To determine the points of discontinuity in a piecewise function, we need to analyze the transition points between the different pieces of the function. In this case, we focus on two critical values: \( x = -2 \) and \( x = 0 \).

Starting with \( x = -2 \), we evaluate the left-sided and right-sided limits. The left-sided limit as \( x \) approaches \(-2\) is derived from the first piece of the function, yielding:

\[\lim_{{x \to -2^-}} f(x) = 5\]

For the right-sided limit, we use the second piece of the function, which is defined as \( x^2 + 2x + 1 \). Plugging in \(-2\) gives:

\[\lim_{{x \to -2^+}} f(x) = (-2)^2 + 2(-2) + 1 = 4 - 4 + 1 = 1\]

Since the left-sided limit (5) does not equal the right-sided limit (1), the overall limit as \( x \) approaches \(-2\) does not exist. Consequently, we find that:

\[\lim_{{x \to -2}} f(x) \neq f(-2)\]

Thus, there is a discontinuity at \( x = -2 \).

Next, we examine \( x = 0 \). Again, we calculate the left-sided and right-sided limits. The left-sided limit as \( x \) approaches \( 0 \) is:

\[\lim_{{x \to 0^-}} f(x) = 0^2 + 2(0) + 1 = 1\]

For the right-sided limit, we use the last piece of the function, which is \( x + 1 \). Evaluating at \( 0 \) gives:

\[\lim_{{x \to 0^+}} f(x) = 0 + 1 = 1\]

Since both the left-sided and right-sided limits are equal, we conclude:

\[\lim_{{x \to 0}} f(x) = 1\]

Now, we check the function value at \( x = 0 \):

\[f(0) = 0^2 + 2(0) + 1 = 1\]

Since the limit equals the function value, the function is continuous at \( x = 0 \).

In summary, the only point of discontinuity in this piecewise function occurs at \( x = -2 \), where the limits do not match. The function is continuous at \( x = 0 \).