The problem is asking for the fewest possible number of zeros of a continuous function $f$ on the interval $[-2, 3]$ based on the table provided. Let's examine the table of values:

To solve the problem, we will apply the Intermediate Value Theorem, which tells us that if a function is continuous on an interval, and the function changes signs between two points, then there is at least one zero in between those points.

Analyzing the table values:

1. From $x = -2$ to $x = -1$, $f(x)$ changes from $-2$ to $5$. Since the sign changes (from negative to positive), there is at least one zero between $x = -2$ and $x = -1$.
2. From $x = -1$ to $x = 0$, $f(x)$ changes from $5$ to $2$. There is no sign change here, so no zero between $x = -1$ and $x = 0$.
3. From $x = 0$ to $x = 1$, $f(x)$ changes from $2$ to $-4$. Since the sign changes (from positive to negative), there is at least one zero between $x = 0$ and $x = 1$.
4. From $x = 1$ to $x = 2$, $f(x)$ changes from $-4$ to $-1$. There is no sign change, so no zero between $x = 1$ and $x = 2$.
5. From $x = 2$ to $x = 3$, $f(x)$ changes from $-1$ to $3$. Since the sign changes (from negative to positive), there is at least one zero between $x = 2$ and $x = 3$.

Conclusion:

There are at least three intervals where a sign change occurs, meaning there are at least three zeros in the interval $[-2, 3]$.

Therefore, the fewest possible number of zeros is three, and the correct answer is D: Three.

Would you like more details or further clarification? Here are five related questions you might find interesting:

1. What is the Intermediate Value Theorem and how is it applied?
2. How can we estimate the location of zeros more precisely based on function values?
3. Can a continuous function have more zeros than suggested by sign changes?
4. How do discontinuities affect the number of zeros in a function?
5. What methods can be used to find the exact zeros of a function?

Tip: Always check for sign changes in a continuous function to estimate the number of zeros between given points!