discontinuous at $x=2.$
(c)
Since each piece of this function is a polynomial, the only $x$-values where
$f$ might be discontinuous here are 0 and 2. We investigate at $x=2$ first. From the left, where $x$-values are less than $2$,
$$
\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(-x^{2}+x+2 \right)=0
$$
From the right, where $x$-values are greater than 2,
$$
\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}\left(1\right)=1
$$
so, $ \lim _{x \rightarrow 2} f(x) $ does not exist since
$$
\lim _{x \rightarrow 2^{+}} f(x) \ne \lim _{x \rightarrow 2^{-}}f(x)
$$
Now, we investigate at $x=0$ first. From the left, where $x$-values are less than $0$,
$$
\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(2 \right)=2
$$
From the right, where $x$-values are greater than 0,
$$
\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(-x^{2}+x+2 \right)=2
$$
then we have
$$
f(0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}}f(x)=2
$$
The graph is continuous at $x=0$
