Answer :
To solve for the zeros and the y-intercept of the function [tex]\( f(x) = -(x+1)(x-3)(x+2) \)[/tex], follow these steps:
### Step-by-Step Solution
#### 1. Finding the Zeros
The zeros of the function are the values of [tex]\( x \)[/tex] that make the function equal to zero.
Given the function:
[tex]\[ f(x) = -(x+1)(x-3)(x+2) \][/tex]
Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -(x+1)(x-3)(x+2) = 0 \][/tex]
This equation will be zero if any of the following factors is zero:
[tex]\[ x + 1 = 0 \][/tex]
[tex]\[ x - 3 = 0 \][/tex]
[tex]\[ x + 2 = 0 \][/tex]
Solving each of these:
[tex]\[ x + 1 = 0 \implies x = -1 \][/tex]
[tex]\[ x - 3 = 0 \implies x = 3 \][/tex]
[tex]\[ x + 2 = 0 \implies x = -2 \][/tex]
Thus, the zeros of the function are [tex]\( -1, 3 \)[/tex], and [tex]\( -2 \)[/tex].
#### 2. Finding the y-intercept
The y-intercept is found by evaluating the function at [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = -(0 + 1)(0 - 3)(0 + 2) \][/tex]
[tex]\[ f(0) = - (1)(-3)(2) \][/tex]
[tex]\[ f(0) = - (1 \cdot -3 \cdot 2) \][/tex]
[tex]\[ f(0) = - (-6) \][/tex]
[tex]\[ f(0) = 6 \][/tex]
Thus, the y-intercept is at [tex]\( (0, 6) \)[/tex].
### Final Answer
The zeros of the function [tex]\( f(x) = -(x+1)(x-3)(x+2) \)[/tex] are [tex]\( -1, 3, \)[/tex] and [tex]\( -2 \)[/tex].
The y-intercept of the function is located at [tex]\( (0, 6) \)[/tex].
So, filling in the blanks:
The zeros of the function [tex]\( f(x)=-(x+1)(x-3)(x+2) \)[/tex] are [tex]\( -1, 3 \)[/tex], and [tex]\(\boxed{-2}\)[/tex]. The [tex]\( y \)[/tex]-intercept of the function is located at [tex]\( (0, \boxed{6}) \)[/tex].
### Step-by-Step Solution
#### 1. Finding the Zeros
The zeros of the function are the values of [tex]\( x \)[/tex] that make the function equal to zero.
Given the function:
[tex]\[ f(x) = -(x+1)(x-3)(x+2) \][/tex]
Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -(x+1)(x-3)(x+2) = 0 \][/tex]
This equation will be zero if any of the following factors is zero:
[tex]\[ x + 1 = 0 \][/tex]
[tex]\[ x - 3 = 0 \][/tex]
[tex]\[ x + 2 = 0 \][/tex]
Solving each of these:
[tex]\[ x + 1 = 0 \implies x = -1 \][/tex]
[tex]\[ x - 3 = 0 \implies x = 3 \][/tex]
[tex]\[ x + 2 = 0 \implies x = -2 \][/tex]
Thus, the zeros of the function are [tex]\( -1, 3 \)[/tex], and [tex]\( -2 \)[/tex].
#### 2. Finding the y-intercept
The y-intercept is found by evaluating the function at [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = -(0 + 1)(0 - 3)(0 + 2) \][/tex]
[tex]\[ f(0) = - (1)(-3)(2) \][/tex]
[tex]\[ f(0) = - (1 \cdot -3 \cdot 2) \][/tex]
[tex]\[ f(0) = - (-6) \][/tex]
[tex]\[ f(0) = 6 \][/tex]
Thus, the y-intercept is at [tex]\( (0, 6) \)[/tex].
### Final Answer
The zeros of the function [tex]\( f(x) = -(x+1)(x-3)(x+2) \)[/tex] are [tex]\( -1, 3, \)[/tex] and [tex]\( -2 \)[/tex].
The y-intercept of the function is located at [tex]\( (0, 6) \)[/tex].
So, filling in the blanks:
The zeros of the function [tex]\( f(x)=-(x+1)(x-3)(x+2) \)[/tex] are [tex]\( -1, 3 \)[/tex], and [tex]\(\boxed{-2}\)[/tex]. The [tex]\( y \)[/tex]-intercept of the function is located at [tex]\( (0, \boxed{6}) \)[/tex].
