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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](0,0)[/tex].


Sagot :

Let's find the missing values step-by-step.

The function given is [tex]\( f(x) = -(x+1)(x-3)(x+2) \)[/tex].

### Finding the Zeros

The zeros of the function are the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex].

To find the zeros, we need to set each factor equal to zero and solve for [tex]\( x \)[/tex]:

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

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

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

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

### Finding the Y-Intercept

The y-intercept of a function is the value of [tex]\( f(x) \)[/tex] when [tex]\( x = 0 \)[/tex].

Substitute [tex]\( x = 0 \)[/tex] into the function and solve for [tex]\( f(0) \)[/tex]:
[tex]\[ f(0) = -( (0 + 1)(0 - 3)(0 + 2) ) \][/tex]

Simplify inside the parentheses first:
[tex]\[ f(0) = -( 1 \cdot -3 \cdot 2 ) \][/tex]

Then simplify further:
[tex]\[ f(0) = -(-6) = 6 \][/tex]

So, the y-intercept of the function is [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], and the y-intercept of the function is located at [tex]\( (0, 6) \)[/tex].

Thus, you should complete the boxes as follows:
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].