To find the zeros of the function \( f(x) = (x - 1)(x + 2)(x - 3) \), we set the function equal to zero and solve for \( x \):
- \( x - 1 = 0 \) ⇒ \( x = 1 \)
- \( x + 2 = 0 \) ⇒ \( x = -2 \)
- \( x - 3 = 0 \) ⇒ \( x = 3 \)
Thus, the zeros of the function \( f(x) \) are \( x = 1, -2, 3 \).
Next, we analyze the end behavior of the function. The polynomial is of degree 3 (since the highest power of \( x \) in the expanded version is \( x^3 \)). The leading coefficient is positive, as it comes from the expansion of the function.
For a polynomial of odd degree with a positive leading coefficient:
- As \( x \to -\infty \), \( f(x) \to -\infty \) (the function goes downward to the left).
- As \( x \to +\infty \), \( f(x) \to +\infty \) (the function goes upward to the right).
Combining this information, we have:
- Zeros: \( 1, -2, 3 \)
- End behavior: continues downward to the left and upward to the right.
The correct choice that matches this analysis is:
1, -2, 3; continues downward to the left and upward to the right.