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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].

Sagot :

Let's solve the problem step-by-step.

1. Identifying the Zeros:
The zeros of the function [tex]\( f(x) = -(x+1)(x-3)(x+2) \)[/tex] are the values of [tex]\( x \)[/tex] that make the function equal to zero. We can determine these zeros by setting each factor inside the parentheses equal to zero:

- For [tex]\( x + 1 = 0 \)[/tex], solving this gives [tex]\( x = -1 \)[/tex].
- For [tex]\( x - 3 = 0 \)[/tex], solving this gives [tex]\( x = 3 \)[/tex].
- For [tex]\( x + 2 = 0 \)[/tex], solving this gives [tex]\( x = -2 \)[/tex].

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

2. Finding the y-intercept:
The y-intercept is the value of [tex]\( y \)[/tex] when [tex]\( x = 0 \)[/tex]. To find it, we substitute [tex]\( x = 0 \)[/tex] into the function:

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

Simplifying inside the parentheses:

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

Multiplying these values together:

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

[tex]\[ f(0) = -( -6 ) \][/tex]

[tex]\[ f(0) = 6 \][/tex]

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

Now, we fill in the correct answers in the given boxes:

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