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The zeros of the function [tex]f(x) = -(x+1)(x-3)(x+2)[/tex] are [tex]-1[/tex], [tex]3[/tex], and [tex]\square[/tex], and the [tex]y[/tex]-intercept of the function is located at [tex](0, \square)[/tex].

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GinnyAnswer
To find the zeros of the function [tex]\( f(x) = -(x + 1)(x - 3)(x + 2) \)[/tex], we set [tex]\( f(x) = 0 \)[/tex] and solve for [tex]\( x \)[/tex]:

[tex]\[ -(x + 1)(x - 3)(x + 2) = 0 \][/tex]

Setting each factor to zero:

[tex]\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \][/tex]
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
[tex]\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \][/tex]

Thus, the zeros of the function [tex]\( f(x) \)[/tex] are [tex]\(-1\)[/tex], [tex]\(3\)[/tex], and [tex]\(-2\)[/tex].

Next, to find the y-intercept, we evaluate the function at [tex]\( x = 0 \)[/tex]:

[tex]\[ f(0) = -(0 + 1)(0 - 3)(0 + 2) \][/tex]
[tex]\[ f(0) = -(1)(-3)(2) \][/tex]
[tex]\[ f(0) = -(-6) \][/tex]
[tex]\[ f(0) = 6 \][/tex]

Therefore, the y-intercept is located at [tex]\((0, 6)\)[/tex].

So, the completed statement is:
The zeros of the function [tex]\(f(x) = -(x + 1)(x - 3)(x + 2)\)[/tex] are [tex]\(-1, 3\)[/tex], and [tex]\(-2\)[/tex], and the [tex]\(y\)[/tex]-intercept of the function is located at [tex]\((0, 6)\)[/tex].
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