Question

The zeros of the function f(x)=-(x+1)(x-3)(x+2) are −1, 3, and ______, and the y-intercept of the function is located at (0, ______ )

Answer 1

The zeros of the function f(x) = -(x+1)(x-3)(x+2) are -1, 3, and -2, and the y-intercept of the function is located at (0, 6).

Step-by-step explanation:

The zeros of the function f(x) = -(x+1)(x-3)(x+2) are -1, 3, and -2, and the y-intercept of the function is located at (0, 6).

To find the zeros of the function, we set f(x) equal to zero and solve for x:
f(x) = 0
-(x+1)(x-3)(x+2) = 0
Setting each factor equal to zero, we get:
x+1 = 0 => x = -1
x-3 = 0 => x = 3
x+2 = 0 => x = -2

The y-intercept of the function occurs when x is zero. Therefore, we substitute x = 0 into the function:
f(0) = -(0+1)(0-3)(0+2) = -(1)(-3)(2) = 6